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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
    }
    

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