<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 s, int start, int end)
{
    while(start <end)
    {
        if(s[start] != s[end])
        {
            return false;
        }
        start++;
        end--;

    }
    return true;
}
    int countSubstrings(string s) {
        // go through each element 
        // then for each element go through it's substring 
        // then check if substring is plaindrome or not 

        int n = s.length();
        int count = 0;

        for(int i =0; i <n; i++)
        {
            for(int j =i; j < n; j++)
            {
                if(isPalindrome(s, i, j))
                {
                    count++;
                }
            }
        }
         return count;
        }
       
};
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