### 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.

### Solution:

Time Complexity : O(n^2)

Space Complexity: O(n^2)

```
var countSubstrings = function(s) {
const n = s.length;
const dp = [...Array(n)].map(() => Array(n).fill(false));
let count = 0;
// Base case: single letter substrings
for(let i = 0; i < n; i++) {
dp[i][i] = true;
count++
}
// Base case: double letter substrings
for(let i = 0; i < n-1; i++) {
dp[i][i+1] = (s[i] === s[i+1]);
dp[i][i+1] && count++;
}
// substrings longer than 2 chars
for(let len = 3; len <= n; len++) {
let start = 0, end = start+len-1;
while(end < n) {
dp[start][end] = (dp[start+1][end-1] && s[start] === s[end]);
dp[start][end] && count++;
start++; end++;
}
}
return count;
};
```

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