# Form minimum number from given sequence

• Difficulty Level : Hard
• Last Updated : 26 Jul, 2022

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 canâ€™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. Below is the implementation of above idea.

## C++

 // C++ program to print minimum number that can be formed// from a given sequence of Is and Ds#include using namespace std; // Prints the minimum number that can be formed from// input sequence of I's and D'svoid 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

## Java

 // Java program to print minimum number that can be formed// from a given sequence of Is and Dsclass 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'sdef 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 codeif __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 Dsusing 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



## Javascript



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

This solution is suggested by Swapnil Trambake.

Alternate Solution:
Let’s observe a few facts in case of 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#includeusing 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    vectorv;     // 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

## Java

 // Java program to print minimum number that can be formed// from a given sequence of Is and Dsimport 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 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 Dsdef 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 codeprintLeast("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 Dsusing 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 al = new List();             // 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 codepublic 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



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

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 using namespace std; // Function to decode the given sequence to construct// minimum number without repeated digitsvoid PrintMinNumberForPattern(string seq){    // result store output string    string result;     // create an empty stack of integers    stack 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 functionint 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 Dsclass 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.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 Dsdef 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 CodePrintMinNumberForPattern("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 Dsusing 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



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 using namespace std;   // Returns minimum number made from given sequence without repeating digitsstring 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 functionint 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 approachimport 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 digitsdef 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 Codeif __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 approachusing System;class GFG{     // Returns minimum number made from given// sequence without repeating digitsstatic 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 Codepublic 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



Output

13254
12
321
123
21435
126543
321654798

Time Complexity: O(N)
Auxiliary Space: O(N)
This solution is suggested by Brij Desai.

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.

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 #include #include #include 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 repr should the min//such numbervector didi_seq_gen(string s){    if (s.size() == 0)        return {};    vector 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 inputs        = { "IDID", "I",     "DD",      "II",            "DIDI", "IIDDD", "DDIDDIID" };    for (auto x : inputs) {        vector ans = didi_seq_gen(x);        for (auto i : ans) {            cout << i;        }        cout << endl;    }    return 0;}

## Java

 // Java program to generate required sequenceimport 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 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 repr    // should the min such number    public static List didi_seq_gen(String s)    {        if (s.length() == 0)            return new ArrayList<>();        List 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 repr 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 callprint(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" ))

Output

13254
12
321
123
21435
126543
321654798

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

#### 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 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



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 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

Output

321654798

Time Complexity: O(n)

Auxiliary Space: O(1)

My Personal Notes arrow_drop_up