README_EN.md

647. Palindromic Substrings

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Description

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.

 

Solutions

Python3

class Solution:
    def countSubstrings(self, s: str) -> int:
        t = '^#' + '#'.join(s) + '#$'
        n = len(t)
        p = [0 for _ in range(n)]
        pos, maxRight = 0, 0
        ans = 0
        for i in range(1, n - 1):
            p[i] = min(maxRight - i, p[2 * pos - i]) if maxRight > i else 1
            while t[i - p[i]] == t[i + p[i]]:
                p[i] += 1
            if i + p[i] > maxRight:
                maxRight = i + p[i]
                pos = i
            ans += p[i] // 2
        return ans

Java

class Solution {
    public int countSubstrings(String s) {
        StringBuilder sb = new StringBuilder("^#");
        for (char ch : s.toCharArray()) {
            sb.append(ch).append('#');
        }
        String t = sb.append('$').toString();
        int n = t.length();
        int[] p = new int[n];
        int pos = 0, maxRight = 0;
        int ans = 0;
        for (int i = 1; i < n - 1; i++) {
            p[i] = maxRight > i ? Math.min(maxRight - i, p[2 * pos - i]) : 1;
            while (t.charAt(i - p[i]) == t.charAt(i + p[i])) {
                p[i]++;
            }
            if (i + p[i] > maxRight) {
                maxRight = i + p[i];
                pos = i;
            }
            ans += p[i] / 2;
        }
        return ans;
    }
}

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