Number of Good Ways to Split a String

Problem

You are given a string s.

A split is called good if you can split s into two non-empty strings sleft and sright where their concatenation is equal to s (i.e., sleft + sright = s) and the number of distinct letters in sleft and sright is 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").

Constraints:

1 <= s.length <= 10^5
s consists of only lowercase English letters.

Solution (Java)

class Solution {
    public int numSplits(String s) {
        HashSet<Character> hs = new HashSet<>();
        int[] dp1 = new int[s.length() - 1];
        int[] dp2 = new int[s.length() - 1];
        for (int i = 0; i < s.length() - 1; i++) {
            char ch = s.charAt(i);
            hs.add(ch);
            dp1[i] = hs.size();
        }
        HashSet<Character> hm = new HashSet<>();
        for (int i = s.length() - 1; i > 0; i--) {
            char ch = s.charAt(i);
            hm.add(ch);
            dp2[i - 1] = hm.size();
        }
        int count = 0;
        for (int i = 0; i < s.length() - 1; i++) {
            if (dp1[i] == dp2[i]) {
                count++;
            }
        }
        return count;
    }
}

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).