查找最多可更改K个位数可形成的最大数字

给定一个表示数字和整数K的字符串str ，任务是找到最大的数字，该数字可以通过更改给定数字中最多K个数字来形成。

例子：

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

Input: str = “5687”, K = 2
Output: 9987

方法：为了获得最大数目，必须将最左边的数字替换为9s。对于从最左边的数字开始的数字的每个数字，如果尚未为9并且K大于0，则将其替换为9，然后将K减1。对于K大于0的每个数字重复这些步骤。最后，打印更新的号码。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
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
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

输出：

999931