# 1525. Number of Good Ways to Split a String

## Description

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 <= 105
• s consists of only lowercase English letters.

## Solutions

• class Solution {
public int numSplits(String s) {
Map<Character, Integer> cnt = new HashMap<>();
for (char c : s.toCharArray()) {
cnt.merge(c, 1, Integer::sum);
}
Set<Character> vis = new HashSet<>();
int ans = 0;
for (char c : s.toCharArray()) {
if (cnt.merge(c, -1, Integer::sum) == 0) {
cnt.remove(c);
}
if (vis.size() == cnt.size()) {
++ans;
}
}
return ans;
}
}

• class Solution {
public:
int numSplits(string s) {
unordered_map<char, int> cnt;
for (char& c : s) {
++cnt[c];
}
unordered_set<char> vis;
int ans = 0;
for (char& c : s) {
vis.insert(c);
if (--cnt[c] == 0) {
cnt.erase(c);
}
ans += vis.size() == cnt.size();
}
return ans;
}
};

• class Solution:
def numSplits(self, s: str) -> int:
cnt = Counter(s)
vis = set()
ans = 0
for c in s:
cnt[c] -= 1
if cnt[c] == 0:
cnt.pop(c)
ans += len(vis) == len(cnt)
return ans


• func numSplits(s string) (ans int) {
cnt := map[rune]int{}
for _, c := range s {
cnt[c]++
}
vis := map[rune]bool{}
for _, c := range s {
vis[c] = true
cnt[c]--
if cnt[c] == 0 {
delete(cnt, c)
}
if len(vis) == len(cnt) {
ans++
}
}
return
}