Lexicographically largest string possible by at most K replacements

Given a string S of length N, consisting of lowercase alphabets, the task is to find the lexicographically longest string that can be obtained by replacing at most K characters from the given string.

Examples:

Input: S = "dbza", K = 1 
Output: zbza 
Explanation: Replace S[0] (= 'd') with 'z' to obtain the lexicographically largest string.

Input: S = "zzzz", K = 2 
Output: zzzz

Approach: The given problem can be solved by using the Greedy Approach. The idea is to traverse the array from left to right and replace all the non z characters with z. Follow the steps below to solve the problem:

• Run a loop for i = 0 to the length of the string:
  • If the current character is not equal to z, and K is not equal to 0:
    • Then, replace the current character with z.
    • K = K - 1.
• Finally, return the modified string.

Below is the implementation of the above approach:

// C++ implementation of
// the above approach

#include <bits/stdc++.h>
using namespace std;

string largestString(string s, int k)
{
    // Traverse each element of the string
    for (int i = 0; i < s.size(); i++) {

        // If the current character
        // can be replaced with 'z'
        if (s[i] != 'z' && k > 0) {

            s[i] = 'z';
            k--;
        }
    }

    // Return the modified string
    return s;
}

// Driver code
int main()
{
    string s = "dbza";
    int k = 1;

    cout << largestString(s, k) << endl;

    return 0;
}

// Java implementation of the above approach
class GFG{
    
static String largestString(String s, int k)
{
    
    // Traverse each element of the string
    for(int i = 0; i < s.length(); i++)
    {
        
        // If the current character
        // can be replaced with 'z'
        if (s.charAt(i) != 'z' && k > 0)
        {
            s = s.replace(s.charAt(i),'z');
            k--;
        }
    }
    
    // Return the modified string
    return s;
}

// Driver code
public static void main(String args[]) 
{
    String s = "dbza";
    int k = 1;
    
    System.out.println(largestString(s, k));
}
}

// This code is contributed by SoumikMondal

# Python3 implementation of
# the above approach
def largestString(s, k):
    
    # Traverse each element of the string
    for i in range(len(s)):
        
        # If the current character
        # can be replaced with 'z'
        if (s[i] != 'z' and k > 0):
            s[i] = 'z'
            k -= 1

    # Return the modified string
    return "".join(s)

# Driver code
if __name__ == '__main__':
    
    s = "dbza"
    k = 1

    print (largestString([i for i in s], k))

# This code is contributed by mohit kumar 29

// C# implementation of
// the above approach
using System;
using System.Collections.Generic;

class GFG{
    
static string largestString(string s, int k)
{
    
    // Traverse each element of the string
    for(int i = 0; i < s.Length; i++)
    {
        
        // If the current character
        // can be replaced with 'z'
        if (s[i] != 'z' && k > 0)
        {
            s = s.Replace(s[i],'z');
            k--;
        }
    }

    // Return the modified string
    return s;
}

// Driver code
public static void Main()
{
    string s = "dbza";
    int k = 1;

    Console.Write(largestString(s, k));
}
}

// This code is contributed by SURENDRA_GANGWAR

<script>
        // JavaScript implementation of
        // the above approach

        function largestString(s, k)
        {
        
            // Traverse each element of the string
            for (let i = 0; i < s.length; i++) {

                // If the current character
                // can be replaced with 'z'
                if (s[i] != 'z' && k > 0) {
                    s = s.substring(0, i) + 'z' + s.substring(i + 1);
                    k--;
                }
            }

            // Return the modified string
            return s;
        }

        // Driver code

        var s = "dbza";
        var k = 1;

        document.write(largestString(s, k));
        
  // This code is contributed by Potta Lokesh
  </script>

zbza

Time Complexity: O(N) 
Auxiliary Space: O(1)