<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:
    unordered_map<int, unordered_map<int, int>> dp;
    int go(string &s, int i, int j){
        if(i>=j) return 1;
        if(dp[i].count(j)) return dp[i][j];

        bool ans=false;

        if(s[i]==s[j]){
            ans=true && go(s, i+1, j-1);
        } else{
            ans=false;
        }
        go(s, i+1, j);
        go(s, i, j-1);
        return dp[i][j] = ans;
    }
    int countSubstrings(string s) {
        go(s, 0, s.size()-1);
        int ans=0;
        for(auto &x: dp){
            for(auto &y: x.second){
                ans+=y.second;
            }
        }
        return ans + s.size();
    }
};
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