<p>You are given an array of <code>n</code> pairs <code>pairs</code> where <code>pairs[i] = [left<sub>i</sub>, right<sub>i</sub>]</code> and <code>left<sub>i</sub> &lt; right<sub>i</sub></code>.</p>

<p>A pair <code>p2 = [c, d]</code> <strong>follows</strong> a pair <code>p1 = [a, b]</code> if <code>b &lt; c</code>. A <strong>chain</strong> of pairs can be formed in this fashion.</p>

<p>Return <em>the length longest chain which can be formed</em>.</p>

<p>You do not need to use up all the given intervals. You can select pairs in any order.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> pairs = [[1,2],[2,3],[3,4]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> The longest chain is [1,2] -&gt; [3,4].
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> pairs = [[1,2],[7,8],[4,5]]
<strong>Output:</strong> 3
<strong>Explanation:</strong> The longest chain is [1,2] -&gt; [4,5] -&gt; [7,8].
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == pairs.length</code></li>
	<li><code>1 &lt;= n &lt;= 1000</code></li>
	<li><code>-1000 &lt;= left<sub>i</sub> &lt; right<sub>i</sub> &lt;= 1000</code></li>
</ul>

class Solution {
public:

    int LIS(int idx, int prevIdx, vector<vector<int>>& pairs, vector<vector<int>> &dp, int n) {
        if (idx == n) return 0;

        // Handle the case when prevIdx == -1 separately
        if (dp[idx][prevIdx + 1] != -1) return dp[idx][prevIdx + 1];

        // TAKE
        int take = 0;
        if (prevIdx == -1 || pairs[idx][0] > pairs[prevIdx][1]) {
            take = 1 + LIS(idx + 1, idx, pairs, dp, n);
        }

        // SKIP
        int skip = LIS(idx + 1, prevIdx, pairs, dp, n);

        // Store the result in dp with adjusted indexing (prevIdx + 1)
        dp[idx][prevIdx + 1] = max(skip, take);

        return dp[idx][prevIdx + 1];
    }

    int findLongestChain(vector<vector<int>>& pairs) {
        int n = pairs.size();

        // Sort pairs based on the first element of each pair
        sort(pairs.begin(), pairs.end());

        // DP table where prevIdx can be -1, so we need a table of size (n x (n+1))
        vector<vector<int>> dp(n, vector<int>(n + 1, -1));

        return LIS(0, -1, pairs, dp, n);
    }
};

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