<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) {
        unordered_map<int, unordered_map<int, bool>>dp;
        int len = s.size();
        int ans = 0;
        for(int i = 0; i < len; i++){
            dp[i][1] = true;
            dp[i][0] = true;
            if(dp[i][1]) ans++;
        }
        cout<<"ans = "<<ans<<endl;
        for(int l = 2; l <= len; l++){
            for(int i = 0; i < len; i++){
                //cout<<"l = "<<l<<" s = "<<s[i+1]<<" "<<s[i+l-1]<<" dp "<<dp[i+1][l - 2]<<" l - 2:"<<l-2<<endl;
                if(i + l - 1 >= len) break;
                dp[i][l] = ( s[i] == s[i + l - 1] && dp[i+1][l - 2]);
                if(dp[i][l]) ans++;
                //cout<<i<<" l = "<<l<<" dp = "<<dp[i][l]<<endl;
            }
            //cout<<"ans = "<<ans<<endl;
        }
        return ans;
    }
};
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