<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 size = s.size();
        vector<vector<int>> count(size, vector<int>(size, 0));
        vector<vector<bool>> dp(size, vector<bool>(size, false));

        for (int i = 0 ; i < size ; ++i) {
            dp[i][i] = true;
            count[i][i] = 1;
        }

        for (int i = 0 ; i < size ; ++i) {  // i: end of substring
            for (int j = i - 1 ; j >= 0 ; --j) {    // j: start of substring
                // [j,i]區間是palindrome有兩種可能
                // 1. [j]==[i] 且 距離只差1(substring長度=2)
                // 2. [j]==[i] 且 [j+1,i-1]也是palindrome
                if (s[j] == s[i] && (i == (j + 1) || dp[j + 1][i - 1])) {
                    dp[j][i] = true;
                    count[j][i] = count[j + 1][i] + count[j][i - 1] - count[j + 1][i - 1] + 1;
                } else {
                    dp[j][i] = false;
                    count[j][i] = count[j + 1][i] + count[j][i - 1] - count[j + 1][i - 1];
                }
            }
        }

        return count[0][size - 1];
    }
};
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