<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 {
private:
    bool isPali(string s, int start, int end) {
        while (start <= end) {
            if (s[start] != s[end]) {
                return false;
            }
            start++;
            end--;
        }
        return true;
    }
public:
    int countSubstrings(string s) {
        vector<int> res(s.size());
        for (int i = 0; i < s.size(); i++) {
            int num = 0;
            for (int j = i; j >=0; j--) {
                if (isPali(s, j, i)) {
                    num++;
                }
            }
            int prev = (i > 0) ? res[i - 1] : 0;
            res[i] = prev + num;
        }
        return res[res.size() - 1];
    }
};
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