使第一个字符串成为第二个字符串的子字符串所需的最小更改

给定两个字符串S1和S2(S1的大小<= S2的大小)。任务是找到字符串S2中要替换的最少字符数，以使字符串S1是S2的子字符串。

例子：

Input : S1 = cdef, S2 = abbdef
Output : 1

Input : S1 = gfg, S2 = fgg
Output : 2 

方法：

1. 遍历字符串S2
   • 从S2中的每个索引中，检查S1长度的子字符串中不匹配的字符数
   • 在ans中存储和更新上一个和当前不匹配的最小值
2. 返回ans。

下面是上述方法的实现：

C++

// CPP program to find the minimum number of
// characters to be replaced in string S2, such
// that S1 is a substring of S2
  
#include 
using namespace std;
  
// Function to find the minimum number of
// characters to be replaced in string S2, such
// that S1 is a substring of S2
int minimumChar(string S1, string S2)
{
    // Get the sizes of both strings
    int n = S1.size(), m = S2.size();
  
    int ans = INT_MAX;
  
    // Traverse the string S2
    for (int i = 0; i < m - n + 1; i++) {
        int minRemovedChar = 0;
  
        // From every index in S2, check the number of
        // mis-matching characters in substring of
        // length of S1
        for (int j = 0; j < n; j++) {
            if (S1[j] != S2[i + j]) {
                minRemovedChar++;
            }
        }
  
        // Take minimum of prev and current mis-match
        ans = min(minRemovedChar, ans);
    }
  
    // return answer
    return ans;
}
  
// Driver Code
int main()
{
    string S1 = "abc";
    string S2 = "paxzk";
  
    cout << minimumChar(S1, S2);
  
    return 0;
}

Java

// Java program to find the minimum 
// number of characters to be 
// replaced in string S2, such
// that S1 is a substring of S2
import java.io.*;
  
class GFG 
{
  
// Function to find the minimum 
// number of characters to be 
// replaced in string S2, such
// that S1 is a substring of S2
static int minimumChar(String S1,
                       String S2)
{
    // Get the sizes of both strings
    int n = S1.length();
    int m = S2.length();
  
    int ans = Integer.MAX_VALUE ;
  
    // Traverse the string S2
    for (int i = 0; i < m - n + 1; i++)
    {
        int minRemovedChar = 0;
  
        // From every index in S2, check 
        // the number of mis-matching 
        // characters in substring of
        // length of S1
        for (int j = 0; j < n; j++) 
        {
            if (S1.charAt(j) != S2.charAt(i + j)) 
            {
                minRemovedChar++;
            }
        }
  
        // Take minimum of prev and 
        // current mis-match
        ans = Math.min(minRemovedChar, ans);
    }
  
    // return answer
    return ans;
}
  
// Driver Code
public static void main (String[] args) 
{
    String S1 = "abc";
    String S2 = "paxzk";
      
    System.out.println(minimumChar(S1, S2));
}
}
  
// This code is contributed by Shashank

Python3

# Python3 program to find the minimum 
# number of characters to be replaced 
# in string S2, such that S1 is a 
# substring of S2
import sys 
  
# Function to find the minimum number of
# characters to be replaced in string S2,
# such that S1 is a substring of S2
def minimumChar(S1, S2):
      
    # Get the sizes of both strings
    n, m = len(S1), len(S2)
  
    ans = sys.maxsize
  
    # Traverse the string S2
    for i in range(m - n + 1):
        minRemovedChar = 0
  
        # From every index in S2, check the 
        # number of mis-matching characters 
        # in substring of length of S1
        for j in range(n):
            if (S1[j] != S2[i + j]):
                minRemovedChar += 1
                  
        # Take minimum of prev and 
        # current mis-match
        ans = min(minRemovedChar, ans)
          
    # return answer
    return ans
      
# Driver Code
if __name__ == '__main__':
    S1 = "abc"
    S2 = "paxzk"
    print(minimumChar(S1, S2))
  
# This code is contributed 
# by PrinciRaj1992

C#

// C# program to find the minimum 
// number of characters to be 
// replaced in string S2, such
// that S1 is a substring of S2
using System;
  
class GFG 
{
  
// Function to find the minimum 
// number of characters to be 
// replaced in string S2, such
// that S1 is a substring of S2
static int minimumChar(String S1,
                       String S2)
{
    // Get the sizes of both strings
    int n = S1.Length;
    int m = S2.Length;
  
    int ans = Int32.MaxValue ;
  
    // Traverse the string S2
    for (int i = 0; i < m - n + 1; i++)
    {
        int minRemovedChar = 0;
  
        // From every index in S2, check 
        // the number of mis-matching 
        // characters in substring of
        // length of S1
        for (int j = 0; j < n; j++) 
        {
            if (S1[j] != S2[i + j]) 
            {
                minRemovedChar++;
            }
        }
  
        // Take minimum of prev and 
        // current mis-match
        ans = Math.Min(minRemovedChar, ans);
    }
  
    // return answer
    return ans;
}
  
// Driver Code
public static void Main() 
{
    String S1 = "abc";
    String S2 = "paxzk";
      
    Console.WriteLine(minimumChar(S1, S2));
}
}
  
// This code is contributed 
// by Akanksha Rai(Abby_akku)

输出：

2

时间复杂度：O(N * M)。