<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 count = 0;
    
    // Finds whether there exists any pailndrome strings between the stated limit i.e 'i' to 's.length()'
    void findSubstrings(string s, int start, int n) {
        for(int end = start; end<n; end++) {
            if(isPalindrome(start, end, s)) {
                count++;
            }
        }
    }
    
    // Checks for a string to be a palindrome
    bool isPalindrome(int start, int end, string s) {
        while(start<=end) {
            if(s[start++]!=s[end--]) return false;
        }
        
        return true;
        
    }
    
    // Driver Function
    int countSubstrings(string s) {
        // Defines limits for checking a string is palindrome or not
        for(int i=0; i<s.length(); i++) {
            findSubstrings(s, i, s.length());
        }
        
        return count;
    }
};
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