<p>Given a string <code>s</code>, return <em>the number of <strong>palindromic substrings</strong> in it</em>.</p>

<p>A string is a <strong>palindrome</strong> when it reads the same backward as forward.</p>

<p>A <strong>substring</strong> is a contiguous sequence of characters within the string.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;abc&quot;
<strong>Output:</strong> 3
<strong>Explanation:</strong> Three palindromic strings: &quot;a&quot;, &quot;b&quot;, &quot;c&quot;.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;aaa&quot;
<strong>Output:</strong> 6
<strong>Explanation:</strong> Six palindromic strings: &quot;a&quot;, &quot;a&quot;, &quot;a&quot;, &quot;aa&quot;, &quot;aa&quot;, &quot;aaa&quot;.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 1000</code></li>
	<li><code>s</code> consists of lowercase English letters.</li>
</ul>

class Solution {
public:
    int countSubstrings(string s) {
        int len = s.length();
        int count = 0;
        vector<vector<bool>> dp(len, vector<bool>(len, false));

        for (int i = 0; i < len; i++) {
            dp[i][i] = 1;
            count++;
        }

        for (int i = 0; i < len-1; i++) {
            dp[i][i+1] = (s[i] == s[i+1]);
            count += dp[i][i+1];
        }

        for (int i = 3; i <= len; i++) {
            for (int j = 0; j + i - 1 < len; j++) {
                if (isPalindrome(s, j, j + i - 1, dp) == true) {
                    count++;
                }
            }
        }

        return count;
    }

private:
    bool isPalindrome(string s, int startIdx, int endIdx, vector<vector<bool>>& dp)
    {
        if (s[startIdx] != s[endIdx]) {
            dp[startIdx][endIdx] = false;
            return false;
        }

        dp[startIdx][endIdx] = dp[startIdx+1][endIdx-1];
        return dp[startIdx][endIdx];
    }
};
398720