646. Maximum Length of Pair Chain

You are given n pairs of numbers. In every pair, the first number is always smaller than the second number.

Now, we define a pair (c, d) can follow another pair (a, b) if and only if b < c. Chain of pairs can be formed in this fashion.

Given a set of pairs, find the length longest chain which can be formed. You needn't use up all the given pairs. You can select pairs in any order.

Example 1:

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

Note:

1. The number of given pairs will be in the range [1, 1000].

// Dynamic Programming
int findLongestChain(vector<vector<int>>& pairs) { // time: O(n^2); space: O(n)
    if (pairs.empty()) return 0;
    int n = pairs.size();
    sort(pairs.begin(), pairs.end());
    vector<int> len(n, 1); // len[i]: longest length of chain considering pairs[0...i]
    for (int i = 1; i < n; ++i) {
        for (int j = 0; j < i; ++j) {
            if (pairs[j][1] < pairs[i][0]) 
                len[i] = max(len[i], len[j] + 1); 
        }
    }
    return len.back();
}

Greedy method只需要將input array根據end value來排序就可以依序找出符合的chain。

// Greedy
int findLongestChain(vector<vector<int>>& pairs) { // time: O(nlogn); space: O(1)
    if (pairs.empty()) return 0;
    sort(pairs.begin(), pairs.end(), [](vector<int>& a, vector<int>& b) {
        return a[1] < b[1] || (a[1] == b[1] && a[0] < b[0]);
    }); // sort based on the end values
    int res = 1, tail = pairs[0][1];
    for (int i = 1; i < pairs.size(); ++i) {
        if (pairs[i][0] > tail) {
            tail = pairs[i][1];
            ++res;
        }
    }
    return res;
}