<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:
    map<pair<int, int>, bool>mmap;

    bool isPalin(string &s, int i, int j) {
        if (j-i < 1) {
            return true;
        }

        auto p = make_pair(i, j);
        if (mmap.find(p) != mmap.end()) {
            return mmap[p];
        }

        if (s[i] != s[j]) {
            mmap[p] = false;
            return false;
        }
        
        auto palin = isPalin(s, i+1, j-1);
        mmap[p] = palin;
        return palin;
    }
public:
    int countSubstrings(string s) {
        int sLen = s.length(), resp = 0;
        
        for (int i = 0; i < sLen; i++) {
            mmap[make_pair(i, i)] = true;

            resp++;
        }
        
        for (int i = 0; i < sLen - 1; i++) {
            for (int j = i+1; j < sLen; j++) {
                if (isPalin(s, i, j)) {
                    resp++;
                }
            }
        }
        
        return resp;
    }
};
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