<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 isPalindrom(string s, int start, int end)
    {
        int i = start, j = end;

        while(i <= j)
        {
            if(s[i] != s[j]) return false;

            i++, j--;
        }
        return true;
    }

    int findPossiblePalindrome(string s, int window)
    {
        int i = 0, j = window - 1, count = 0;

        while(j < s.size())
        {
            if(isPalindrom(s, i, j)) count++;

            i++, j++;
        }

        return count;
    }


    int countSubstrings(string s) {
        int ans = s.size();

        for(int i = 2; i <= s.size(); i++)
        {
            ans += findPossiblePalindrome(s, i);
        }

        return ans;
    }
};
364453