<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 sz = s.size();
        vector<int> prev1(sz);
        iota(prev1.begin(), prev1.end(), 0);
        vector<int> prev2(prev1.begin(), prev1.end());
        bool p1 = true;
        bool p2 = true;

        
        int count = sz;
        for(int i = 2; i <= sz; ++i) {
            vector<int> cur;
            cur.reserve(sz);
            //cout << i << endl;
            bool found = false;
            for(auto n : prev1) {
                int l = n - 1;
                int r = n + i - 2;
                
                if(l < 0 || r >= sz) {
                    continue;
                }
                //cout << l << " " << r << endl;
                if(s[l] == s[r]) {
                    //cout << "in" << endl;
                    cur.push_back(l);
                    ++count;
                    found = true;
                }
            }
            
            prev1 = prev2;
            prev2 = cur;
            p1 = p2;
            p2 = found;
            if(!p1 && !p2) {
                break;
            }
        }
        return count;
    }
};
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