647. Palindromic Substrings (Medium)

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:

1. The input string length won't exceed 1000.

Solution 1: O(n^2); O(1)

    int count = 1;

    public int countSubstrings(String s) {
        if (s == null || s.length() == 0) {
            return 0;
        }

        for (int i = 0; i < s.length() - 1; i++) {
            checkPalindrome(s, i, i);     //To check the palindrome of odd length palindromic sub-string
            checkPalindrome(s, i, i + 1);   //To check the palindrome of even length palindromic sub-string
        }
        return count;
    }

    private void checkPalindrome(String s, int i, int j) {
        while (i >= 0 && j < s.length() && s.charAt(i) == s.charAt(j)) {    //Check for the palindrome string
            count++;    //Increment the count if palindrome substring found
            i--;    //To trace string in left direction
            j++;    //To trace string in right direction
        }
    }

Solution 2: O(n^2); O(n^2)

    public int countSubstringsB(String s) {
        int n = s.length();
        int res = 0;
        boolean[][] dp = new boolean[n][n];
        for (int i = n - 1; i >= 0; i--) {
            for (int j = i; j < n; j++) {
                dp[i][j] = s.charAt(i) == s.charAt(j) && (j - i < 3 || dp[i + 1][j - 1]);
                if (dp[i][j]) {
                    ++res;
                }
            }
        }
        return res;
    }