Minimum steps of manufacturing letter ectopic words - C language solution

The minimum number of steps to make heterotopic words -c Language solution

Give you two strings of equal length s and t. In each step , You can choose to t Medium Any character Replace with Another character .

Return to make t Become s The minimum number of steps for an alphabetic ectopic word .

Letter heterotopic word Means the same letters , But the arrangement is different （ It could be the same ） String .

Example 1：

Output ：s = “bab”, t = “aba”
Output ：1
Tips ： use ‘b’ Replace t The first of ‘a’,t = “bba” yes s A letter heterotopic word of .

Example 2：

Output ：s = “leetcode”, t = “practice”
Output ：5
Tips ： Replace... With appropriate characters t Medium ‘p’, ‘r’, ‘a’, ‘i’ and ‘c’, send t become s Letter heteronym of .

Example 3：

Output ：s = “anagram”, t = “mangaar”
Output ：0
Tips ：“anagram” and “mangaar” Itself is a group of letter ectopic words .

Example 4：

Output ：s = “xxyyzz”, t = “xxyyzz”
Output ：0

Example 5：

Output ：s = “friend”, t = “family”
Output ：4

This problem is not difficult , We count the number of different letters of two words , Then we can get the result according to the difference .

int minSteps(char * s, char * t){
    
    int s_t[26];
    int s_t2[26];
    int i,j;
    int n=0;
    for(i=0;i<26;i++){
    
        s_t[i]=0;
         s_t2[i]=0;
    }
    i=0;
    while(s[i]!='\0'){
    
        s_t[s[i]-97]++;
        s_t2[t[i]-97]++;
        i++;
    }
      for(i=0;i<26;i++){
    
        if(s_t[i]!=0){
    
            if(s_t[i]-s_t2[i]<0)
            n=n+abs(s_t[i]-s_t2[i]);
        }
        if(s_t2[i]!=0&&s_t[i]==0){
    
             n=n+abs(s_t[i]-s_t2[i]);
        }
       
    }
    return n;

}