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Formatted question description: https://leetcode.ca/all/1525.html

1525. Number of Good Ways to Split a String (Medium)

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

Related Topics:
String, Bit Manipulation

Solution 1.

Simply use a linear scan and increment the answer if the left unique count equals the right unique count. If the left unique count is greater than the left one, break the loop.

// OJ: https://leetcode.com/problems/number-of-good-ways-to-split-a-string/
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int numSplits(string s) {
        int total[26] = {}, left[26] = {}, ans = 0;
        for (char c : s) total[c - 'a']++;
        for (char c : s) {
            ++left[c - 'a'];
            int L = 0, R = 0;
            for (int i = 0; i < 26; ++i) {
                if (left[i]) ++L;
                if (total[i] - left[i]) ++R;
            }
            if (L == R) ++ans;
            if (L > R) break;
        }
        return ans;
    }
};

Another implementation that doesn’t require the O(26) check

// OJ: https://leetcode.com/problems/number-of-good-ways-to-split-a-string/
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int numSplits(string s) {
        int right[26] = {}, left[26] = {}, ans = 0, L = 0, R = 0;
        for (char c : s ) R += right[c - 'a']++ == 0;
        for (char c : s) {
            L += left[c - 'a']++ == 0;
            R -= right[c - 'a']-- == 1;
            ans += L == R;
            if (L > R) break;
        }
        return ans;
    }
};

• Java
• C++
• Python
• Go

• class Solution {
      public int numSplits(String s) {
          Map<Character, Integer> map1 = new HashMap<Character, Integer>();
          Map<Character, Integer> map2 = new HashMap<Character, Integer>();
          int length = s.length();
          for (int i = 0; i < length; i++) {
              char c = s.charAt(i);
              int count = map2.getOrDefault(c, 0) + 1;
              map2.put(c, count);
          }
          int goodSplits = 0;
          for (int i = 0; i < length - 1; i++) {
              char c = s.charAt(i);
              int count1 = map1.getOrDefault(c, 0) + 1;
              map1.put(c, count1);
              int count2 = map2.get(c) - 1;
              if (count2 > 0)
                  map2.put(c, count2);
              else
                  map2.remove(c);
              if (map1.size() == map2.size())
                  goodSplits++;
              else if (map1.size() > map2.size())
                  break;
          }
          return goodSplits;
      }
  }
  
  ############
  
  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;
      }
  }
  

• // OJ: https://leetcode.com/problems/number-of-good-ways-to-split-a-string/
  // Time: O(N)
  // Space: O(1)
  class Solution {
  public:
      int numSplits(string s) {
          int total[26] = {}, left[26] = {}, ans = 0;
          for (char c : s) total[c - 'a']++;
          for (char c : s) {
              ++left[c - 'a'];
              int L = 0, R = 0;
              for (int i = 0; i < 26; ++i) {
                  if (left[i]) ++L;
                  if (total[i] - left[i]) ++R;
              }
              if (L == R) ++ans;
              if (L > R) break;
          }
          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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