<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) {
        vector< vector<int> > dp(1000, vector<int>(1000, 1));
        int ans = s.length();
        for(int i = 0; i < s.length() - 1; i++){
            dp[i][i+1] = (s[i] == s[i+1]);
            ans += dp[i][i+1];
        }
        
        for(int d = 2; d < s.length(); d++){
            int i = 0, j = d;
            while(i < s.length() && j < s.length()){
                if(s[i] == s[j]){
                    dp[i][j] = dp[i + 1][j - 1];
                }
                else{
                    dp[i][j] = 0;
                }
                ans += dp[i][j];
                i++;
                j++;
            }
        }
        return ans;
    }
};
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