1525. Number of Good Ways to Split a String

You are given a string s, a split is called good if you can split s into 2 non-empty strings p and q where its concatenation is equal to s and the number of distinct letters in p and q are the same.

Return the number of good splits you can make in s.

Example 1:

Input: s = "aacaba"
Output: 2
Explanation: There are 5 ways to split "aacaba" and 2 of them are good. 
("a", "acaba") Left string and right string contains 1 and 3 different letters respectively.
("aa", "caba") Left string and right string contains 1 and 3 different letters respectively.
("aac", "aba") Left string and right string contains 2 and 2 different letters respectively (good split).
("aaca", "ba") Left string and right string contains 2 and 2 different letters respectively (good split).
("aacab", "a") Left string and right string contains 3 and 1 different letters respectively.

Example 2:

Input: s = "abcd"
Output: 1
Explanation: Split the string as follows ("ab", "cd").

Example 3:

Input: s = "aaaaa"
Output: 4
Explanation: All possible splits are good.

Example 4:

Input: s = "acbadbaada"
Output: 2

Constraints:

• s contains only lowercase English letters.

• 1 <= s.length <= 10^5

Solution

Intuitive :

We need to know for each spit point i, L [0...i], R[i+1...n], the character count of each sides.

So just collect the character and their count on a map. and then move the split index to check the count;

    public int numSplits(String s) {
        int[] fromLeft = new int[26];
        int countL = 0, countR = 0;
        int[] fromRight = new int[26];
        
        for(int i = 0; i < s.length(); i++){
            fromLeft[s.charAt(i) - 'a']++;
            if(fromLeft[s.charAt(i) - 'a'] == 1){
                countL++;
            }
        }
        int ret = 0;
        for(int i = 0; i < s.length(); i++){
            
            fromLeft[s.charAt(i)-'a']--;
            if(fromLeft[s.charAt(i)-'a'] == 0){
                countL--;
            }
            fromRight[s.charAt(i) - 'a']++;
            if(fromRight[s.charAt(i) - 'a'] == 1){
                countR++;
            }
            if(countL == countR){
                ret++;
            }
            
        }
        
        return ret;
        
    }