# Find Maximum number possible by doing at-most K swaps

Last Updated : 17 Apr, 2024

Given two positive integers M and K, find the maximum integer possible by doing at-most K swap operations on its digits.

Examples:

Input: M = 254, K = 1
Output: 524
Explanation: Swap 5 with 2 so number becomes 524

Input: M = 254, K = 2
Output: 542
Explanation: Swap 5 with 2 so number becomes 524, Swap 4 with 2 so number becomes 542

Input: M = 68543, K = 1
Output: 86543
Explanation: Swap 8 with 6 so number becomes 86543

Input: M = 7599, K = 2
Output: 9975
Explanation: Swap 9 with 5 so number becomes 7995, Swap 9 with 7 so number becomes 9975

Input: M = 76543, K = 1
Output: 76543
Explanation: No swap is required.

Input: M = 129814999, K = 4
Output: 999984211
Explanation: Swap 9 with 1 so number becomes 929814991, Swap 9 with 2 so number becomes 999814291, Swap 9 with 8 so number becomes 999914281, Swap 1 with 8 so number becomes 999984211

## Naive solution for the Largest number in K swaps:

The idea is to consider every digit and swap it with digits following it one at a time and see if it leads to the maximum number. The process is repeated K times. The code can be further optimized, if the current digit is swapped with a digit less than the following digit.

Follow the below steps to Implement the idea:

• Create a global variable that will store the maximum string or number.
• Define a recursive function that takes the string as a number and value of k
• Run a nested loop, the outer loop from 0 to the length of string -1, and the inner loop from i+1 to the end of the string.
• Swap the ith and jth characters and check if the string is now maximum and update the maximum string.
• Call the function recursively with parameters: string and k-1.
• Now again swap back the ith and jth character.

Below is the Implementation of the above approach:

C++ ```// C++ program to find maximum // integer possible by doing // at-most K swap operations // on its digits. #include <bits/stdc++.h> using namespace std; // Function to find maximum // integer possible by // doing at-most K swap // operations on its digits void findMaximumNum( string str, int k, string& max) { // Return if no swaps left if (k == 0) return; int n = str.length(); // Consider every digit for (int i = 0; i < n - 1; i++) { // Compare it with all digits after it for (int j = i + 1; j < n; j++) { // if digit at position i // is less than digit // at position j, swap it // and check for maximum // number so far and recurse // for remaining swaps if (str[i] < str[j]) { // swap str[i] with str[j] swap(str[i], str[j]); // If current num is more // than maximum so far if (str.compare(max) > 0) max = str; // recurse of the other k - 1 swaps findMaximumNum(str, k - 1, max); // Backtrack swap(str[i], str[j]); } } } } // Driver code int main() { string str = "129814999"; int k = 4; string max = str; findMaximumNum(str, k, max); cout << max << endl; return 0; } ``` Java ```// Java program to find maximum // integer possible by doing // at-most K swap operations // on its digits. import java.util.*; class GFG{ static String max; // Function to find maximum // integer possible by // doing at-most K swap // operations on its digits static void findMaximumNum(char[] str, int k) { // Return if no swaps left if (k == 0) return; int n = str.length; // Consider every digit for (int i = 0; i < n - 1; i++) { // Compare it with all digits // after it for (int j = i + 1; j < n; j++) { // if digit at position i // is less than digit // at position j, swap it // and check for maximum // number so far and recurse // for remaining swaps if (str[i] < str[j]) { // swap str[i] with // str[j] char t = str[i]; str[i] = str[j]; str[j] = t; // If current num is more // than maximum so far if (String.valueOf(str).compareTo(max) > 0) max = String.valueOf(str); // recurse of the other // k - 1 swaps findMaximumNum(str, k - 1); // Backtrack char c = str[i]; str[i] = str[j]; str[j] = c; } } } } // Driver code public static void main(String[] args) { String str = "129814999"; int k = 4; max = str; findMaximumNum(str.toCharArray(), k); System.out.print(max + "\n"); } } // This code is contributed by 29AjayKumar ``` Python3 ```# Python3 program to find maximum # integer possible by doing at-most # K swap operations on its digits. # utility function to swap two # characters of a string def swap(string, i, j): return (string[:i] + string[j] + string[i + 1:j] + string[i] + string[j + 1:]) # function to find maximum integer # possible by doing at-most K swap # operations on its digits def findMaximumNum(string, k, maxm): # return if no swaps left if k == 0: return n = len(string) # consider every digit for i in range(n - 1): # and compare it with all digits after it for j in range(i + 1, n): # if digit at position i is less than # digit at position j, swap it and # check for maximum number so far and # recurse for remaining swaps if string[i] < string[j]: # swap string[i] with string[j] string = swap(string, i, j) # If current num is more than # maximum so far if string > maxm[0]: maxm[0] = string # recurse of the other k - 1 swaps findMaximumNum(string, k - 1, maxm) # backtrack string = swap(string, i, j) # Driver Code if __name__ == "__main__": string = "129814999" k = 4 maxm = [string] findMaximumNum(string, k, maxm) print(maxm[0]) # This code is contributed # by vibhu4agarwal ``` C# ```// C# program to find maximum // integer possible by doing // at-most K swap operations // on its digits. using System; class GFG{ static String max; // Function to find maximum // integer possible by // doing at-most K swap // operations on its digits static void findMaximumNum(char[] str, int k) { // Return if no swaps left if (k == 0) return; int n = str.Length; // Consider every digit for (int i = 0; i < n - 1; i++) { // Compare it with all digits // after it for (int j = i + 1; j < n; j++) { // if digit at position i // is less than digit // at position j, swap it // and check for maximum // number so far and recurse // for remaining swaps if (str[i] < str[j]) { // swap str[i] with // str[j] char t = str[i]; str[i] = str[j]; str[j] = t; // If current num is more // than maximum so far if (String.Join("", str).CompareTo(max) > 0) max = String.Join("", str); // recurse of the other // k - 1 swaps findMaximumNum(str, k - 1); // Backtrack char c = str[i]; str[i] = str[j]; str[j] = c; } } } } // Driver code public static void Main(String[] args) { String str = "129814999"; int k = 4; max = str; findMaximumNum(str.ToCharArray(), k); Console.Write(max + "\n"); } } // This code is contributed by gauravrajput1 ``` Javascript ```<script> // Javascript program to find maximum // integer possible by doing // at-most K swap operations // on its digits. let max; // Function to find maximum // integer possible by // doing at-most K swap // operations on its digits function findMaximumNum(str,k) { // Return if no swaps left if (k == 0) return; let n = str.length; // Consider every digit for (let i = 0; i < n - 1; i++) { // Compare it with all digits // after it for (let j = i + 1; j < n; j++) { // if digit at position i // is less than digit // at position j, swap it // and check for maximum // number so far and recurse // for remaining swaps if (str[i] < str[j]) { // swap str[i] with // str[j] let t = str[i]; str[i] = str[j]; str[j] = t; // If current num is more // than maximum so far if ((str).join("")>(max) ) max = (str).join(""); // recurse of the other // k - 1 swaps findMaximumNum(str, k - 1); // Backtrack let c = str[i]; str[i] = str[j]; str[j] = c; } } } } // Driver code let str = "129814999"; let k = 4; max = str; findMaximumNum(str.split(""), k); document.write(max + "<br>"); // This code is contributed by unknown2108 </script> ```

Output
`999984211`

Time Complexity: O((N2)k). For every digit, N2 recursive calls are generated until the value of k is 0 Thus O((N2)k).
Auxiliary Space: O(N). This is the space required to store the output string.

## Find the Maximum number possible by doing at-most K swaps by swapping with the maximum element on the right:

It can be observed that to make the maximum string, the maximum digit is shifted to the front. So, instead of trying all pairs, try only those pairs where one of the elements is the maximum digit that is not yet swapped to the front.

Follow the below steps to Implement the idea::

1. Create a global variable that will store the maximum string or number.
2. Define a recursive function that takes the string as a number, the value of k, and the current index.
3. Find the index of the maximum element in the range current index to end.
4. if the index of the maximum element is not equal to the current index then decrement the value of k.
5. Run a loop from the current index to the end of the array
6. If the ith digit is equal to the maximum element
7. Swap the ith and element at the current index and check if the string is now maximum and update the maximum string.
8. Call the function recursively with parameters: string and k.
9. Now again swap back the ith and element at the current index.

Below is the Implementation of the above approach:

C++ ```// C++ program to find maximum // integer possible by doing // at-most K swap operations on // its digits. #include <bits/stdc++.h> using namespace std; // Function to find maximum // integer possible by // doing at-most K swap operations // on its digits void findMaximumNum( string str, int k, string& max, int ctr) { // return if no swaps left if (k == 0) return; int n = str.length(); // Consider every digit after // the cur position char maxm = str[ctr]; for (int j = ctr + 1; j < n; j++) { // Find maximum digit greater // than at ctr among rest if (maxm < str[j]) maxm = str[j]; } // If maxm is not equal to str[ctr], // decrement k if (maxm != str[ctr]) --k; // search this maximum among the rest from behind //first swap the last maximum digit if it occurs more than 1 time //example str= 1293498 and k=1 then max string is 9293418 instead of 9213498 for (int j = n-1; j >=ctr; j--) { // If digit equals maxm swap // the digit with current // digit and recurse for the rest if (str[j] == maxm) { // swap str[ctr] with str[j] swap(str[ctr], str[j]); // If current num is more than // maximum so far if (str.compare(max) > 0) max = str; // recurse other swaps after cur findMaximumNum(str, k, max, ctr + 1); // Backtrack swap(str[ctr], str[j]); } } } // Driver code int main() { string str = "129814999"; int k = 4; string max = str; findMaximumNum(str, k, max, 0); cout << max << endl; return 0; } ``` Java ```// Java program to find maximum // integer possible by doing // at-most K swap operations on // its digits. import java.io.*; class Res { static String max = ""; } class Solution { // Function to set highest possible digits at given // index. public static void findMaximumNum(char ar[], int k, Res r) { if (k == 0) return; int n = ar.length; for (int i = 0; i < n - 1; i++) { for (int j = i + 1; j < n; j++) { // if digit at position i is less than digit // at position j, we swap them and check for // maximum number so far. if (ar[j] > ar[i]) { char temp = ar[i]; ar[i] = ar[j]; ar[j] = temp; String st = new String(ar); // if current number is more than // maximum so far if (r.max.compareTo(st) < 0) { r.max = st; } // calling recursive function to set the // next digit. findMaximumNum(ar, k - 1, r); // backtracking temp = ar[i]; ar[i] = ar[j]; ar[j] = temp; } } } } // Function to find the largest number after k swaps. public static void main(String[] args) { String str = "129814999"; int k = 4; Res r = new Res(); r.max = str; findMaximumNum(str.toCharArray(), k, r); //Print the answer stored in res class System.out.println(r.max); } } ``` Python3 ```# Python3 program to find maximum # integer possible by doing at-most # K swap operations on its digits. # function to find maximum integer # possible by doing at-most K swap # operations on its digits def findMaximumNum(string, k, maxm, ctr): # return if no swaps left if k == 0: return n = len(string) # Consider every digit after # the cur position mx = string[ctr] for i in range(ctr+1,n): # Find maximum digit greater # than at ctr among rest if int(string[i]) > int(mx): mx=string[i] # If maxm is not equal to str[ctr], # decrement k if(mx!=string[ctr]): k=k-1 # search this maximum among the rest from behind # first swap the last maximum digit if it occurs more than 1 time # example str= 1293498 and k=1 then max string is 9293418 instead of 9213498 for i in range(ctr,n): # If digit equals maxm swap # the digit with current # digit and recurse for the rest if(string[i]==mx): # swap str[ctr] with str[j] string[ctr], string[i] = string[i], string[ctr] new_str = "".join(string) # If current num is more than # maximum so far if int(new_str) > int(maxm[0]): maxm[0] = new_str # recurse of the other k - 1 swaps findMaximumNum(string, k , maxm, ctr+1) # backtrack string[ctr], string[i] = string[i], string[ctr] # Driver Code if __name__ == "__main__": string = "129814999" k = 4 maxm = [string] string = [char for char in string] findMaximumNum(string, k, maxm, 0) print(maxm[0]) # This code is contributed Aarti_Rathi ``` C# ```// C# program to find maximum // integer possible by doing // at-most K swap operations on // its digits. using System; class Res { public String max = ""; } public class Solution { // Function to set highest possible digits at given // index. static void findMaximumNum(char []ar, int k, Res r) { if (k == 0) return; int n = ar.Length; for (int i = 0; i < n - 1; i++) { for (int j = i + 1; j < n; j++) { // if digit at position i is less than digit // at position j, we swap them and check for // maximum number so far. if (ar[j] > ar[i]) { char temp = ar[i]; ar[i] = ar[j]; ar[j] = temp; String st = new String(ar); // if current number is more than // maximum so far if (r.max.CompareTo(st) < 0) { r.max = st; } // calling recursive function to set the // next digit. findMaximumNum(ar, k - 1, r); // backtracking temp = ar[i]; ar[i] = ar[j]; ar[j] = temp; } } } } // Function to find the largest number after k swaps. public static void Main(String[] args) { String str = "129814999"; int k = 4; Res r = new Res(); r.max = str; findMaximumNum(str.ToCharArray(), k, r); // Print the answer stored in res class Console.WriteLine(r.max); } } // This code is contributed by shikhasingrajput ``` Javascript ```function findMaximumNum(string, k, maxm, ctr) { // return if no swaps left if (k == 0) { return; } const n = string.length; // Consider every digit after // the cur position let mx = string[ctr]; for (let i = ctr + 1; i < n; i++) { // Find maximum digit greater // than at ctr among rest if (parseInt(string[i]) > parseInt(mx)) { mx = string[i]; } } // If maxm is not equal to str[ctr], // decrement k if (mx != string[ctr]) { k = k - 1; } // search this maximum among the rest from behind // first swap the last maximum digit if it occurs more than 1 time // example str= 1293498 and k=1 then max string is 9293418 instead of 9213498 for (let i = ctr; i < n; i++) { // If digit equals maxm swap // the digit with current // digit and recurse for the rest if (string[i] == mx) { // swap str[ctr] with str[j] [string[ctr], string[i]] = [string[i], string[ctr]]; const new_str = string.join(""); // If current num is more than // maximum so far if (parseInt(new_str) > parseInt(maxm[0])) { maxm[0] = new_str; } // recurse of the other k - 1 swaps findMaximumNum(string, k, maxm, ctr + 1); // backtrack [string[ctr], string[i]] = [string[i], string[ctr]]; } } } // Driver Code const string = "129814999"; const k = 4; const maxm = [string]; const strArr = string.split(""); findMaximumNum(strArr, k, maxm, 0); console.log(maxm[0]); ```

Output
`999984211`

Time Complexity: O(Nk), For every recursive call N recursive calls are generated until the value of k is 0, Thus O((Nk).
Auxiliary Space: O(N). The space required to store the output string.

Exercise:

Previous Article
Next Article
Article Tags :
Practice Tags :