最小化使两个给定字符串排列所需的给定操作的计数

给定两个字符串str1和str2 ，任务是计算使str1和str2彼此排列所需的两个字符串之一的以下三种类型的最小操作次数：

1. 在字符串插入一个字符。
2. 从字符串删除一个字符。
3. 用字符串的另一个字符替换一个字符。

注意：以上所有操作都是等价的。

例子：

Input: str1 = “geeksforgeeks”, str2 = “geeksforcoder”
Output: 4
Explanation: Rearrange the string str2 to “geeksforcedor”
Replace the value of str1[8] to ‘c’.
Replace the value of str1[10] to ‘d’.
Replace the value of str1[11] to ‘o’.
Replace the value of str1[12] to ‘r’.
Therefore, the required output is 4.

Input: str1 = “geeks”, str2 = “keeg”
Output: 1

方法：使用Hashing存储字符串中每个字符出现的频率可以解决这个问题。以下是解决问题的意见：

X = Number of characters which are present in both string, str1 and str2.
N1 – X = Number of characters present only in str1.
N2 – X = Number of characters present only in str2.
Total number of replacement operations = min(N1 – X, N2 – X)
Total number of insert/rRemove operations = max(N1 – X, N2 – X) – min(N1 – X, N2 – X).
Therefore, total number of operations = max(N1 – X, N2 – X),

请按照以下步骤解决问题：

1. 初始化两个数组，比如freq1[]和freq2[]分别存储str1和str2的所有字符的频率。
2. 遍历这两个字符串和两个所述字符串的各字符的频率分别[]存储在阵列FREQ1 []和FREQ2。
3. 遍历数组freq1[]和freq2[] 。
4. 对于每个^第i^个字符，如果freq1[i] 超过 freq2[i] ，则将freq1[i]替换为freq1[i] – freq2[i]并设置freq2[i] = 0 ，反之亦然。
5. 最后，计算数组freq1[]和freq2[]的总和，并打印它们之间的最大值作为答案

以下是上述方法的实现：

C++

// C++ program to implement
// the above approach
 
#include 
using namespace std;
 
// Function to minimize the count of
// operations to make str1 and str2
// permutations of each other
int ctMinEdits(string str1, string str2)
{
    int N1 = str1.length();
    int N2 = str2.length();
 
    // Store the frequency of
    // each character of str1
    int freq1[256] = { 0 };
    for (int i = 0; i < N1; i++) {
        freq1[str1[i]]++;
    }
 
    // Store the frequency of
    // each character of str2
    int freq2[256] = { 0 };
    for (int i = 0; i < N2; i++) {
        freq2[str2[i]]++;
    }
 
    // Traverse the freq1[] and freq2[]
    for (int i = 0; i < 256; i++) {
 
        // If frequency of character in
        // str1 is greater than str2
        if (freq1[i] > freq2[i]) {
            freq1[i] = freq1[i]
                       - freq2[i];
            freq2[i] = 0;
        }
 
        // Otherwise
        else {
            freq2[i] = freq2[i]
                       - freq1[i];
            freq1[i] = 0;
        }
    }
 
    // Store sum of freq1[]
    int sum1 = 0;
 
    // Store sum of freq2[]
    int sum2 = 0;
 
    for (int i = 0; i < 256; i++) {
        sum1 += freq1[i];
        sum2 += freq2[i];
    }
 
    return max(sum1, sum2);
}
 
// Driver Code
int main()
{
    string str1 = "geeksforgeeks";
    string str2 = "geeksforcoder";
    cout << ctMinEdits(str1, str2);
}

Java

// Java program to implement
// the above approach
import java.util.*;
import java.io.*;
import java.lang.Math;
 
class GFG{
     
// Function to minimize the count of
// operations to make str1 and str2
// permutations of each other
static int ctMinEdits(String str1, String str2)
{
    int N1 = str1.length();
    int N2 = str2.length();
   
    // Store the frequency of
    // each character of str1
    int freq1[] = new int[256];
    Arrays.fill(freq1, 0);
     
    for(int i = 0; i < N1; i++)
    {
        freq1[str1.charAt(i)]++;
    }
   
    // Store the frequency of
    // each character of str2
    int freq2[] = new int[256];
    Arrays.fill(freq2, 0);
     
    for(int i = 0; i < N2; i++)
    {
        freq2[str2.charAt(i)]++;
    }
   
    // Traverse the freq1[] and freq2[]
    for(int i = 0; i < 256; i++)
    {
         
        // If frequency of character in
        // str1 is greater than str2
        if (freq1[i] > freq2[i])
        {
            freq1[i] = freq1[i] - freq2[i];
            freq2[i] = 0;
        }
   
        // Otherwise
        else
        {
            freq2[i] = freq2[i] - freq1[i];
            freq1[i] = 0;
        }
    }
   
    // Store sum of freq1[]
    int sum1 = 0;
   
    // Store sum of freq2[]
    int sum2 = 0;
   
    for(int i = 0; i < 256; i++)
    {
        sum1 += freq1[i];
        sum2 += freq2[i];
    }
   
    return Math.max(sum1, sum2);
}
 
// Driver Code
public static void main(final String[] args)
{
    String str1 = "geeksforgeeks";
    String str2 = "geeksforcoder";
     
    System.out.println(ctMinEdits(str1, str2));
}
}
 
// This code is contributed by bikram2001jha

Python3

# Python3 program to implement
# the above approach
  
# Function to minimize the count of
# operations to make str1 and str2
# permutations of each other
def ctMinEdits(str1, str2):
     
    N1 = len(str1)
    N2 = len(str2)
  
    # Store the frequency of
    # each character of str1
    freq1 =  [0] * 256
    for i in range(N1):
        freq1[ord(str1[i])] += 1
     
    # Store the frequency of
    # each character of str2
    freq2 = [0] * 256
    for i in range(N2):
        freq2[ord(str2[i])] += 1
     
    # Traverse the freq1[] and freq2[]
    for i in range(256):
  
        # If frequency of character in
        # str1 is greater than str2
        if (freq1[i] > freq2[i]):
            freq1[i] = freq1[i] - freq2[i]
            freq2[i] = 0
  
        # Otherwise
        else:
            freq2[i] = freq2[i] - freq1[i]
            freq1[i] = 0
         
    # Store sum of freq1[]
    sum1 = 0
  
    # Store sum of freq2[]
    sum2 = 0
  
    for i in range(256):
        sum1 += freq1[i]
        sum2 += freq2[i]
     
    return max(sum1, sum2)
 
# Driver Code
str1 = "geeksforgeeks"
str2 = "geeksforcoder"
 
print(ctMinEdits(str1, str2))
 
# This code is contributed by code_hunt

C#

// C# program to implement
// the above approach
using System;
 
class GFG{
      
// Function to minimize the count of
// operations to make str1 and str2
// permutations of each other
static int ctMinEdits(string str1, string str2)
{
    int N1 = str1.Length;
    int N2 = str2.Length;
    
    // Store the frequency of
    // each character of str1
    int[] freq1 = new int[256];
    freq1[0] = str1[0];
      
    for(int i = 0; i < N1; i++)
    {
        freq1[str1[i]]++;
    }
    
    // Store the frequency of
    // each character of str2
    int[] freq2 = new int[256];
    freq2[0] = str2[0];
      
    for(int i = 0; i < N2; i++)
    {
        freq2[str2[i]]++;
    }
    
    // Traverse the freq1[] and freq2[]
    for(int i = 0; i < 256; i++)
    {
         
        // If frequency of character in
        // str1 is greater than str2
        if (freq1[i] > freq2[i])
        {
            freq1[i] = freq1[i] - freq2[i];
            freq2[i] = 0;
        }
    
        // Otherwise
        else
        {
            freq2[i] = freq2[i] - freq1[i];
            freq1[i] = 0;
        }
    }
    
    // Store sum of freq1[]
    int sum1 = 0;
    
    // Store sum of freq2[]
    int sum2 = 0;
    
    for(int i = 0; i < 256; i++)
    {
        sum1 += freq1[i];
        sum2 += freq2[i];
    }
    return Math.Max(sum1, sum2);
}
  
// Driver Code
public static void Main()
{
    string str1 = "geeksforgeeks";
    string str2 = "geeksforcoder";
      
    Console.WriteLine(ctMinEdits(str1, str2));
}
}
 
// This code is contributed by code_hunt

Javascript

 
// Javascript program to implement
// the above approach
 
// Function to minimize the count of
// operations to make str1 and str2
// permutations of each other
function ctMinEdits(str1, str2)
{
    let N1 = str1.length;
    let N2 = str2.length;
    
    // Store the frequency of
    // each character of str1
    let freq1 = new Array(256).fill(0);
      
    for(let i = 0; i < N1; i++)
    {
        freq1[str1[i].charCodeAt()]++;
    }
    
    // Store the frequency of
    // each character of str2
    let freq2 = new Array(256).fill(0);
      
    for(let i = 0; i < N2; i++)
    {
        freq2[str2[i].charCodeAt()]++;
    }
    
    // Traverse the freq1[] and freq2[]
    for(let i = 0; i < 256; i++)
    {
          
        // If frequency of character in
        // str1 is greater than str2
        if (freq1[i] > freq2[i])
        {
            freq1[i] = freq1[i] - freq2[i];
            freq2[i] = 0;
        }
    
        // Otherwise
        else
        {
            freq2[i] = freq2[i] - freq1[i];
            freq1[i] = 0;
        }
    }
    
    // Store sum of freq1[]
    let sum1 = 0;
    
    // Store sum of freq2[]
    let sum2 = 0;
    
    for(let i = 0; i < 256; i++)
    {
        sum1 += freq1[i];
        sum2 += freq2[i];
    }
    
    return Math.max(sum1, sum2);
}
 
// Driver Code
 
    let str1 = "geeksforgeeks";
    let str2 = "geeksforcoder";
      
    document.write(ctMinEdits(str1.split(''), str2.split('')));
      

输出：

4

时间复杂度： O(N)
辅助空间： O(1)

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