Find Maximum number possible by doing at-most K swaps

Given a positive integer, find maximum integer possible by doing at-most K swap operations on its digits.

Examples:

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

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

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

Input: M = 7599, K = 2
Output: 9975
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
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:

Approach: 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.



Algorithm:

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

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
}

chevron_right






                        filter_none









Python3














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















# 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 



















                        chevron_right







                        filter_none











Output:


999984211





Complexity Analysis:


    

  • Time Complexity: O((n^2)^k).
    
For every recursive call n^2 recursive calls is generated until the value of k is 0. So total recursive calls are O((n^2)^k).
    • 

  • Space Complexity:O(n).
    
This is the space required to store the output string.
    • 



Efficient Solution:


Approach: The above approach traverses the whole string at each recursive call which is highly inefficient and unnecessary. Also, pre-computing the maximum digit after the current at a recursive call avoids unnecessary exchanges with each digit. 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 element is the maximum digit which is not yet swapped to the front.

There is an improvement by 27580 microseconds for each test case.


Algorithm:


    

  1. Create a global variable which will store the maximum string or number.
    2. 

  2. Define a recursive function that takes the string as number, value of k, and current index.
    3. 

  3. Find the index of maximum element in range current index to end.
    4. 

  4. if the index of maximum element is not equal to current index then decrement value of k.
    5. 

  5. Run a loop from current index to the end of array
    6. 

  6. If the ith digit is equal to the maximum element
    7. 

  7. Swap the ith and element at current index and check if the string is now maximum and update the maximum string.
    8. 

  8. Call the function recursively with parameters: string and k.
    9. 

  9. Now again swap back the ith and element at current index.
    10. 





C++














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















// 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 k, 


    // decrement k 


    if (maxm != str[ctr]) 


        --k; 


  


    // search this maximum among the rest 


    for (int j = ctr; j < n; 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; 


} 



















                        chevron_right







                        filter_none











Output:


999984211





Complexity Analysis:


    

  • Time Complexity: O(n^k).
    
For every recursive call n recursive calls is generated until the value of k is 0. So total recursive calls are O((n)^k).
    • 

  • Space Complexity: O(n).
    
The space required to store the output string.
    • 



Exercise:


    

  1. Find minimum integer possible by doing at-least K swap operations on its digits.
    2. 

  2. Find maximum/minimum integer possible by doing exactly K swap operations on its digits.
    3. 



This article is contributed by Aditya Goel.If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.


Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.


Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.





    

        My Personal Notes
        arrow_drop_up
    

    

        
        

            
            
        

    


Recommended Posts: