<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:
    
    bool isPalindrome(string& temp,unordered_map<string,int>& dp) {
      int s=0,e=temp.size()-1;
      if(dp.find(temp)!=dp.end()){
        if(dp[temp]==-1) return false;
        return true;
      }
      while(s<e){
        if(temp[s]!=temp[e]){
            dp[temp]=-1;
            return false;
        }
        s++;
        e--;
      }
      dp[temp]=1;
      return true;
    }

   
    void countPalindromicSubstrings(int index, const string& s, int& totalCount,unordered_map<string,int>& dp ) {
        if (index >= s.size()) return;
        string temp="";
        for (int i = index; i< s.size(); i++) {
            temp+=s[i];
            if (isPalindrome(temp, dp)) {
                totalCount++;
            }
        }
        countPalindromicSubstrings(index + 1, s, totalCount, dp);
    }
    
    int countSubstrings(const string& s) {
        int totalCount = 0;
        int n = s.size();
        unordered_map<string,int> dp;
        countPalindromicSubstrings(0, s, totalCount, dp);
        return totalCount;
    }
};

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