Find the largest number that can be formed by changing at most K digits

0 0 3 minutes read

Given string str representing a number and an integer K, the task is to find the largest number that can be formed by changing at most K digits in the given number.

Examples:

Input: str = “569431”, K = 3
Output: 999931
Replace first, second and fourth digits with 9.
Input: str = “5687”, K = 2
Output: 9987

Approach: In order to get the maximum number possible, leftmost digits must be replaced with 9s. For every digit of the number starting from the leftmost digit, if it is not already 9 and K is greater than 0 then replace it with 9 and decrement K by 1. Repeat these steps for every digit while K is greater than 0. Finally, print the updated number.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <iostream>
using namespace std;
 
// Function to return the maximum number
// that can be formed by changing
// at most k digits in str
string findMaximumNum(string str, int n, int k)
{
 
    // For every digit of the number
    for (int i = 0; i < n; i++) {
 
        // If no more digits can be replaced
        if (k < 1)
            break;
 
        // If current digit is not already 9
        if (str[i] != '9') {
 
            // Replace it with 9
            str[i] = '9';
 
            // One digit has been used
            k--;
        }
    }
 
    return str;
}
 
// Driver code
int main()
{
    string str = "569431";
    int n = str.length();
    int k = 3;
 
    cout << findMaximumNum(str, n, k);
 
    return 0;
}

Java

// Java implementation of the approach
import java.io.*;
class GFG
{
    // Function to return the maximum number
    // that can be formed by changing
    // at most k digits in str
    static StringBuilder findMaximumNum(StringBuilder str, 
                                             int n, int k)
    {
     
        // For every digit of the number
        for (int i = 0; i < n; i++) 
        {
     
            // If no more digits can be replaced
            if (k < 1)
                break;
     
            // If current digit is not already 9
            if (str.charAt(i) != '9') 
            {
     
                // Replace it with 9
                str.setCharAt(i, '9');
     
                // One digit has been used
                k--;
            }
        }
        return str;
    }
     
    // Driver code
    public static void main (String [] args)
    {
        StringBuilder str = new StringBuilder("569431");
         
        int n = str.length();
        int k = 3;
     
        System.out.println(findMaximumNum(str, n, k));
    }
}
 
// This code is contributed by ihritik

Python3

# Python3 implementation of the approach 
 
# Function to return the maximum number 
# that can be formed by changing 
# at most k digits in str 
def findMaximumNum(st, n, k):
 
    # For every digit of the number 
    for i in range(n): 
 
        # If no more digits can be replaced 
        if (k < 1): 
            break
 
        # If current digit is not already 9 
        if (st[i] != '9'): 
 
            # Replace it with 9 
            st = st[0:i] + '9' + st[i + 1:] 
 
            # One digit has been used 
            k -= 1
 
    return st
 
# Driver code 
st = "569431"
n = len(st) 
k = 3
print(findMaximumNum(st, n, k))
 
# This code is contributed by
# divyamohan123

C#

// C# implementation of the approach
using System;
using System.Text;
 
class GFG
{
    // Function to return the maximum number
    // that can be formed by changing
    // at most k digits in str
    static StringBuilder findMaximumNum(StringBuilder str, 
                                        int n, int k)
    {
     
        // For every digit of the number
        for (int i = 0; i < n; i++) 
        {
     
            // If no more digits can be replaced
            if (k < 1)
                break;
     
            // If current digit is not already 9
            if (str[i] != '9')
            {
     
                // Replace it with 9
                str[i] = '9';
     
                // One digit has been used
                k--;
            }
        }
        return str;
    }
     
    // Driver code
    public static void Main ()
    {
        StringBuilder str = new StringBuilder("569431");
         
        int n = str.Length;
        int k = 3;
     
        Console.WriteLine(findMaximumNum(str, n, k));
    }
}
 
// This code is contributed by ihritik

Javascript

<script>
      // JavaScript implementation of the approach
      // Function to return the maximum number
      // that can be formed by changing
      // at most k digits in str
      function findMaximumNum(str, n, k) {
        // For every digit of the number
        for (var i = 0; i < n; i++) {
          // If no more digits can be replaced
          if (k < 1) break;
 
          // If current digit is not already 9
          if (str[i] !== "9") {
            // Replace it with 9
            str[i] = "9";
 
            // One digit has been used
            k--;
          }
        }
        return str.join("");
      }
 
      // Driver code
      var str = "569431";
      var n = str.length;
      var k = 3;
 
      document.write(findMaximumNum(str.split(""), n, k));
    </script>

Output:

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

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!