646

646. Maximum Length of Pair Chain (Medium)

You are given an array of n pairs pairs where pairs[i] = [lefti, righti] and lefti < righti.

A pair p2 = [c, d] follows a pair p1 = [a, b] if b < c. A chain of pairs can be formed in this fashion.

Return the length longest chain which can be formed.

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

 

Example 1:

Input: pairs = [[1,2],[2,3],[3,4]]
Output: 2
Explanation: The longest chain is [1,2] -> [3,4].

Example 2:

Input: pairs = [[1,2],[7,8],[4,5]]
Output: 3
Explanation: The longest chain is [1,2] -> [4,5] -> [7,8].

 

Constraints:

• n == pairs.length
• 1 <= n <= 1000
• -1000 <= lefti < righti <= 1000

Companies: Amazon, Google, Flipkart

Related Topics:
Array, Dynamic Programming, Greedy, Sorting

Similar Questions:

• Longest Increasing Subsequence (Medium)
• Non-decreasing Subsequences (Medium)
• Longest Non-decreasing Subarray From Two Arrays (Medium)

Solution 1. DP

First sort the array in ascending order.

Let dp[i] be the length of maximum chain formed using a subsequence of A[0..i] where A[i] must be used.

dp[i] = max(1, max(1 + dp[j] | pair j can go after pair i ))

// OJ: https://leetcode.com/problems/maximum-length-of-pair-chain/
// Author: github.com/lzl124631x
// Time: O(N^2)
// Space: O(N)
class Solution {
public:
    int findLongestChain(vector<vector<int>>& A) {
        sort(begin(A), end(A));
        int N = A.size();
        vector<int> dp(N, 1);
        for (int i = 1; i < N; ++i) {
            for (int j = 0; j < i; ++j) {
                if (A[j][1] >= A[i][0]) continue;
                dp[i] = max(dp[i], 1 + dp[j]);
            }
        }
        return *max_element(begin(dp), end(dp));
    }
};

Solution 3. Interval Scheduling Maximization (ISM)

// OJ: https://leetcode.com/problems/maximum-length-of-pair-chain/
// Author: github.com/lzl124631x
// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
    int findLongestChain(vector<vector<int>>& A) {
        sort(begin(A), end(A), [](auto &a, auto &b) { return a[1] < b[1]; });
        int e = INT_MIN, ans = 0;
        for (auto &v : A) {
            if (e >= v[0]) continue;
            e = v[1];
            ++ans;
        }
        return ans;
    }
};