<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 check(string& s,unordered_map<string,int>& mp){
        int i=0,l=s.length();
        if(s==""){
            return 0;
        }
        else if(mp[s]>0){
            return mp[s]-1;
        }

        for(i=0;i<l/2;i++){
            if(s[i]==s[l-i-1]){
                //continue...
            }
            else{
                mp[s]=1;
                return 0;
            }
        }

        mp[s]=2;
        return 1;
    }

    int countSubstrings(string s) {
        int ans=0;
        unordered_map<string,int>mp;
        for(int i=0;i<s.length();i++){
            string t;
            for(int j=i;j<s.length();j++){
                t.push_back(s[j]);
                ans+=check(t,mp);
            }
        }

        return ans;
    }
};
164559