确保两个字符串具有相同字符所需的最少操作次数

给定两个字符串返回确保两个字符串具有相同字符所需的最少操作次数的值，即两个字符串成为彼此的字谜。

例子：

Input : s1 = "aab"
        s2 = "aba" 
Output :  2
Explanation : string 1 contains 2 a's and 1 b, 
also string 2 contains same characters

Input : s1 = "abc"
        s2 = "cdd"
Output : 2
Explanation : string 1 contains 1 a, 1 b, 1 c
while string 2 contains 1 c and 2 d's
so there are 2 different characters

问题来源： Yatra.com 面试经历 |设置 7

这个想法是分别为两个字符串创建一个额外的计数数组，然后计算字符的差异。

C++

// C++ program to count least number
// of manipulations to have two strings
// set of same characters
#include 
using namespace std;
 
const int MAX_CHAR = 26;
 
// return the count of manipulations
// required
int leastCount(string s1, string s2, int n)
{
    int count1[MAX_CHAR] = { 0 };
    int count2[MAX_CHAR] = { 0 };
 
    // count the number of different
    // characters in both strings
    for (int i = 0; i < n; i++) {
        count1[s1[i] - 'a'] += 1;
        count2[s2[i] - 'a'] += 1;
    }
 
    // check the difference in characters
    // by comparing count arrays
    int res = 0;
    for (int i = 0; i < MAX_CHAR; i++) {
        if (count1[i] != 0) {
            res += abs(count1[i] - count2[i]);
        }
    }
    return res;
}
 
// driver program
int main()
{
    string s1 = "abc";
    string s2 = "cdd";
    int len = s1.length();
    int res = leastCount(s1, s2, len);
    cout << res << endl;
    return 0;
}

Java

// Java program to count least number
// of manipulations to have two
// strings set of same characters
import java.io.*;
 
public class GFG {
 
    static int MAX_CHAR = 26;
 
    // return the count of manipulations
    // required
    static int leastCount(String s1,
                        String s2, int n)
    {
         
        int[] count1 = new int[MAX_CHAR];
        int[] count2 = new int[MAX_CHAR];
 
        // count the number of different
        // characters in both strings
        for (int i = 0; i < n; i++)
        {
            count1[s1.charAt(i) - 'a'] += 1;
            count2[s2.charAt(i) - 'a'] += 1;
        }
 
        // check the difference in characters
        // by comparing count arrays
        int res = 0;
         
        for (int i = 0; i < MAX_CHAR; i++)
        {
            if (count1[i] != 0) {
                res += Math.abs(count1[i]
                                 - count2[i]);
            }
        }
         
        return res;
    }
 
    // driver program
    static public void main(String[] args)
    {
        String s1 = "abc";
        String s2 = "cdd";
        int len = s1.length();
        int res = leastCount(s1, s2, len);
         
        System.out.println(res);
    }
}
 
// This code is contributed by vt_m.

Python3

# Python3 program to count least number
# of manipulations to have two strings
# set of same characters
MAX_CHAR = 26
 
# return the count of manipulations
# required
def leastCount(s1, s2, n):
 
    count1 = [0] * MAX_CHAR
    count2 = [0] * MAX_CHAR
 
    # count the number of different
    # characters in both strings
    for i in range ( n):
        count1[ord(s1[i]) - ord('a')] += 1
        count2[ord(s2[i]) - ord('a')] += 1
 
    # check the difference in characters
    # by comparing count arrays
    res = 0
    for i in range (MAX_CHAR):
        if (count1[i] != 0):
            res += abs(count1[i] - count2[i])
     
    return res
 
# Driver Code
if __name__ == "__main__":
 
    s1 = "abc"
    s2 = "cdd"
    l = len(s1)
    res = leastCount(s1, s2, l)
    print (res)
 
# This code is contributed by ita_c

C#

// C# program to count least number
// of manipulations to have two strings
// set of same characters
using System;
 
public class GFG {
 
    static int MAX_CHAR = 26;
 
    // return the count of manipulations
    // required
    static int leastCount(string s1,
                        string s2, int n)
    {
         
        int[] count1 = new int[MAX_CHAR];
        int[] count2 = new int[MAX_CHAR];
 
        // count the number of different
        // characters in both strings
        for (int i = 0; i < n; i++)
        {
            count1[s1[i] - 'a'] += 1;
            count2[s2[i] - 'a'] += 1;
        }
 
        // check the difference in characters
        // by comparing count arrays
        int res = 0;
        for (int i = 0; i < MAX_CHAR; i++)
        {
            if (count1[i] != 0) {
                res += Math.Abs(count1[i]
                                - count2[i]);
            }
        }
         
        return res;
    }
 
    // driver program
    static public void Main()
    {
        string s1 = "abc";
        string s2 = "cdd";
        int len = s1.Length;
        int res = leastCount(s1, s2, len);
        Console.WriteLine(res);
    }
}
 
// This code is contributed by vt_m.

Javascript

输出：