<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 dp[1005][4000];
    int func(int i, int prev, vector<vector<int>>& pairs) {
        if(i == pairs.size()) return 0;
        if(dp[i][prev] != -1) return dp[i][prev];
        int ans = func(i + 1, prev, pairs);
        if(prev < pairs[i][0]) {
            ans = max(ans, 1 + func(i + 1, pairs[i][1], pairs));
        }
        return dp[i][prev] = ans;
    }
    int findLongestChain(vector<vector<int>>& pairs) {
        memset(dp, -1, sizeof(dp));
        sort(pairs.begin(), pairs.end());
        for(int i = 0; i < pairs.size(); ++i) {
            pairs[i][1] += 1001;
            pairs[i][0] += 1001;
        }
        return func(0, 0, pairs);
    }
};
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