<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>

int paldrome(string s, int sz, int stt, int ed, int og){
    if(og==0) return 0;
    if(stt>0 && ed<sz-1){
        if(s[stt-1]==s[ed+1]) return paldrome(s, sz, stt-1, ed+1, og+1);
        else return og;
    }
    else return og;
}
class Solution {
public:
    int countSubstrings(string s) {
        int sz = s.size();
        int ans = 0;
        vector<int> dp_odd(sz, 1);
        for(int i=0; i<sz; i++){
            dp_odd[i] = paldrome(s, sz, i, i, 1);
            ans += dp_odd[i];
        }
        vector<int> dp_even(sz-1, 0);
        for(int i=0; i<sz-1;i++){
            dp_even[i] = (s[i]==s[i+1]);
            dp_even[i] = paldrome(s, sz, i, i+1, dp_even[i]);
            ans += dp_even[i];
        } 

        return ans;
    }
};
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