Find largest number smaller than N with same set of digits

Difficulty Level : Medium
Last Updated : 28 Jul, 2021

Given a number N in the form of string. The task is to find the greatest number that has same set of digits as N and is smaller than N. If it is not possible to find any such numebr, then print “not possible”.

Examples:

Input:  N = "218765"
Output: 218756

Input:  N = "1234"
Output: Not Possible

Input: N = "262345"
Output: 256432

Following are few observations about the greatest number smaller than N:

If all digits sorted in ascending order, then output is always “Not Possible”. For example, 1234.
If all digits are sorted in descending order, then we need to swap last two digits. For example, 4321.
For other cases, we need to process the number from rightmost side (why? because we need to find the greatest of all smaller numbers).

Algorithm:

Traverse the given number from rightmost digit, keep traversing till you find a digit which is greater than the previously traversed digit. For example, if the input number is “262345”, we stop at 6 because 6 is greater than previous digit 5. If we do not find such a digit, then output is “Not Possible”.
Now search the right side of above found digit ‘d’ for the greatest digit smaller than ‘d’. For “262345?, the right side of 6 contains “2345”. The greatest digit smaller than 6 is 5.
Swap the above found two digits, we get “252346” in above example.
Now sort all digits from position next to ‘d’ to the end of number in descending order. The number that we get after sorting is the output. For above example, we sort digits in bold 252346. We get 256432 which is the previous smaller number for input 262345.

Below is the implementation of the above approach:

C++

// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find previous number
void findPrevious(string number, int n)
{
    int i, j;
 
    // I) Start from the right most digit
    // and find the first digit that is
    // smaller than the digit next to it.
    for (i = n - 1; i > 0; i--)
        if (number[i] < number[i - 1])
            break;
 
    // If no such digit is found
    // then all digits are in ascending order
    // means there cannot be a smallest number
    // with same set of digits
    if (i == 0) {
        cout << "Previous number is not possible";
        return;
    }
 
    // II) Find the greatest digit on
    // right side of (i-1)'th digit that is
    // smaller than number[i-1]
    int x = number[i - 1], greatest = i;
    for (j = i; j < n; j++)
        if (number[j] < x && number[j] > number[greatest])
            greatest = j;
 
    // III) Swap the above found smallest digit with number[i-1]
    swap(number[greatest], number[i - 1]);
 
    // IV) Sort the digits after (i-1) in descending order
    sort(number.begin() + i, number.begin() + n, greater<char>());
 
    cout << "Greatest smaller number with same set of digits is " << number;
 
    return;
}
 
// Driver code
int main()
{
    string digits = "262345";
    int n = digits.length();
 
    findPrevious(digits, n);
 
    return 0;
}

Java

// Java implementation of the above approach
import java.util.*;
 
class GFG
{
 
    // Function to find previous number
    static void findPrevious(char[] number, int n)
    {
        int i, j;
 
        // I) Start from the right most digit
        // and find the first digit that is
        // smaller than the digit next to it.
        for (i = n - 1; i > 0; i--)
        {
            if (number[i] < number[i - 1])
            {
                break;
            }
        }
 
        // If no such digit is found
        // then all digits are in ascending order
        // means there cannot be a smallest number
        // with same set of digits
        if (i == 0)
        {
            System.out.print("Previous number is not possible");
            return;
        }
 
        // II) Find the greatest digit on
        // right side of (i-1)'th digit that is
        // smaller than number[i-1]
        int x = number[i - 1], greatest = i;
        for (j = i; j < n; j++)
        {
            if (number[j] < x && number[j] > number[greatest])
            {
                greatest = j;
            }
        }
 
        // III) Swap the above found smallest digit with number[i-1]
        swap(number, greatest, i - 1);
 
        // IV) Sort the digits after (i-1) in descending order
        Arrays.sort(number, i, n);
        reverse(number, i, n - 1);
        System.out.print("Greatest smaller number with" +
              "same set of digits is " + String.valueOf(number));
 
        return;
    }
 
    static String swap(char[] ch, int i, int j)
    {
        char temp = ch[i];
        ch[i] = ch[j];
        ch[j] = temp;
        return String.valueOf(ch);
    }
 
    static void reverse(char str[], int start, int end)
    {
 
        // Temporary variable to store character
        char temp;
        while (start <= end)
        {
            // Swapping the first and last character
            temp = str[start];
            str[start] = str[end];
            str[end] = temp;
            start++;
            end--;
        }
    }
     
    // Driver code
    public static void main(String[] args)
    {
        String digits = "262345";
        int n = digits.length();
 
        findPrevious(digits.toCharArray(), n);
    }
}
 
// This code has been contributed by 29AjayKumar

Python3

# Python3 implementation of the above approach
 
# Function to find previous number
def findPrevious(number, n):
 
    # This is necessary as strings
    # do not support item assignment
    number = list(number)
 
    i, j = -1, -1
 
    # I) Start from the right most digit
    # and find the first digit that is
    # smaller than the digit next to it.
    for i in range(n - 1, 0, -1):
        if number[i] < number[i - 1]:
            break
     
    # If no such digit is found
    # then all digits are in ascending order
    # means there cannot be a smallest number
    # with same set of digits
    if i == 0:
        print("Previous number is not possible")
        return
     
    x, greatest = number[i - 1], i
 
    # II) Find the greatest digit on
    # right side of(i-1)'th digit that is
    # smaller than number[i-1]
    for j in range(i, n):
        if (number[j] < x and
            number[j] > number[greatest]):
            greatest = j
     
    # III) Swap the above found smallest digit
    # with number[i-1]
    (number[greatest],
     number[i - 1]) = (number[i - 1],
                       number[greatest])
 
    l = number[i:]
    del number[i:]
 
    # IV) Sort the digits after(i-1)
    # in descending order
    l.sort(reverse = True)
 
    number += l
 
    # Again join the list to make it string
    number = '' . join(number)
    print("Greatest smaller number with",
          "same set of digits is", number)
 
    return
 
# Driver Code
if __name__ == "__main__":
    digits = "262345"
    n = len(digits)
 
    findPrevious(digits, n)
 
# This code is contributed by sanjeev2552

C#

// C# implementation of the approach
using System;
using System.Collections.Generic;
 
class GFG
{
 
    // Function to find previous number
    static void findPrevious(char[] number, int n)
    {
        int i, j;
 
        // I) Start from the right most digit
        // and find the first digit that is
        // smaller than the digit next to it.
        for (i = n - 1; i > 0; i--)
        {
            if (number[i] < number[i - 1])
            {
                break;
            }
        }
 
        // If no such digit is found
        // then all digits are in ascending order
        // means there cannot be a smallest number
        // with same set of digits
        if (i == 0)
        {
            Console.Write("Previous number is not possible");
            return;
        }
 
        // II) Find the greatest digit on
        // right side of (i-1)'th digit that is
        // smaller than number[i-1]
        int x = number[i - 1], greatest = i;
        for (j = i; j < n; j++)
        {
            if (number[j] < x && number[j] > number[greatest])
            {
                greatest = j;
            }
        }
 
        // III) Swap the above found
        // smallest digit with number[i-1]
        swap(number, greatest, i - 1);
 
        // IV) Sort the digits after (i-1) in descending order
        Array.Sort(number, i, n-i);
        reverse(number, i, n - 1);
        Console.Write("Greatest smaller number with" +
            "same set of digits is " + String.Join("",number));
 
        return;
    }
 
    static String swap(char[] ch, int i, int j)
    {
        char temp = ch[i];
        ch[i] = ch[j];
        ch[j] = temp;
        return String.Join("",ch);
    }
 
    static void reverse(char []str, int start, int end)
    {
 
        // Temporary variable to store character
        char temp;
        while (start <= end)
        {
            // Swapping the first and last character
            temp = str[start];
            str[start] = str[end];
            str[end] = temp;
            start++;
            end--;
        }
    }
     
    // Driver code
    public static void Main(String[] args)
    {
        String digits = "262345";
        int n = digits.Length;
 
        findPrevious(digits.ToCharArray(), n);
    }
}
 
/* This code contributed by PrinciRaj1992 */

Javascript

<script>
 
// Javascript implementation of the above approach
 
// Function to find previous number
function findPrevious(number, n)
{
    let i, j;
     
    // I) Start from the right most digit
    // and find the first digit that is
    // smaller than the digit next to it.
    for(i = n - 1; i > 0; i--)
    {
        if (number[i] < number[i - 1])
        {
            break;
        }
    }
 
    // If no such digit is found then all
    // digits are in ascending order means
    // there cannot be a smallest number
    // with same set of digits
    if (i == 0)
    {
        document.write("Previous number " +
                       "is not possible");
        return;
    }
 
    // II) Find the greatest digit on right
    // side of (i-1)'th digit that is smaller
    // than number[i-1]
    let x = number[i - 1], greatest = i;
    for(j = i; j < n; j++)
    {
        if (number[j] < x && number[j] >
            number[greatest])
        {
            greatest = j;
        }
    }
 
    // III) Swap the above found smallest
    // digit with number[i-1]
    swap(number, greatest, i - 1);
 
    // IV) Sort the digits after (i-1)
    // in descending order
    number = number.slice(0, i).concat(
             number.slice(i, n).sort(
                 function(a, b){return b - a;}));
     
    document.write("Greatest smaller number with " +
                   "same set of digits is " +
                   (number).join(""));
 
    return;
}
 
function swap(ch, i, j)
{
    let temp = ch[i];
    ch[i] = ch[j];
    ch[j] = temp;
    return String.valueOf(ch);
}
 
function reverse(str, start, end)
{
     
    // Temporary variable to store character
    let temp;
     
    while (start <= end)
    {
         
        // Swapping the first and last character
        temp = str[start];
        str[start] = str[end];
        str[end] = temp;
        start++;
        end--;
    }
}
 
// Driver code
let digits = "262345";
let n = digits.length;
 
findPrevious(digits.split(""), n);
 
// This code is contributed by rag2127
 
</script>

Output:

Greatest smaller number with same set of digits is 256432

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