<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 checkPalindrome(string str,int i, int j){
    while(i<=j){
        if(str[i]!=str[j])
        return false;

        i++;j--;
    }

    return true;
}


int palindromicSubStrings(string str, int i, int j){
    int count=0;

    while(i>=0 && j<str.length()){
        if(checkPalindrome(str,i,j)){
            i--;j++;
            count++;
        }

        else
        break;
    }

    return count;
}

    int countSubstrings(string s) {
        int count=0;

        for(int i=0;i<s.length();i++){
        //odd 
        count+=palindromicSubStrings(s,i,i);

        //even
        if(i+1<s.length())
        count+=palindromicSubStrings(s,i,i+1);
    }

    return count; 

    }
};
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