3223: Minimum-Length-of-String-After-Operations
Medium

table of contents

This question looks a lot more complicated than it actually is.

The trick is that, in string s, if a character appears:

(you can easily demonstrate this using an example)

Since there are only 26 letters in the alphabet, we can keep track of this data in an integer array length 26 while we iterate through the string s.

code

class Solution {
public:
    int minimumLength(string s) {
        int alphabet[26] = {0};
        for (char c : s) {
            if (alphabet[c-'a'] == 0) {
                ++alphabet[c-'a'];
            } else {
                alphabet[c-'a'] ^= 3;       
            };
        }  
        int ans = 0;
        for (int i : alphabet) {
            ans += i;
        }
        return ans;
    }
};

complexity