Form minimum number from given sequence

Auxiliary Given a pattern containing only I’s and D’s. I for increasing and D for decreasing. Device an algorithm to print the minimum number following that pattern. Digits from 1-9 and digits can’t repeat.

Examples: 

   Input: D        Output: 21
   Input: I        Output: 12
   Input: DD       Output: 321
   Input: II       Output: 123
   Input: DIDI     Output: 21435
   Input: IIDDD    Output: 126543
   Input: DDIDDIID Output: 321654798

Source: Amazon Interview Question

We strongly recommend that you click here and practice it, before moving on to the solution.

Below are some important observations

Since digits can’t repeat, there can be at most 9 digits in output. 

Also, number of digits in output is one more than number of characters in input. Note that the first character of input corresponds to two digits in output.

Idea is to iterate over input array and keep track of last printed digit and maximum digit printed so far. 

Steps were to follow to solve this problem:

• Create a static function named “PrintMinNumberForPattern” that takes a string array arr as input.
• Create two variables “curr_max” and “last_entry” and initialize them to 0.
• Traverse through the input array arr using a for a loop.
• Create a variable “noOfNextD” and initialize it to 0.
• If the character at index i is ‘I’, perform the following steps:
  • Find the number of next consecutive D’s available by iterating over the array from the next index until a non-‘D’                                 character is encountered.
  • If i is 0, set curr_max to noOfNextD+2 and print the incremented sequence from 1.
  • If i is not 0, set curr_max to curr_max + noOfNextD + 1 and print the digit for ‘I’.
  • For all next consecutive ‘D’ characters, print the decremented sequence.
• If the character at index i is ‘D’, perform the following steps:
  •  If i is 0, find the number of the next consecutive ‘D’ characters available and set curr_max to noOfNextD+2. Print the first                        two digits (curr_max and curr_max-1).
  •  If i is not 0, print the decremented value of last_entry.
• Print a newline character after the for loop completes.

Below is the implementation of the above idea:

C++

// C++ program to print minimum number that can be formed
// from a given sequence of Is and Ds
#include <bits/stdc++.h>
using namespace std;

// Prints the minimum number that can be formed from
// input sequence of I's and D's
void PrintMinNumberForPattern(string arr)
{
    // Initialize current_max (to make sure that
    // we don't use repeated character
    int curr_max = 0;

    // Initialize last_entry (Keeps track for
    // last printed digit)
    int last_entry = 0;

    int j;

    // Iterate over input array
    for (int i=0; i<arr.length(); i++)
    {
        // Initialize 'noOfNextD' to get count of
        // next D's available
        int noOfNextD = 0;

        switch(arr[i])
        {
        case 'I':
            // If letter is 'I'

            // Calculate number of next consecutive D's
            // available
            j = i+1;
            while (arr[j] == 'D' && j < arr.length())
            {
                noOfNextD++;
                j++;
            }

            if (i==0)
            {
                curr_max = noOfNextD + 2;

                // If 'I' is first letter, print incremented
                // sequence from 1
                cout << " " << ++last_entry;
                cout << " " << curr_max;

                // Set max digit reached
                last_entry = curr_max;
            }
            else
            {
                // If not first letter

                // Get next digit to print
                curr_max = curr_max + noOfNextD + 1;

                // Print digit for I
                last_entry = curr_max;
                cout << " " << last_entry;
            }

            // For all next consecutive 'D' print
            // decremented sequence
            for (int k=0; k<noOfNextD; k++)
            {
                cout << " " << --last_entry;
                i++;
            }
            break;

        // If letter is 'D'
        case 'D':
            if (i == 0)
            {
                // If 'D' is first letter in sequence
                // Find number of Next D's available
                j = i+1;
                while (arr[j] == 'D' && j < arr.length())
                {
                    noOfNextD++;
                    j++;
                }

                // Calculate first digit to print based on
                // number of consecutive D's
                curr_max = noOfNextD + 2;

                // Print twice for the first time
                cout << " " << curr_max << " " << curr_max - 1;

                // Store last entry
                last_entry = curr_max - 1;
            }
            else
            {
                // If current 'D' is not first letter

                // Decrement last_entry
                cout << " " << last_entry - 1;
                last_entry--;
            }
            break;
        }
    }
    cout << endl;
}

// Driver program to test above
int main()
{
    PrintMinNumberForPattern("IDID");
    PrintMinNumberForPattern("I");
    PrintMinNumberForPattern("DD");
    PrintMinNumberForPattern("II");
    PrintMinNumberForPattern("DIDI");
    PrintMinNumberForPattern("IIDDD");
    PrintMinNumberForPattern("DDIDDIID");
    return 0;
}

Java

// Java program to print minimum number that can be formed
// from a given sequence of Is and Ds
class GFG
{

    // Prints the minimum number that can be formed from
    // input sequence of I's and D's
    static void PrintMinNumberForPattern(String arr)
    {
        // Initialize current_max (to make sure that
        // we don't use repeated character
        int curr_max = 0;

        // Initialize last_entry (Keeps track for
        // last printed digit)
        int last_entry = 0;

        int j;

        // Iterate over input array
        for (int i = 0; i < arr.length(); i++)
        {
            // Initialize 'noOfNextD' to get count of
            // next D's available
            int noOfNextD = 0;

            switch (arr.charAt(i))
            {
                case 'I':
                    // If letter is 'I'

                    // Calculate number of next consecutive D's
                    // available
                    j = i + 1;
                    while (j < arr.length() && arr.charAt(j) == 'D')
                    {
                        noOfNextD++;
                        j++;
                    }

                    if (i == 0)
                    {
                        curr_max = noOfNextD + 2;

                        // If 'I' is first letter, print incremented
                        // sequence from 1
                        System.out.print(" " + ++last_entry);
                        System.out.print(" " + curr_max);

                        // Set max digit reached
                        last_entry = curr_max;
                    }
                    else
                    {
                        // If not first letter

                        // Get next digit to print
                        curr_max = curr_max + noOfNextD + 1;

                        // Print digit for I
                        last_entry = curr_max;
                        System.out.print(" " + last_entry);
                    }

                    // For all next consecutive 'D' print
                    // decremented sequence
                    for (int k = 0; k < noOfNextD; k++)
                    {
                        System.out.print(" " + --last_entry);
                        i++;
                    }
                    break;

                // If letter is 'D'
                case 'D':
                    if (i == 0)
                    {
                        // If 'D' is first letter in sequence
                        // Find number of Next D's available
                        j = i + 1;
                        while (j < arr.length()&&arr.charAt(j) == 'D')
                        {
                            noOfNextD++;
                            j++;
                        }

                        // Calculate first digit to print based on
                        // number of consecutive D's
                        curr_max = noOfNextD + 2;

                        // Print twice for the first time
                        System.out.print(" " + curr_max + " " + (curr_max - 1));

                        // Store last entry
                        last_entry = curr_max - 1;
                    }
                    else
                    {
                        // If current 'D' is not first letter

                        // Decrement last_entry
                        System.out.print(" " + (last_entry - 1));
                        last_entry--;
                    }
                    break;
            }
        }
        System.out.println();
    }

    // Driver code
    public static void main(String[] args)
    {
        PrintMinNumberForPattern("IDID");
        PrintMinNumberForPattern("I");
        PrintMinNumberForPattern("DD");
        PrintMinNumberForPattern("II");
        PrintMinNumberForPattern("DIDI");
        PrintMinNumberForPattern("IIDDD");
        PrintMinNumberForPattern("DDIDDIID");
    }
}

// This code is contributed by Princi Singh

Python3

# Python3 program to print minimum number that
# can be formed from a given sequence of Is and Ds

# Prints the minimum number that can be formed from
# input sequence of I's and D's
def PrintMinNumberForPattern(arr):

    # Initialize current_max (to make sure that
    # we don't use repeated character
    curr_max = 0

    # Initialize last_entry (Keeps track for
    # last printed digit)
    last_entry = 0
    i = 0

    # Iterate over input array
    while i < len(arr):

        # Initialize 'noOfNextD' to get count of
        # next D's available
        noOfNextD = 0
        if arr[i] == "I":

            # If letter is 'I'

            # Calculate number of next consecutive D's
            # available
            j = i + 1
            while j < len(arr) and arr[j] == "D":
                noOfNextD += 1
                j += 1
            if i == 0:
                curr_max = noOfNextD + 2
                last_entry += 1

                # If 'I' is first letter, print incremented
                # sequence from 1
                print("", last_entry, end = "")
                print("", curr_max, end = "")

                # Set max digit reached
                last_entry = curr_max
            else:

                # If not first letter

                # Get next digit to print
                curr_max += noOfNextD + 1

                # Print digit for I
                last_entry = curr_max
                print("", last_entry, end = "")

            # For all next consecutive 'D' print
            # decremented sequence
            for k in range(noOfNextD):
                last_entry -= 1
                print("", last_entry, end = "")
                i += 1

        # If letter is 'D'
        elif arr[i] == "D":
            if i == 0:

                # If 'D' is first letter in sequence
                # Find number of Next D's available
                j = i + 1
                while j < len(arr) and arr[j] == "D":
                    noOfNextD += 1
                    j += 1

                # Calculate first digit to print based on
                # number of consecutive D's
                curr_max = noOfNextD + 2

                # Print twice for the first time
                print("", curr_max, curr_max - 1, end = "")

                # Store last entry
                last_entry = curr_max - 1
            else:

                # If current 'D' is not first letter

                # Decrement last_entry
                print("", last_entry - 1, end = "")
                last_entry -= 1
        i += 1
    print()

# Driver code
if __name__ == "__main__":
    PrintMinNumberForPattern("IDID")
    PrintMinNumberForPattern("I")
    PrintMinNumberForPattern("DD")
    PrintMinNumberForPattern("II")
    PrintMinNumberForPattern("DIDI")
    PrintMinNumberForPattern("IIDDD")
    PrintMinNumberForPattern("DDIDDIID")

# This code is contributed by
# sanjeev2552

C#

// C# program to print minimum number that can be formed
// from a given sequence of Is and Ds
using System;

class GFG
{

    // Prints the minimum number that can be formed from
    // input sequence of I's and D's
    static void PrintMinNumberForPattern(String arr)
    {
        // Initialize current_max (to make sure that
        // we don't use repeated character
        int curr_max = 0;

        // Initialize last_entry (Keeps track for
        // last printed digit)
        int last_entry = 0;

        int j;

        // Iterate over input array
        for (int i = 0; i < arr.Length; i++)
        {
            // Initialize 'noOfNextD' to get count of
            // next D's available
            int noOfNextD = 0;

            switch (arr[i])
            {
                case 'I':
                    // If letter is 'I'

                    // Calculate number of next consecutive D's
                    // available
                    j = i + 1;
                    while (j < arr.Length && arr[j] == 'D')
                    {
                        noOfNextD++;
                        j++;
                    }

                    if (i == 0)
                    {
                        curr_max = noOfNextD + 2;

                        // If 'I' is first letter, print incremented
                        // sequence from 1
                        Console.Write(" " + ++last_entry);
                        Console.Write(" " + curr_max);

                        // Set max digit reached
                        last_entry = curr_max;
                    }
                    else
                    {
                        // If not first letter

                        // Get next digit to print
                        curr_max = curr_max + noOfNextD + 1;

                        // Print digit for I
                        last_entry = curr_max;
                        Console.Write(" " + last_entry);
                    }

                    // For all next consecutive 'D' print
                    // decremented sequence
                    for (int k = 0; k < noOfNextD; k++)
                    {
                        Console.Write(" " + --last_entry);
                        i++;
                    }
                    break;

                // If letter is 'D'
                case 'D':
                    if (i == 0)
                    {
                        // If 'D' is first letter in sequence
                        // Find number of Next D's available
                        j = i + 1;
                        while (j < arr.Length&&arr[j] == 'D')
                        {
                            noOfNextD++;
                            j++;
                        }

                        // Calculate first digit to print based on
                        // number of consecutive D's
                        curr_max = noOfNextD + 2;

                        // Print twice for the first time
                        Console.Write(" " + curr_max + " " + (curr_max - 1));

                        // Store last entry
                        last_entry = curr_max - 1;
                    }
                    else
                    {
                        // If current 'D' is not first letter

                        // Decrement last_entry
                        Console.Write(" " + (last_entry - 1));
                        last_entry--;
                    }
                    break;
            }
        }
        Console.WriteLine();
    }

    // Driver code
    public static void Main(String[] args)
    {
        PrintMinNumberForPattern("IDID");
        PrintMinNumberForPattern("I");
        PrintMinNumberForPattern("DD");
        PrintMinNumberForPattern("II");
        PrintMinNumberForPattern("DIDI");
        PrintMinNumberForPattern("IIDDD");
        PrintMinNumberForPattern("DDIDDIID");
    }
}

// This code is contributed by Princi Singh

PHP

<?php
// PHP program to print minimum
// number that can be formed
// from a given sequence of
// Is and Ds

// Prints the minimum number
// that can be formed from
// input sequence of I's and D's
function PrintMinNumberForPattern($arr)
{
    // Initialize current_max
    // (to make sure that
    // we don't use repeated
    // character
    $curr_max = 0;

    // Initialize last_entry
    // (Keeps track for
    // last printed digit)
    $last_entry = 0;

    $j;

    // Iterate over
    // input array
    for ($i = 0; $i < strlen($arr); $i++)
    {
        // Initialize 'noOfNextD'
        // to get count of
        // next D's available
        $noOfNextD = 0;

        switch($arr[$i])
        {
        case 'I':
            // If letter is 'I'

            // Calculate number of
            // next consecutive D's
            // available
            $j = $i + 1;
            while ($arr[$j] == 'D' &&
                   $j < strlen($arr))
            {
                $noOfNextD++;
                $j++;
            }

            if ($i == 0)
            {
                $curr_max = $noOfNextD + 2;

                // If 'I' is first letter,
                // print incremented
                // sequence from 1
                echo " " , ++$last_entry;
                echo " " , $curr_max;

                // Set max
                // digit reached
                $last_entry = $curr_max;
            }
            else
            {
                // If not first letter

                // Get next digit
                // to print
                $curr_max = $curr_max +
                            $noOfNextD + 1;

                // Print digit for I
                $last_entry = $curr_max;
                echo " " , $last_entry;
            }

            // For all next consecutive 'D'
            // print decremented sequence
            for ($k = 0; $k < $noOfNextD; $k++)
            {
                echo " " , --$last_entry;
                $i++;
            }
            break;

        // If letter is 'D'
        case 'D':
            if ($i == 0)
            {
                // If 'D' is first letter
                // in sequence. Find number
                // of Next D's available
                $j = $i+1;
                while (($arr[$j] == 'D') &&
                       ($j < strlen($arr)))
                {
                    $noOfNextD++;
                    $j++;
                }

                // Calculate first digit
                // to print based on
                // number of consecutive D's
                $curr_max = $noOfNextD + 2;

                // Print twice for
                // the first time
                echo " " , $curr_max ,
                     " " ,$curr_max - 1;

                // Store last entry
                $last_entry = $curr_max - 1;
            }
            else
            {
                // If current 'D'
                // is not first letter

                // Decrement last_entry
                echo " " , $last_entry - 1;
                $last_entry--;
            }
            break;
        }
    }

echo "\n";
}

// Driver Code
PrintMinNumberForPattern("IDID");
PrintMinNumberForPattern("I");
PrintMinNumberForPattern("DD");
PrintMinNumberForPattern("II");
PrintMinNumberForPattern("DIDI");
PrintMinNumberForPattern("IIDDD");
PrintMinNumberForPattern("DDIDDIID");

// This code is contributed by aj_36
?>

Javascript

<script>
// Javascript program to print minimum number that can be formed
// from a given sequence of Is and Ds

// Prints the minimum number that can be formed from
    // input sequence of I's and D's
function PrintMinNumberForPattern(arr)
{
    // Initialize current_max (to make sure that
        // we don't use repeated character
        let curr_max = 0;

        // Initialize last_entry (Keeps track for
        // last printed digit)
        let last_entry = 0;

        let j;

        // Iterate over input array
        for (let i = 0; i < arr.length; i++)
        {
            // Initialize 'noOfNextD' to get count of
            // next D's available
            let noOfNextD = 0;

            switch (arr[i])
            {
                case 'I':
                    // If letter is 'I'

                    // Calculate number of next consecutive D's
                    // available
                    j = i + 1;
                    while (j < arr.length && arr[j] == 'D')
                    {
                        noOfNextD++;
                        j++;
                    }

                    if (i == 0)
                    {
                        curr_max = noOfNextD + 2;

                        // If 'I' is first letter, print incremented
                        // sequence from 1
                        document.write(" " + ++last_entry);
                        document.write(" " + curr_max);

                        // Set max digit reached
                        last_entry = curr_max;
                    }
                    else
                    {
                        // If not first letter

                        // Get next digit to print
                        curr_max = curr_max + noOfNextD + 1;

                        // Print digit for I
                        last_entry = curr_max;
                        document.write(" " + last_entry);
                    }

                    // For all next consecutive 'D' print
                    // decremented sequence
                    for (let k = 0; k < noOfNextD; k++)
                    {
                        document.write(" " + --last_entry);
                        i++;
                    }
                    break;

                // If letter is 'D'
                case 'D':
                    if (i == 0)
                    {
                        // If 'D' is first letter in sequence
                        // Find number of Next D's available
                        j = i + 1;
                        while (j < arr.length && arr[j] == 'D')
                        {
                            noOfNextD++;
                            j++;
                        }

                        // Calculate first digit to print based on
                        // number of consecutive D's
                        curr_max = noOfNextD + 2;

                        // Print twice for the first time
                        document.write(" " + curr_max + " " + (curr_max - 1));

                        // Store last entry
                        last_entry = curr_max - 1;
                    }
                    else
                    {
                        // If current 'D' is not first letter

                        // Decrement last_entry
                        document.write(" " + (last_entry - 1));
                        last_entry--;
                    }
                    break;
            }
        }
        document.write("<br>");
}

// Driver code
PrintMinNumberForPattern("IDID");
PrintMinNumberForPattern("I");
PrintMinNumberForPattern("DD");
PrintMinNumberForPattern("II");
PrintMinNumberForPattern("DIDI");
PrintMinNumberForPattern("IIDDD");
PrintMinNumberForPattern("DDIDDIID");

// This code is contributed by ab2127
</script>

Output

 1 3 2 5 4
 1 2
 3 2 1
 1 2 3
 2 1 4 3 5
 1 2 6 5 4 3
 3 2 1 6 5 4 7 9 8

Time Complexity: O(N^2), overall time complexity. Where, N is the length of the string.
Auxiliary Space: O(1).

This solution is suggested by Swapnil Trambake.

Alternate Solution:
Let’s observe a few facts in case of a minimum number: 

• The digits can’t repeat hence there can be 9 digits at most in output.
• To form a minimum number , at every index of the output, we are interested in the minimum number which can be placed at that index.

The idea is to iterate over the entire input array , keeping track of the minimum number (1-9) which can be placed at that position of the output.

The tricky part of course occurs when ‘D’ is encountered at index other than 0. In such a case we have to track the nearest ‘I’ to the left of ‘D’ and increment each number in the output vector by 1 in between ‘I’ and ‘D’. 
We cover the base case as follows: 

• If the first character of input is ‘I’ then we append 1 and 2 in the output vector and the minimum available number is set to 3 .The index of most recent ‘I’ is set to 1.
• If the first character of input is ‘D’ then we append 2 and 1 in the output vector and the minimum available number is set to 3, and the index of most recent ‘I’ is set to 0.

Now we iterate the input string from index 1 till its end and: 

• If the character scanned is ‘I’ , a minimum value that has not been used yet is appended to the output vector .We increment the value of minimum no. available and index of most recent ‘I’ is also updated.
• If the character scanned is ‘D’ at index i of input array, we append the ith element from output vector in the output and track the nearest ‘I’ to the left of ‘D’ and increment each number in the output vector by 1 in between ‘I’ and ‘D’.

Following is the program for the same: 

C++

// C++ program to print minimum number that can be formed
// from a given sequence of Is and Ds
#include<bits/stdc++.h>
using namespace std;

void printLeast(string arr)
{
    // min_avail represents the minimum number which is
    // still available for inserting in the output vector.
    // pos_of_I keeps track of the most recent index
    // where 'I' was encountered w.r.t the output vector
    int min_avail = 1, pos_of_I = 0;

    //vector to store the output
    vector<int>v;

    // cover the base cases
    if (arr[0]=='I')
    {
        v.push_back(1);
        v.push_back(2);
        min_avail = 3;
        pos_of_I = 1;
    }
    else
    {
        v.push_back(2);
        v.push_back(1);
        min_avail = 3;
        pos_of_I = 0;
    }

    // Traverse rest of the input
    for (int i=1; i<arr.length(); i++)
    {
        if (arr[i]=='I')
        {
            v.push_back(min_avail);
            min_avail++;
            pos_of_I = i+1;
        }
        else
        {
            v.push_back(v[i]);
            for (int j=pos_of_I; j<=i; j++)
                v[j]++;

            min_avail++;
        }
    }

    // print the number
    for (int i=0; i<v.size(); i++)
        cout << v[i] << " ";
    cout << endl;
}

// Driver program to check the above function
int main()
{
    printLeast("IDID");
    printLeast("I");
    printLeast("DD");
    printLeast("II");
    printLeast("DIDI");
    printLeast("IIDDD");
    printLeast("DDIDDIID");
    return 0;
}

Java

// Java program to print minimum number that can be formed
// from a given sequence of Is and Ds
import java.io.*;
import java.util.*;
public class GFG {

       static void printLeast(String arr)
       {
              // min_avail represents the minimum number which is
              // still available for inserting in the output vector.
              // pos_of_I keeps track of the most recent index
              // where 'I' was encountered w.r.t the output vector
              int min_avail = 1, pos_of_I = 0;

              //vector to store the output
              ArrayList<Integer> al = new ArrayList<>();

              // cover the base cases
              if (arr.charAt(0) == 'I')
              {
                  al.add(1);
                  al.add(2);
                  min_avail = 3;
                  pos_of_I = 1;
              }

              else
              {
                  al.add(2);
                  al.add(1);
                  min_avail = 3;
                  pos_of_I = 0;
              }

              // Traverse rest of the input
              for (int i = 1; i < arr.length(); i++)
              {
                   if (arr.charAt(i) == 'I')
                   {
                       al.add(min_avail);
                       min_avail++;
                       pos_of_I = i + 1;
                   }
                   else
                   {
                       al.add(al.get(i));
                       for (int j = pos_of_I; j <= i; j++)
                            al.set(j, al.get(j) + 1);

                       min_avail++;
                   }
              }

              // print the number
              for (int i = 0; i < al.size(); i++)
                   System.out.print(al.get(i) + " ");
              System.out.println();
       }


       // Driver code
       public static void main(String args[])
       {
              printLeast("IDID");
              printLeast("I");
              printLeast("DD");
              printLeast("II");
              printLeast("DIDI");
              printLeast("IIDDD");
              printLeast("DDIDDIID");
       }
}
// This code is contributed by rachana soma

Python3

# Python3 program to print minimum number
# that can be formed from a given sequence
# of Is and Ds
def printLeast(arr):

    # min_avail represents the minimum
    # number which is still available
    # for inserting in the output vector.
    # pos_of_I keeps track of the most
    # recent index where 'I' was
    # encountered w.r.t the output vector
    min_avail = 1
    pos_of_I = 0

    # Vector to store the output
    v = []

    # Cover the base cases
    if (arr[0] == 'I'):
        v.append(1)
        v.append(2)

        min_avail = 3
        pos_of_I = 1
    else:
        v.append(2)
        v.append(1)

        min_avail = 3
        pos_of_I = 0

    # Traverse rest of the input
    for i in range(1, len(arr)):
        if (arr[i] == 'I'):
            v.append(min_avail)
            min_avail += 1
            pos_of_I = i + 1
        else:
            v.append(v[i])
            for j in range(pos_of_I, i + 1):
                v[j] += 1
            min_avail += 1

    # Print the number
    print(*v, sep = ' ')

# Driver code
printLeast("IDID")
printLeast("I")
printLeast("DD")
printLeast("II")
printLeast("DIDI")
printLeast("IIDDD")
printLeast("DDIDDIID")

# This code is contributed by avanitrachhadiya2155

C#

// C# program to print minimum number that can be formed
// from a given sequence of Is and Ds
using System;
using System.Collections.Generic;

class GFG
{

static void printLeast(String arr)
{
    // min_avail represents the minimum number which is
    // still available for inserting in the output vector.
    // pos_of_I keeps track of the most recent index
    // where 'I' was encountered w.r.t the output vector
    int min_avail = 1, pos_of_I = 0;

    //vector to store the output
    List<int> al = new List<int>();

    // cover the base cases
    if (arr[0] == 'I')
    {
        al.Add(1);
        al.Add(2);
        min_avail = 3;
        pos_of_I = 1;
    }

    else
    {
        al.Add(2);
        al.Add(1);
        min_avail = 3;
        pos_of_I = 0;
    }

    // Traverse rest of the input
    for (int i = 1; i < arr.Length; i++)
    {
        if (arr[i] == 'I')
        {
            al.Add(min_avail);
            min_avail++;
            pos_of_I = i + 1;
        }
        else
        {
            al.Add(al[i]);
            for (int j = pos_of_I; j <= i; j++)
                al[j] = al[j] + 1;

            min_avail++;
        }
    }

    // print the number
    for (int i = 0; i < al.Count; i++)
        Console.Write(al[i] + " ");
    Console.WriteLine();
}


// Driver code
public static void Main(String []args)
{
    printLeast("IDID");
    printLeast("I");
    printLeast("DD");
    printLeast("II");
    printLeast("DIDI");
    printLeast("IIDDD");
    printLeast("DDIDDIID");
}
}

// This code is contributed by Rajput-Ji

Javascript

<script>

    // Javascript program to print
    // minimum number that can be formed
    // from a given sequence of Is and Ds

    function printLeast(arr)
    {
        // min_avail represents the
        // minimum number which is
        // still available for inserting
        // in the output vector.
        // pos_of_I keeps track of the
        // most recent index
        // where 'I' was encountered
        // w.r.t the output vector
        let min_avail = 1, pos_of_I = 0;

        //vector to store the output
        let al = [];

        // cover the base cases
        if (arr[0] == 'I')
        {
            al.push(1);
            al.push(2);
            min_avail = 3;
            pos_of_I = 1;
        }

        else
        {
            al.push(2);
            al.push(1);
            min_avail = 3;
            pos_of_I = 0;
        }

        // Traverse rest of the input
        for (let i = 1; i < arr.length; i++)
        {
            if (arr[i] == 'I')
            {
                al.push(min_avail);
                min_avail++;
                pos_of_I = i + 1;
            }
            else
            {
                al.push(al[i]);
                for (let j = pos_of_I; j <= i; j++)
                    al[j] = al[j] + 1;

                min_avail++;
            }
        }

        // print the number
        for (let i = 0; i < al.length; i++)
            document.write(al[i] + " ");
        document.write("</br>");
    }

    printLeast("IDID");
    printLeast("I");
    printLeast("DD");
    printLeast("II");
    printLeast("DIDI");
    printLeast("IIDDD");
    printLeast("DDIDDIID");

</script>

Output

1 3 2 5 4 
1 2 
3 2 1 
1 2 3 
2 1 4 3 5 
1 2 6 5 4 3 
3 2 1 6 5 4 7 9 8 

Time Complexity: O(N²) ,here N is length of string .
Auxiliary Space: O(N)  since N extra space has been taken.

This solution is suggested by Ashutosh Kumar.

Method 3 
We can that when we encounter I, we got numbers in increasing order but if we encounter ‘D’, we want to have numbers in decreasing order. Length of the output string is always one more than the input string. So the loop is from 0 to the length of the string. We have to take numbers from 1-9 so we always push (i+1) to our stack. Then we check what is the resulting character at the specified index.So, there will be two cases which are as follows:- 

Case 1: If we have encountered I or we are at the last character of input string, then pop from the stack and add it to the end of the output string until the stack gets empty. 

Case 2: If we have encountered D, then we want the numbers in decreasing order. so we just push (i+1) to our stack.

C++

// C++ program to print minimum number that can be formed
// from a given sequence of Is and Ds
#include <bits/stdc++.h>
using namespace std;

// Function to decode the given sequence to construct
// minimum number without repeated digits
void PrintMinNumberForPattern(string seq)
{
    // result store output string
    string result;

    // create an empty stack of integers
    stack<int> stk;

    // run n+1 times where n is length of input sequence
    for (int i = 0; i <= seq.length(); i++)
    {
        // push number i+1 into the stack
        stk.push(i + 1);

        // if all characters of the input sequence are
        // processed or current character is 'I'
        // (increasing)
        if (i == seq.length() || seq[i] == 'I')
        {
            // run till stack is empty
            while (!stk.empty())
            {
                // remove top element from the stack and
                // add it to solution
                result += to_string(stk.top());
                result += " ";
                stk.pop();
            }
        }
    }

    cout << result << endl;
}

// main function
int main()
{
    PrintMinNumberForPattern("IDID");
    PrintMinNumberForPattern("I");
    PrintMinNumberForPattern("DD");
    PrintMinNumberForPattern("II");
    PrintMinNumberForPattern("DIDI");
    PrintMinNumberForPattern("IIDDD");
    PrintMinNumberForPattern("DDIDDIID");
    return 0;
}

Java

import java.util.Stack;

// Java program to print minimum number that can be formed
// from a given sequence of Is and Ds
class GFG {

// Function to decode the given sequence to construct
// minimum number without repeated digits
    static void PrintMinNumberForPattern(String seq) {
        // result store output string
        String result = "";

        // create an empty stack of integers
        Stack<Integer> stk = new Stack<Integer>();

        // run n+1 times where n is length of input sequence
        for (int i = 0; i <= seq.length(); i++) {
            // push number i+1 into the stack
            stk.push(i + 1);

            // if all characters of the input sequence are
            // processed or current character is 'I'
            // (increasing)
            if (i == seq.length() || seq.charAt(i) == 'I') {
                // run till stack is empty
                while (!stk.empty()) {
                    // remove top element from the stack and
                    // add it to solution
                    result += String.valueOf(stk.peek());
                    result += " ";
                    stk.pop();
                }
            }
        }

        System.out.println(result);
    }

// main function
    public static void main(String[] args) {
        PrintMinNumberForPattern("IDID");
        PrintMinNumberForPattern("I");
        PrintMinNumberForPattern("DD");
        PrintMinNumberForPattern("II");
        PrintMinNumberForPattern("DIDI");
        PrintMinNumberForPattern("IIDDD");
        PrintMinNumberForPattern("DDIDDIID");
    }
}
// This code is contributed by PrinciRaj1992

Python3

# Python3 program to print minimum
# number that can be formed from a
# given sequence of Is and Ds
def PrintMinNumberForPattern(Strr):

    # Take a List to work as Stack
    stack = []

    # String for storing result
    res = ''

    # run n+1 times where n is length
    # of input sequence, As length of
    # result string is always 1 greater
    for i in range(len(Strr) + 1):

        # Push number i+1 into the stack
        stack.append(i + 1)

        # If all characters of the input
        # sequence are processed or current
        # character is 'I
        if (i == len(Strr) or Strr[i] == 'I'):

            # Run While Loop Until stack is empty
            while len(stack) > 0:

                # pop the element on top of stack
                # And store it in result String
                res += str(stack.pop())
                res += ' '

    # Print the result
    print(res)

# Driver Code
PrintMinNumberForPattern("IDID")
PrintMinNumberForPattern("I")
PrintMinNumberForPattern("DD")
PrintMinNumberForPattern("II")
PrintMinNumberForPattern("DIDI")
PrintMinNumberForPattern("IIDDD")
PrintMinNumberForPattern("DDIDDIID")

# This code is contributed by AyushManglani

C#

// C# program to print minimum number that can be formed
// from a given sequence of Is and Ds
using System;
using System.Collections;
public class GFG {

// Function to decode the given sequence to construct
// minimum number without repeated digits
    static void PrintMinNumberForPattern(String seq) {
        // result store output string
        String result = "";

        // create an empty stack of integers
        Stack stk = new Stack();

        // run n+1 times where n is length of input sequence
        for (int i = 0; i <= seq.Length; i++) {
            // push number i+1 into the stack
            stk.Push(i + 1);

            // if all characters of the input sequence are
            // processed or current character is 'I'
            // (increasing)
            if (i == seq.Length || seq[i] == 'I') {
                // run till stack is empty
                while (stk.Count!=0) {
                    // remove top element from the stack and
                    // add it to solution
                    result += String.Join("",stk.Peek());
                    result += " ";
                    stk.Pop();
                }
            }
        }

        Console.WriteLine(result);
    }

// main function
    public static void Main() {
        PrintMinNumberForPattern("IDID");
        PrintMinNumberForPattern("I");
        PrintMinNumberForPattern("DD");
        PrintMinNumberForPattern("II");
        PrintMinNumberForPattern("DIDI");
        PrintMinNumberForPattern("IIDDD");
        PrintMinNumberForPattern("DDIDDIID");
    }
}
// This code is contributed by 29AjayKumar

Javascript

<script>

    // Javascript program to print
    // minimum number that can be formed
    // from a given sequence of Is and Ds

    // Function to decode the given
    // sequence to construct
    // minimum number without repeated digits
    function PrintMinNumberForPattern(seq)
    {
        // result store output string
        let result = "";

        // create an empty stack of integers
        let stk = [];

        // run n+1 times where n is length
        // of input sequence
        for (let i = 0; i <= seq.length; i++)
        {
            // push number i+1 into the stack
            stk.push(i + 1);

            // if all characters of the input
            // sequence are
            // processed or current character is 'I'
            // (increasing)
            if (i == seq.length || seq[i] == 'I')
            {
                // run till stack is empty
                while (stk.length!=0) {
                // remove top element from
               // the stack and
              // add it to solution
                 result +=
                 (stk[stk.length - 1]).toString();
                 result += " ";
                 stk.pop();
                }
            }
        }

        document.write(result + "</br>");
    }

    PrintMinNumberForPattern("IDID");
    PrintMinNumberForPattern("I");
    PrintMinNumberForPattern("DD");
    PrintMinNumberForPattern("II");
    PrintMinNumberForPattern("DIDI");
    PrintMinNumberForPattern("IIDDD");
    PrintMinNumberForPattern("DDIDDIID");

</script>

Output

1 3 2 5 4 
1 2 
3 2 1 
1 2 3 
2 1 4 3 5 
1 2 6 5 4 3 
3 2 1 6 5 4 7 9 8 

Time Complexity: O(n) 
Auxiliary Space: O(n),  since n extra space has been taken.
This method is contributed by Roshni Agarwal. 

Method 4 (Using two pointers)
Observation

1. Since we have to find a minimum number without repeating digits, maximum length of output can be 9 (using each 1-9 digits once)
2. Length of the output will be exactly one greater than input length.
3. The idea is to iterate over the string and do the following if current character is ‘I’ or string is ended. 
   1. Assign count in increasing order to each element from current-1 to the next left index of ‘I’ (or starting index is reached).
   2. Increase the count by 1.

Input  :  IDID
Output : 13254

Input  :  I
Output : 12

Input  :  DD
Output : 321

Input  :  II
Output : 123

Input  :  DIDI
Output : 21435

Input  :  IIDDD
Output : 126543

Input  :  DDIDDIID
Output : 321654798

Below is the implementation of above approach:

C++

// C++ program of above approach
#include <bits/stdc++.h>
using namespace std;

// Returns minimum number made from given sequence without repeating digits
string getMinNumberForPattern(string seq)
{
    int n = seq.length();

    if (n >= 9)
        return "-1";

    string result(n+1, ' ');

    int count = 1;

    // The loop runs for each input character as well as
    // one additional time for assigning rank to remaining characters
    for (int i = 0; i <= n; i++)
    {
        if (i == n || seq[i] == 'I')
        {
            for (int j = i - 1 ; j >= -1 ; j--)
            {
                result[j + 1] = '0' + count++;
                if(j >= 0 && seq[j] == 'I')
                    break;
            }
        }
    }
    return result;
}

// main function
int main()
{
    string inputs[] = {"IDID", "I", "DD", "II", "DIDI", "IIDDD", "DDIDDIID"};

    for (string input : inputs)
    {
        cout << getMinNumberForPattern(input) << "\n";
    }
    return 0;
}

Java

// Java program of above approach
import java.io.IOException;

public class Test
{
    // Returns minimum number made from given sequence without repeating digits
    static String getMinNumberForPattern(String seq)
    {
        int n = seq.length();

        if (n >= 9)
            return "-1";

        char result[] = new char[n + 1];

        int count = 1;

        // The loop runs for each input character as well as
        // one additional time for assigning rank to each remaining characters
        for (int i = 0; i <= n; i++)
        {
            if (i == n || seq.charAt(i) == 'I')
            {
                for (int j = i - 1; j >= -1; j--)
                {
                    result[j + 1] = (char) ((int) '0' + count++);
                    if (j >= 0 && seq.charAt(j) == 'I')
                        break;
                }
            }
        }
        return new String(result);
    }

    public static void main(String[] args) throws IOException
    {
        String inputs[] = { "IDID", "I", "DD", "II", "DIDI", "IIDDD", "DDIDDIID" };

        for(String input : inputs)
        {
            System.out.println(getMinNumberForPattern(input));
        }
    }
}

Python3

# Python3 program of above approach

# Returns minimum number made from
# given sequence without repeating digits
def getMinNumberForPattern(seq):
    n = len(seq)

    if (n >= 9):
        return "-1"

    result = [None] * (n + 1)

    count = 1

    # The loop runs for each input character
    # as well as one additional time for
    # assigning rank to remaining characters
    for i in range(n + 1):
        if (i == n or seq[i] == 'I'):
            for j in range(i - 1, -2, -1):
                result[j + 1] = int('0' + str(count))
                count += 1
                if(j >= 0 and seq[j] == 'I'):
                    break
    return result

# Driver Code
if __name__ == '__main__':
    inputs = ["IDID", "I", "DD", "II",
              "DIDI", "IIDDD", "DDIDDIID"]
    for Input in inputs:
        print(*(getMinNumberForPattern(Input)))

# This code is contributed by PranchalK

C#

// C# program of above approach
using System;
class GFG
{

// Returns minimum number made from given
// sequence without repeating digits
static String getMinNumberForPattern(String seq)
{
    int n = seq.Length;

    if (n >= 9)
        return "-1";

    char []result = new char[n + 1];

    int count = 1;

    // The loop runs for each input character
    // as well as one additional time for
    // assigning rank to each remaining characters
    for (int i = 0; i <= n; i++)
    {
        if (i == n || seq[i] == 'I')
        {
            for (int j = i - 1; j >= -1; j--)
            {
                result[j + 1] = (char) ((int) '0' + count++);
                if (j >= 0 && seq[j] == 'I')
                    break;
            }
        }
    }
    return new String(result);
}

// Driver Code
public static void Main()
{
    String []inputs = { "IDID", "I", "DD", "II",
                        "DIDI", "IIDDD", "DDIDDIID" };

    foreach(String input in inputs)
    {
        Console.WriteLine(getMinNumberForPattern(input));
    }
}
}

// This code is contributed by Rajput-Ji

Javascript

<script>

    // Javascript program of above approach

    // Returns minimum number made from given
    // sequence without repeating digits
    function getMinNumberForPattern(seq)
    {
        let n = seq.length;

        if (n >= 9)
            return "-1";

        let result = new Array(n + 1);

        let count = 1;

        // The loop runs for each input character
        // as well as one additional time for
        // assigning rank to each remaining characters
        for (let i = 0; i <= n; i++)
        {
            if (i == n || seq[i] == 'I')
            {
                for (let j = i - 1; j >= -1; j--)
                {
                    result[j + 1] =
                    String.fromCharCode('0'.charCodeAt() +
                    count++);
                    if (j >= 0 && seq[j] == 'I')
                        break;
                }
            }
        }
        return result.join("");
    }

    let inputs = [ "IDID", "I", "DD", "II", "DIDI",
    "IIDDD", "DDIDDIID" ];

    for(let input = 0; input < inputs.length; input++)
    {
        document.write(
        getMinNumberForPattern(inputs[input]) + "</br>"
);
    }

</script>

Output

13254
12
321
123
21435
126543
321654798

Time Complexity: O(N) 
Auxiliary Space: O(N), since N extra space has been taken.
This solution is suggested by Brij Desai.

Method 5 (Start with the Smallest)

Start with the smallest number as the answer and keep shifting the digits when we encounter a D. 
There is no need to traverse back for the index.
Follow the below steps,

1. Start with the smallest number for len(s)+1 (say for DI, start with “123”)
2. Now, starting with the second digit (index 1) and first character (D), iterate until end of the digits list, keeping track of the first D in a sequence of Ds
   1. When we encounter a D
      move the digit at current index to the first D in the sequence
   2. When we encounter an I
      reset the last known location of D. Nothing to move as the digit is correctly placed (as of now…)

Below is the implementation of the above approach:

C++

// c++ program to generate required sequence
#include <iostream>
#include <stdlib.h>
#include <string>
#include <vector>
using namespace std;

//:param s: a seq consisting only of 'D' and 'I' chars. D is
//for decreasing and I for increasing :return: digits from
//1-9 that fit the str. The number they represent should the min
//such number
vector<string> didi_seq_gen(string s)
{
    if (s.size() == 0)
        return {};
    vector<string> base_list = { "1" };
    for (int i = 2; i < s.size() + 2; i++)
        base_list.push_back(to_string(i));
    int last_D = -1;
    for (int i = 1; i < base_list.size(); i++) {
        if (s[i - 1] == 'D') {
            if (last_D < 0)
                last_D = i - 1;
            string v = base_list[i];
            base_list.erase(base_list.begin() + i);
            base_list.insert(base_list.begin() + last_D, v);
        }
        else
            last_D = -1;
    }
    return base_list;
}

int main()
{
    vector<string> inputs
        = { "IDID", "I",     "DD",      "II",
            "DIDI", "IIDDD", "DDIDDIID" };
    for (auto x : inputs) {
        vector<string> ans = didi_seq_gen(x);
        for (auto i : ans) {
            cout << i;
        }
        cout << endl;
    }
    return 0;
}

Java

// Java program to generate required sequence
import java.util.*;

public class Main {
    public static void main(String[] args)
    {
        String[] inputs
            = { "IDID", "I",     "DD",      "II",
                "DIDI", "IIDDD", "DDIDDIID" };
        for (String x : inputs) {
            List<String> ans = didi_seq_gen(x);
            for (String i : ans) {
                System.out.print(i);
            }
            System.out.println();
        }
    }
    //:param s: a seq consisting only of 'D' and 'I' chars.
    //D is for decreasing and I for increasing :return:
    // digits from 1-9 that fit the str. The number they represent
    // should the min such number
    public static List<String> didi_seq_gen(String s)
    {
        if (s.length() == 0)
            return new ArrayList<>();
        List<String> base_list
            = new ArrayList<>(Arrays.asList("1"));
        for (int i = 2; i < s.length() + 2; i++)
            base_list.add(Integer.toString(i));
        int last_D = -1;
        for (int i = 1; i < base_list.size(); i++) {
            if (s.charAt(i - 1) == 'D') {
                if (last_D < 0)
                    last_D = i - 1;
                String v = base_list.get(i);
                base_list.remove(i);
                base_list.add(last_D, v);
            }
            else {
                last_D = -1;
            }
        }
        return base_list;
    }
}

// This code is contributed by Tapesh (tapeshdua420)

Python3

# Python implementation of the above approach

def didi_seq_gen(s: str):
    '''
    :param s: a seq consisting only of 'D'
    and 'I' chars. D is for decreasing and
    I for increasing
    :return: digits from 1-9 that fit the str.
    The number they represent should the min
    such number
    :rtype: str
    example : for seq DII -> 2134
    '''
    if not s or len(s) <= 0:
        return ""
    base_list = ["1"]
    for i in range(1, len(s) + 1):
        base_list.append(f'{i + 1}')

    last_D = -1
    for i in range(1, len(base_list)):
        if s[i - 1] == 'D':
            if last_D < 0:
                last_D = i - 1
            v = base_list[i]
            del base_list[i]
            base_list.insert(last_D, v)
        else:
            last_D = -1

    return base_list

# Driver Code
# Function call
print(didi_seq_gen("IDID"))
print(didi_seq_gen("I"))
print(didi_seq_gen("DD"))
print(didi_seq_gen("II"))
print(didi_seq_gen("DIDI"))
print(didi_seq_gen("IIDDD"))
print(didi_seq_gen("DDIDDIID" ))

C#

// Include namespace system
using System;
using System.Collections.Generic;

using System.Linq;
using System.Collections;

public class GFG
{
  public static void Main(String[] args)
  {
    String[] inputs = {"IDID", "I", "DD", "II", "DIDI", "IIDDD", "DDIDDIID"};
    foreach (String x in inputs)
    {
      var ans = GFG.didi_seq_gen(x);
      foreach (String i in ans)
      {
        Console.Write(i);
      }
      Console.WriteLine();
    }
  }
  // :param s: a seq consisting only of 'D' and 'I' chars.
  // D is for decreasing and I for increasing :return:
  // digits from 1-9 that fit the str. The number they represent
  // should the min such number
  public static List<String> didi_seq_gen(String s)
  {
    if (s.Length == 0)
    {
      return new List<String>();
    }
    var base_list = new List<String>();
    base_list.Add("1");
    for (int i = 2; i < s.Length + 2; i++)
    {
      base_list.Add(Convert.ToString(i));
    }
    var last_D = -1;
    for (int i = 1; i < base_list.Count; i++)
    {
      if (s[i - 1] == 'D')
      {
        if (last_D < 0)
        {
          last_D = i - 1;
        }
        var v = base_list[i];
        base_list.RemoveAt(i);
        base_list.Insert(last_D,v);
      }
      else
      {
        last_D = -1;
      }
    }
    return base_list;
  }
}

// This code is contributed by aadityaburujwale.

Javascript

// JavaScript implementation of the above approach

function didi_seq_gen(s)
{

    if (!s || s.length <= 0)
        return ""

    let base_list = ["1"]
    for (var i = 1; i <= s.length; i++)
        base_list.push((i + 1).toString())

    let last_D = -1
    for (var i = 1; i < base_list.length; i++)
    {
        if (s[i - 1] == 'D')
        {
            if (last_D < 0)
                last_D = i - 1
            v = base_list[i]
            base_list.splice(i, 1)
            base_list.splice(last_D, 0, v)
        }
        else
            last_D = -1
    }
    return base_list.join("")
}

// Driver Code
// Function call
console.log(didi_seq_gen("IDID"))
console.log(didi_seq_gen("I"))
console.log(didi_seq_gen("DD"))
console.log(didi_seq_gen("II"))
console.log(didi_seq_gen("DIDI"))
console.log(didi_seq_gen("IIDDD"))
console.log(didi_seq_gen("DDIDDIID" ))

// This code is contributed by poojaagarwal2.

Output

13254
12
321
123
21435
126543
321654798

Time Complexity: O(N) 
Auxiliary Space: O(N)

Method 6 : (Space Optimized and modular code of Method 1)

Examples:

Input: "DDDD"
Output: "432156"

For input 1, pattern is like,     D -> D -> D -> D
                               5   4    3    2    1
                               
Input: "DDDII"
Output: "432156"

For input 2, pattern is like,     D -> D -> D -> I -> I
                               4   3    2     1      5       6


Input: "IIDIDIII"
Output: "124365789"

For input 3, pattern is like,     I -> I -> D -> I -> D -> I -> I -> I
                               1    2   4    3    6    5    7    8    9    

Approach:

• Think if the string contains only characters ‘I’ increasing, then there isn’t any problem you can just print and keep incrementing.
• Now think if the string contains only characters ‘D’ increasing, then you somehow have to get the number ‘D’ characters present from initial point, so that you can start from total count of ‘D’ and print by decrementing.
• The problem is when you encounter character ‘D’ after character ‘I’. Here somehow you have to get count of ‘D’ to get the next possible decremental start for ‘D’ and then print by decrementing until you have encountered all of ‘D’.
• Here in this approach the code has been made more modular compared to method 1 of space optimized version.

C++

// This code illustrates to find minimum number following
// pattern with optimized space and modular code.
#include <bits/stdc++.h>
using namespace std;


// This function returns minimum number following
// pattern of increasing or decreasing sequence.
string findMinNumberPattern(string str)
{
    string ans = ""; // Minimum number following pattern

    int i = 0;
    int cur = 1; // cur val following pattern
    int dCount = 0; // Count of char 'D'
    while (i < str.length()) {

        char ch = str[i];

        // If 1st ch == 'I', incr and add to ans
        if (i == 0 && ch == 'I') {
            ans += to_string(cur);
            cur++;
        }

        // If cur char == 'D',
        // incr dCount as well, since we always
        // start counting for dCount from i+1
        if (ch == 'D') {
            dCount++;
        }

        int j = i + 1; // Count 'D' from i+1 index
        while (j < str.length()
               && str[j] == 'D') {
            dCount++;
            j++;
        }

        int k = dCount;  // Store dCount
        while (dCount >= 0) {
            ans += to_string(cur + dCount);
            dCount--;
        }

        cur += (k + 1); // Manages next cur val
        dCount = 0;
        i = j;
    }

    return ans;
}

int main()
{
    cout << (findMinNumberPattern("DIDID")) << endl;
    cout << (findMinNumberPattern("DIDIII")) << endl;
    cout << (findMinNumberPattern("DDDIIDI")) << endl;
    cout << (findMinNumberPattern("IDIDIID")) << endl;
    cout << (findMinNumberPattern("DIIDIDD")) << endl;
    cout << (findMinNumberPattern("IIDIDDD")) << endl;

    return 0;
}

// This code is contributed by suresh07.

Java

/*package whatever //do not write package name here */

// This code illustrates to find minimum number following
// pattern with optimized space and modular code.

import java.io.*;

class GFG {

    // This function returns minimum number following
    // pattern of increasing or decreasing sequence.
    public static String findMinNumberPattern(String str)
    {
        String ans = ""; // Minimum number following pattern

        int i = 0;
        int cur = 1; // cur val following pattern
        int dCount = 0; // Count of char 'D'
        while (i < str.length()) {

            char ch = str.charAt(i);

            // If 1st ch == 'I', incr and add to ans
            if (i == 0 && ch == 'I') {
                ans += cur;
                cur++;
            }

            // If cur char == 'D',
            // incr dCount as well, since we always
            // start counting for dCount from i+1
            if (ch == 'D') {
                dCount++;
            }

            int j = i + 1; // Count 'D' from i+1 index
            while (j < str.length()
                   && str.charAt(j) == 'D') {
                dCount++;
                j++;
            }

            int k = dCount;  // Store dCount
            while (dCount >= 0) {
                ans += (cur + dCount);
                dCount--;
            }

            cur += (k + 1); // Manages next cur val
            dCount = 0;
            i = j;
        }

        return ans;
    }
    public static void main(String[] args)
    {
        System.out.println(findMinNumberPattern("DIDID"));
        System.out.println(findMinNumberPattern("DIDIII"));
        System.out.println(findMinNumberPattern("DDDIIDI"));
        System.out.println(findMinNumberPattern("IDIDIID"));
        System.out.println(findMinNumberPattern("DIIDIDD"));
        System.out.println(findMinNumberPattern("IIDIDDD"));
    }
}

// This code is contributed by Arun M

Python3

# This code illustrates to find minimum number following
# pattern with optimized space and modular code.

# This function returns minimum number following
# pattern of increasing or decreasing sequence.
def findMinNumberPattern(Str):

    ans = "" # Minimum number following pattern

    i = 0
    cur = 1 # cur val following pattern
    dCount = 0 # Count of char 'D'
    while (i < len(Str)) :

        ch = Str[i]

        # If 1st ch == 'I', incr and add to ans
        if (i == 0 and ch == 'I') :
            ans += str(cur)
            cur+=1

        # If cur char == 'D',
        # incr dCount as well, since we always
        # start counting for dCount from i+1
        if (ch == 'D') :
            dCount+=1


        j = i + 1 # Count 'D' from i+1 index
        while (j < len(Str) and Str[j] == 'D') :
            dCount+=1
            j+=1


        k = dCount  # Store dCount
        while (dCount >= 0) :
            ans += str(cur + dCount)
            dCount-=1


        cur += (k + 1) # Manages next cur val
        dCount = 0
        i = j

    return ans

print(findMinNumberPattern("DIDID"))
print(findMinNumberPattern("DIDIII"))
print(findMinNumberPattern("DDDIIDI"))
print(findMinNumberPattern("IDIDIID"))
print(findMinNumberPattern("DIIDIDD"))
print(findMinNumberPattern("IIDIDDD"))

# This code is contributed by mukesh07.

C#

// This code illustrates to find minimum number following
// pattern with optimized space and modular code.
using System;
class GFG {

    // This function returns minimum number following
    // pattern of increasing or decreasing sequence.
    public static string findMinNumberPattern(string str)
    {
        string ans = ""; // Minimum number following pattern

        int i = 0;
        int cur = 1; // cur val following pattern
        int dCount = 0; // Count of char 'D'
        while (i < str.Length) {

            char ch = str[i];

            // If 1st ch == 'I', incr and add to ans
            if (i == 0 && ch == 'I') {
                ans += cur;
                cur++;
            }

            // If cur char == 'D',
            // incr dCount as well, since we always
            // start counting for dCount from i+1
            if (ch == 'D') {
                dCount++;
            }

            int j = i + 1; // Count 'D' from i+1 index
            while (j < str.Length
                   && str[j] == 'D') {
                dCount++;
                j++;
            }

            int k = dCount;  // Store dCount
            while (dCount >= 0) {
                ans += (cur + dCount);
                dCount--;
            }

            cur += (k + 1); // Manages next cur val
            dCount = 0;
            i = j;
        }

        return ans;
    }

  static void Main() {
    Console.WriteLine(findMinNumberPattern("DIDID"));
    Console.WriteLine(findMinNumberPattern("DIDIII"));
    Console.WriteLine(findMinNumberPattern("DDDIIDI"));
    Console.WriteLine(findMinNumberPattern("IDIDIID"));
    Console.WriteLine(findMinNumberPattern("DIIDIDD"));
    Console.WriteLine(findMinNumberPattern("IIDIDDD"));
  }
}

// This code is contributed by mukesh07.

Javascript

<script>
// This code illustrates to find minimum number following
// pattern with optimized space and modular code.

// This function returns minimum number following
    // pattern of increasing or decreasing sequence.
function findMinNumberPattern(str)
{
    let ans = ""; // Minimum number following pattern

        let i = 0;
        let cur = 1; // cur val following pattern
        let dCount = 0; // Count of char 'D'
        while (i < str.length) {

            let ch = str[i];

            // If 1st ch == 'I', incr and add to ans
            if (i == 0 && ch == 'I') {
                ans += cur;
                cur++;
            }

            // If cur char == 'D',
            // incr dCount as well, since we always
            // start counting for dCount from i+1
            if (ch == 'D') {
                dCount++;
            }

            let j = i + 1; // Count 'D' from i+1 index
            while (j < str.length
                   && str[j] == 'D') {
                dCount++;
                j++;
            }

            let k = dCount;  // Store dCount
            while (dCount >= 0) {
                ans += (cur + dCount);
                dCount--;
            }

            cur += (k + 1); // Manages next cur val
            dCount = 0;
            i = j;
        }

        return ans;
}

document.write(findMinNumberPattern("DIDID")+"<br>");
document.write(findMinNumberPattern("DIDIII")+"<br>");
document.write(findMinNumberPattern("DDDIIDI")+"<br>");
document.write(findMinNumberPattern("IDIDIID")+"<br>");
document.write(findMinNumberPattern("DIIDIDD")+"<br>");
document.write(findMinNumberPattern("IIDIDDD")+"<br>");


// This code is contributed by unknown2108
</script>

Output

214365
2143567
43215768
13254687
21354876
12438765

Time Complexity : O(n)

Auxiliary Space : O(1)

Method 7: (Substring Reversals)

The idea is to take the smallest number with len(s)+1 and perform reversals for every substring containing only ‘D’.

Follow below steps to solve the problem:

1. Create the smallest possible number of length len(s)+1.

2. Traverse the string (say i).

3. Find the first and last occurrence of ‘D’ for every substring containing only ‘D’.

4. Reverse every such substring and reinitialize first and last occurrence.

C++14

#include <bits/stdc++.h>
using namespace std;

string get_num_seq(string& str_seq)
{
    int n=str_seq.length(),start=-1,end=-1;
    string ans;

    for(int i=1;i<=n+1;i++)
    ans.push_back(i+48);

    for(int i=0;i<n;i++)
    {
        if(str_seq[i]=='D')
        {
            if(start==-1)
            start=i;
            end=i;
        }
        else {
            if(start!=-1)
            reverse(ans.begin()+start,ans.begin()+end+2);
            start=-1;
            end=-1;
        }
    }

    if(start!=-1)
        reverse(ans.begin()+start,ans.begin()+end+2);

    return ans;
}

// driver's code
int main()
{
    string str_seq="DDIDDIID";
    cout<<get_num_seq(str_seq);

    return 0;
}
// this code is contributed by prophet1999

Java

// Java code for the above approach
import java.io.*;

class GFG {

  public static String getNumSeq(String strSeq)
  {
    int n = strSeq.length();
    int start = -1;
    int end = -1;
    String ans = "";

    for (int i = 1; i <= n + 1; i++) {
      ans += String.valueOf(i);
    }

    for (int i = 0; i < n; i++) {
      if (strSeq.charAt(i) == 'D') {
        if (start == -1) {
          start = i;
        }
        end = i;
      }
      else {
        if (start != -1) {
          ans = reverse(ans, start, end + 2);
        }
        start = -1;
        end = -1;
      }
    }

    if (start != -1) {
      ans = reverse(ans, start, end + 2);
    }

    return ans;
  }

  public static String reverse(String str, int start,
                               int end)
  {
    char[] arr = str.toCharArray();
    for (int i = start, j = end - 1; i < j; i++, j--) {
      char temp = arr[i];
      arr[i] = arr[j];
      arr[j] = temp;
    }
    return new String(arr);
  }

  public static void main(String[] args)
  {
    String strSeq = "DDIDDIID";
    System.out.println(getNumSeq(strSeq));
  }
}

// This code is contributed by lokesh.

Python3

def get_num_seq( str_seq):

    n= len(str_seq)
    start = -1
    end = -1;
    ans = "";

    for i in range(1, n + 2):
        ans += str(i)

    for i in range(n):

        if(str_seq[i] == 'D'):

            if(start == -1):
                start=i;
            end=i;

        else:
            if(start != -1):
                ans = ans[:start] + ans[start:end+2][::-1] + ans[end+2:]
                start = -1;
            end = -1;

    if(start != -1):
        ans =  ans[:start] + ans[start:end+2][::-1] + ans[end+2:]

    return ans;

# driver's code
str_seq="DDIDDIID";
print(get_num_seq(str_seq))

# this code is contributed by phasing17

C#

// C# code for the above approach
using System;

public class GFG {

  public static string GetNumSeq(string strSeq)
  {
    int n = strSeq.Length;
    int start = -1;
    int end = -1;
    string ans = "";

    for (int i = 1; i <= n + 1; i++) {
      ans += i.ToString();
    }

    for (int i = 0; i < n; i++) {
      if (strSeq[i] == 'D') {
        if (start == -1) {
          start = i;
        }
        end = i;
      }
      else {
        if (start != -1) {
          ans = Reverse(ans, start, end + 2);
        }
        start = -1;
        end = -1;
      }
    }

    if (start != -1) {
      ans = Reverse(ans, start, end + 2);
    }

    return ans;
  }

  public static string Reverse(string str, int start,
                               int end)
  {
    char[] arr = str.ToCharArray();
    for (int i = start, j = end - 1; i < j; i++, j--) {
      char temp = arr[i];
      arr[i] = arr[j];
      arr[j] = temp;
    }
    return new string(arr);
  }

  static public void Main()
  {

    // Code
    string strSeq = "DDIDDIID";
    Console.WriteLine(GetNumSeq(strSeq));
  }
}

// This code is contributed by lokeshmvs21.

Javascript

function get_num_seq(strSeq) {
  let n = strSeq.length;
  let start = -1;
  let end = -1;
  let ans = "";

  for (let i = 1; i <= n + 1; i++) {
      ans += String.fromCharCode(i+48);
  }

  for (let i = 0; i < n; i++) {
    if (strSeq[i] === 'D') {
      if (start === -1) {
        start = i;
      }
      end = i;
    } else {
      if (start !== -1) {
        ans = ans.slice(0, start) + ans.slice(start, end + 2).split("").reverse().join("") + ans.slice(end + 2);
      }
      start = -1;
      end = -1;
    }
  }

  if (start !== -1) {
    ans = ans.slice(0, start) + ans.slice(start, end + 2).split("").reverse().join("") + ans.slice(end + 2);
  }

  return ans;
}
//driver's code
  let str_seq="DDIDDIID";
document.write(get_num_seq(str_seq));

Output

321654798

Time Complexity: O(n)
Auxiliary Space: O(1)