1525. Number of Good Ways to Split a String.md

1525. Number of Good Ways to Split a String

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the third one in Biweekly Contest 31.

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Difficulty : Medium

Related Topics : String、Bit Manipulation

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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

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Solution

• mine
  • Java
    • Runtime: 23 ms, faster than 51.50%, Memory Usage: 40.8 MB, less than 52.46% of Java online submissions

      // O(N)time
      // O(D)space  D is the count of distinct char in s
      public int numSplits(String s) {
          char[] arr = s.toCharArray();
          Map<Character, Integer> map = new HashMap<>();
          Set<Character> set = new HashSet<>();
          for (char c : arr) {
              map.put(c, map.getOrDefault(c, 0) + 1);
          }
          int res = 0;
          for (int i = 0; i < arr.length; i++) {
              char c = arr[i];
              set.add(c);
              int t = map.getOrDefault(c, 0);
              if (t > 1) {
                  map.put(c, t - 1);
              } else {
                  map.remove(c);
              }
              if (set.size() == map.size()) {
                  res++;
              }
          }
          return res;
      }
      

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• the most votes

• Sliding Window Runtime: 5 ms, faster than 94.30%, Memory Usage: 39.4 MB, less than 94.54% of Java online submissions

  // O(N)time
  // O(1)space
  public int numSplits(String str) {
      int[] l = new int[26];
      int[] r = new int[26];
      int d_l = 0, d_r = 0;
      int res = 0;
      char[] s = str.toCharArray();
      for (char ch : s) {
          d_r += ++r[ch - 'a'] == 1 ? 1 : 0;
      }
      for (int i = 0; i < s.length; ++i) {
          d_l += ++l[s[i] - 'a'] == 1 ? 1 : 0;
          d_r -= --r[s[i] - 'a'] == 0 ? 1 : 0;
          res += d_l == d_r ? 1 : 0;
      }
      return res;
  }
  

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