# 647. Palindromic Substrings (Medium)

https://leetcode.com/problems/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".
```

Note:

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

## Solutions

``````class Solution {

public int countSubstrings(String s) {
Set<String> validPalindromes = new HashSet<>();
Set<String> invalidPalindromes = new HashSet<>();

int ans = 0;

if (s == null || s.isEmpty()) {
return ans;
}

int len = s.length();

// i is palindrome length
for (int i = 1; i <= len; i++) {

// possible palindrome of length i starts from j
for (int j = 0; j < len - i + 1; j++) {
String subStr = s.substring(j, j + i);

if (invalidPalindromes.contains(subStr)) {
continue;
}

if (validPalindromes.contains(subStr)) {
ans++;

continue;
}

// of length 1, all the single chars are palindrome
if (i == 1) {
ans++;

continue;
}

// of length 2
if (i == 2) {
// if first chart == second char
if (subStr.charAt(0) == subStr.charAt(1)) {

ans++;

} else {
}

// others with different chars are not palindrome
continue;
}

// for length >= 2, expend to both side by 1 char

// FIXME Following mistake should be avoid.
// ignore cases that overstep the boundary
// if (j == 0 || j + i == len) {
//     continue;
// }

if (!validPalindromes.contains(subStr.substring(1, subStr.length() - 1))) {
continue;
}

if (subStr.charAt(0) != subStr.charAt(subStr.length() - 1)) {
} else {
ans++;

}
}
}

return ans;
}
}
``````