<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(int i,int j,string s){
        while(i<=j){
            if(s[i] != s[j]) return false;
            i++, j--;
        }
        return true;
    }
    int solve(int i,string s,vector<int>&dp){
        if(i>=s.size()){
            return 0;
        }
        if(dp[i] != -1) return dp[i];
        int cnt=solve(i+1,s,dp);
        for(int j=i;j<s.size();j++){
            if(isPalindrome(i,j,s)){
                cnt++;
            }
        }
        return dp[i]=cnt;
    }
    int countSubstrings(string s) {
        // vector<int>dp(s.size()+1,-1);
       // return solve(0,s,dp);
        
        //Tabulation
        //aaa   
        vector<int>dp(s.size()+1,0);
        int n=s.size();
        for(int i=n-1;i>=0;i--){
            int cnt=0;
            for(int j=i;j<s.size();j++){
                if(isPalindrome(i,j,s)){
                    cnt++;
                }
            }
            dp[i]=cnt;
        }
        int sum=0;
        for(int i=0;i<dp.size();i++){
            sum+=dp[i];
        }
        return sum;
    }
};
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