<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.size();
        if(n == 1) {
            return 1;
        }

        int res = 1;
        vector<int> prev(n, 0);
        
        for(int i=n-2;i>=0;i--) {
            vector<int> curr(n, 0);
            for(int j=i;j<n;j++) {
                if(j-i == 0) {
                    res++;
                    curr[j] = 1;
                } else if(j - i == 1){
                    if(s[j] == s[i]) {
                        res++;
                        curr[j] = 1;
                    }
                } else {
                    if(prev[j-1] && s[j] == s[i]) {
                        res++;
                        curr[j] = 1;
                    }
                }
            }
            prev = curr;
        }

        return res;
    }
};
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