<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 check(string s)
    {
        int l = 0, r = s.size() - 1;
        while (l <= r)
        {
            if (s[l++] != s[r--]) return false;
        }
        return true;
    }

    int countSubstrings(string s) {
        map<char, vector<int>> mp;
        for (int i = 0; i < s.size(); ++i)
        {
            mp[s[i]].push_back(i);
        }

        int res = s.size();
        for (auto it : mp)
        {
            int n = it.second.size();
            for (int i = 0; i < n; ++i)
            {
                for (int j = i+1; j < n; ++j)
                {
                    if (check(s.substr(it.second[i], it.second[j] - it.second[i]+1))) ++res;
                }
            }
        }

        return res;
    }
};
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