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 strstring 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 codeint main(){ string str = "569431"; int n = str.length(); int k = 3; cout << findMaximumNum(str, n, k); return 0;} |
Java
// Java implementation of the approachimport 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 = 3print(findMaximumNum(st, n, k))# This code is contributed by# divyamohan123 |
C#
// C# implementation of the approachusing 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:
999931
Time Complexity: O(n)
Auxiliary Space: O(1)
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