<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) {
        int n = s.length();
        int res = n;
        queue<pair<int, int>> p_sub;
        for (int i = 0; i < n; ++i) {
            p_sub.push(make_pair(i, i));
            if (i < n - 1 && s[i] == s[i + 1]) {
                res++;
                p_sub.push(make_pair(i, i + 1));
            }
        }
        while (!p_sub.empty()) {
            auto subs = p_sub.front();
            p_sub.pop();
            if (subs.first > 0 && subs.second < n - 1 && s[subs.first - 1] == s[subs.second + 1]) {
                res++;
                p_sub.push(make_pair(subs.first - 1, subs.second + 1));
            }
        }

        return res;
    }
};
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