<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:
    map<pair<int,int>, int> dp;

void rec(string &s, int i, int j){
        //cout<<i<<" "<<j<<endl;
        int n = s.size();

        if(i==j){
            dp[{i,j}] = 1;
            return;
        }

        if(dp.find({i,j})!=dp.end()) return;
        
        // cout<<"1\n";
        if(j>=i+1 and (i+1<n and j>=0) and dp.find({i+1,j})==dp.end())
            rec(s, i+1, j);
        if(j-1>=i and (i<n and j-1>=0) and dp.find({i,j-1})==dp.end())
            rec(s, i, j-1);
        if(j-1>=i+1 and (i+1<n and j-1>=0) and dp.find({i+1,j-1})==dp.end())
            rec(s, i+1, j-1);

        if(s[i]==s[j]){
            if(i+1<=j-1){
                if(dp.find({i+1,j-1})!=dp.end() and dp[{i+1,j-1}]==1)
                    dp[{i,j}]=1;
                else 
                    dp[{i,j}]=0;
            }
            else{
                dp[{i,j}]=1;
            }
        }
        else 
            dp[{i,j}]=0;
        
    }
    int countSubstrings(string s) {
        int n = s.size();
    // rec1(s);
        rec(s, 0, n-1);
        int ans=0;
        
        for(auto x: dp){
            //cout<<x.first.first<<" "<<x.first.second<<" "<<x.second<<endl;
            if(x.second==1)++ans;
        }
        return ans;
    }
};
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