# Find largest number smaller than N with same set of digits

• Difficulty Level : Medium
• Last Updated : 13 Jun, 2022

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 number, 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:

1. If all digits sorted in ascending order, then output is always “Not Possible”. For example, 1234.
2. If all digits are sorted in descending order, then we need to swap last two digits. For example, 4321.
3. 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 ``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

 ``

Output:

`Greatest smaller number with same set of digits is 256432`

Time Complexity: O(nlogn), where n represents the length of the given string.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

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