确保两个字符串具有相同字符所需的最少操作次数
原文:https://www . geesforgeks . org/最少数字操作-需要-确保-两个字符串-相同-字符/
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。
示例:
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 采访心得|第七集
这个想法是为两个字符串分别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 <bits/stdc++.h>
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 语言(一种计算机语言，尤用于创建网站)
// 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.
Python 3
# 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.
java 描述语言
<script>
// Javascript program to count least number
// of manipulations to have two
// strings set of same characters
let MAX_CHAR = 26;
// Return the count of manipulations
// required
function leastCount(s1, s2, n)
{
let count1 = new Array(MAX_CHAR);
let count2 = new Array(MAX_CHAR);
for(let i = 0; i < MAX_CHAR; i++)
{
count1[i] = 0;
count2[i] = 0;
}
// Count the number of different
// characters in both strings
for(let i = 0; i < n; i++)
{
count1[s1[i].charCodeAt(0) -
'a'.charCodeAt(0)] += 1;
count2[s2[i].charCodeAt(0) -
'a'.charCodeAt(0)] += 1;
}
// Check the difference in characters
// by comparing count arrays
let res = 0;
for(let i = 0; i < MAX_CHAR; i++)
{
if (count1[i] != 0)
{
res += Math.abs(count1[i] - count2[i]);
}
}
return res;
}
// Driver Code
let s1 = "abc";
let s2 = "cdd";
let len = s1.length;
let res = leastCount(s1, s2, len);
document.write(res);
// This code is contributed by avanitrachhadiya2155
</script>
输出:
2
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