<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 countSubstrings(string s) {
        if (s.size() == 1) {
            return 1;
        }

        // every letter is a palindromic substring by itself
        // check 2 letter polyndromes
        // check 3 letter polyndromes...
        // etc

        int count = 0;
        std::vector<std::pair<int, int>> prevs;
        prevs.push_back({});
        // fill up 1 and 2 letter polyndromes.
        for (size_t i = 0; i < s.size(); ++i) {
            prevs.push_back({i, i+1});
            count++;
            if (i > 0 && s[i-1] == s[i]) {
                prevs.push_back({i-1, i+1});
                count++;
            }
        }

        while (!prevs.empty()) {
            std::vector<std::pair<int, int>> curr;
            for (const auto& p: prevs) {
                if (p.first == 0 || p.second >= s.size()) continue;
                if (s[p.first - 1] != s[p.second]) continue;
                count++;
                curr.push_back({p.first - 1, p.second + 1});
            }
            prevs = curr;
        }
        return count;
    }
};
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