Welcome to Subscribe On Youtube

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()) {
            vis.add(c);
            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:
            vis.add(c)
            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
}

All Problems

All Solutions