<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 countSubstrings(string s) {
        vector<vector<int>>dp(s.length()+1,vector<int>(s.length()+1,0));
        int ans = s.length();
        int x=1,y=2;
        for(int i=0;i<=s.length();i++){
            dp[0][i] = 0;
            dp[i][0] = 0;
        }
        for(int i=1;i<=s.length();i++){
            dp[i][i] = 1;
        }
        
        for(int i=0;i<s.length()-2+1;i++){
            if(s[y-1] == s[x-1])dp[y][x] = 1;
            y++;
            x++;
        }
        x=1;
        y=2;

        int windowSize = 2;

        string res="";
        while(windowSize<=s.length()){
            y=1;
            x=windowSize;
            res += s[y-1];
            string res2 = res;
            for(int i=0;i<s.length()-windowSize+1;i++){
                res+= s[x-1];
                if(s[y-1] == s[x-1]){
                    if(dp[y+1][x-1] == 0){
                        dp[y][x] = 0;
                    } else {
                        dp[y][x] = 2 + dp[y+1][x-1];
                    }
                }
                if(dp[y][x]){
                    ans++;
                }
                y++;
                x++;
            }
            windowSize++;
        }
        

        return ans;
    }
};
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