646. Maximum Length of Pair Chain

Posted 2020-09-29 蓝色地中海

Problem statement

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].

Solution one: Greedy algorithm

The first solution is greedy algorithm. 

• Sort the pairs by the increasing order of end value.
• Keep a value of minimum end value.
• Update the minimum end value if current pair and next pair are overlapped.
• count  number + 1 if the start value of next pairis greater than minimum end and update the minimum end.

Time complexity is O(n). space complexity is O(1).

class Solution {
public:
    int findLongestChain(vector<vector<int>>& pairs) {
        sort(pairs.begin(), pairs.end(), [](vector<int> l, vector<int> r) 
             { if(l[0] == r[0]){ return l[1] < r[1];}
                 return l[0] < r[0];});
        int len = 1;
        int min_end = pairs[0][1];
        for(int i = 1; i < pairs.size(); i++){
            if(min_end >= pairs[i][1]){
                min_end = pairs[i][1];
            } else if(min_end < pairs[i][0]){
                len++;
                min_end = pairs[i][1];
            }
        }
        return len;
    }
};

Another concise version of solution one.

• Since all pairs are sorted, we just need to find the first non-overlapped pair with current pair. 
• Count number + 1 and update the current pair.

Time complexity is O(n). Space complexity is O(1).

class Solution {
public:
    int findLongestChain(vector<vector<int>>& pairs) {
        // sort the pair by increasing order according to the end
        sort(pairs.begin(), pairs.end(), [](vector<int> l, vector<int> r) { return l[1] < r[1]; });
        int len = 1;
        for(int i = 0, j = 1; j < pairs.size(); j++){
            if(pairs[i][1] < pairs[j][0]){
                len++;
                i = j;
            }
        }
        return len;
    }
};

Solution two: DP.

It is similar with 300. Longest Increasing Subsequence. 

For each element, dp[i], we need loop back from all previous elements to find the solution.

Meanwhile, update the dp[i] and find the maximum length.

Time complexity is O(n * n), space complexity is O(n). Obviously dp solution for this problem is not optimal, but it is still worth to learn how to deduce the dp formula.

class Solution {
public:
    int findLongestChain(vector<vector<int>>& pairs) {
        // sort the pair by increasing order according to the end
        sort(pairs.begin(), pairs.end(), [](vector<int> l, vector<int> r) { return l[1] < r[1]; });
        int size = pairs.size();
        int dp[size] = {1};
        int max_len = 1;
        for(int i = 1; i < pairs.size(); i++){
            for(int j = 0; j < i; j++){
                if(pairs[j][1] < pairs[i][0]){
                    dp[i] = max(dp[i], dp[j] + 1);
                }
            }
            max_len = max(max_len, dp[i]);
        }
        return max_len;
    }
};