647. Palindromic Substrings

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".

Solution

Approach 1: DP in O(N^2) -- start searching for palindromes from short to long ones, using O(N^2) to store existing palindromes found.

Approach 2: expanding from a center, O(1) space and O(N^2) time.

Approach 3: Manacher's Algorithm (?)

Code (Python)

Approach 1:

    def countSubstrings(self, s: str) -> int:
        # DP in O(N^2) -- start searching for palindromes from short to long ones, using O(N^2) to store existing palindromes found
        if not s:
            return 0
        palindromes = set([char for char in s])
        count = len(s)
        for span in range(1, len(s)):
            for start in range(len(s) - span): # while start + span < len(s)
                if s[start] == s[start+span] and (span == 1 or s[start+1:start+span] in palindromes):
                    count += 1
                    palindromes.add(s[start:start+span+1])
        return count

Approach 2:

    def countSubstrings(self, s: str) -> int:
        # expanding from a center, O(1) space and O(N^2) time
        if not s:
            return 0
        count = 0
        for center in range(2 * len(s) - 1): # each letter and each space in between them
            left = center // 2
            right = left + center % 2 # right == left if center is on the letter, otherwise right == left + 1
            while left >= 0 and right < len(s) and s[left] == s[right]:
                count += 1
                left -= 1
                right += 1
        return count

Code (C++)

Approach 1:

Approach 2: