647. Palindromic Substrings.md

647. Palindromic Substrings

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leetcode-cn Daily Challenge on August 19th, 2020.

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Difficulty : Medium

Related Topics : String、Dynamic Programming

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Given a string, your task is to count how many palindromic substrings in this string.

The substrings with different start indexes or end indexes are counted as different substrings even they consist of same characters.

Example 1:

Input: "abc"
Output: 3
Explanation: Three palindromic strings: "a", "b", "c".

Example 2:

Input: "aaa"
Output: 6
Explanation: Six palindromic strings: "a", "a", "a", "aa", "aa", "aaa".

Note:

• The input string length won't exceed 1000.

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Solution

• mine
  • Java
    • Runtime: 1 ms, faster than 99.95%, Memory Usage: 37.5 MB, less than 88.33% of Java online submissions

      //O(N^2)time
      //O(N)space
      public int countSubstrings(String s) {
          char[] arr = s.toCharArray();
          int res = 0;
          for(int i = 0; i < arr.length; i++){
              res += check(arr, i, i);
              res += check(arr, i, i + 1);
          }
          return res;
      }
      
      int check(char[] arr, int s, int e){
          int res = 0;
          while(s >= 0 && e < arr.length && arr[s--] == arr[e++]){
              res++;
          }
          return res;
      }
      

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• the most votes

• Manacher's Algorithm Runtime: 1 ms, faster than 99.95%, Memory Usage: 37.2 MB, less than 97.04% of Java online submissions

  // O(N)time
  // O(N)space
  public int countSubstrings(String S) {
      char[] A = new char[2 * S.length() + 3];
      A[0] = '@';
      A[1] = '#';
      A[A.length - 1] = '$';
      int t = 2;
      for (char c: S.toCharArray()) {
          A[t++] = c;
          A[t++] = '#';
      }
  
      int[] Z = new int[A.length];
      int center = 0, right = 0;
      for (int i = 1; i < Z.length - 1; ++i) {
          if (i < right)
              Z[i] = Math.min(right - i, Z[2 * center - i]);
          while (A[i + Z[i] + 1] == A[i - Z[i] - 1])
              Z[i]++;
          if (i + Z[i] > right) {
              center = i;
              right = i + Z[i];
          }
      }
      int ans = 0;
      for (int v: Z) ans += (v + 1) / 2;
      return ans;
  }
  

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