<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 t[1001][1001];

    int solve(vector<vector<int>>&pairs,int idx,int lastIdx) {

        if(idx >= pairs.size()) return 0;

        if(lastIdx != -1 && t[idx][lastIdx] != -1) return t[idx][lastIdx];

        int skip = solve(pairs,idx+1,lastIdx);

        int take = 0;

        if(lastIdx == -1 || (pairs[idx][0] > pairs[lastIdx][1])) {

            take = 1 + solve(pairs,idx+1,idx);

        }

        if(lastIdx != -1)
            return t[idx][lastIdx] = max(skip,take);
        
        return max(skip,take);

    }
    int findLongestChain(vector<vector<int>>& pairs) {

        int n = pairs.size();

        vector<int>t(1001,1); 

        int ans = 0;

        // memset(t,-1,sizeof(t));

        sort(begin(pairs),end(pairs));

        // return solve(pairs,0,-1);

        for(int i=0;i<n;i++) {

            for(int j=0;j<i;j++) {

                if(pairs[j][1] < pairs[i][0]) {

                    t[i] = max(t[j]+1,t[i]);

                }

            }

        }

        for(auto elem : t) ans = max(ans,elem);

        return ans;
        
    }
};
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