646. Maximum Length of Pair Chain

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Description

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

Solutions

Solution 1: Sorting + Greedy

We sort all pairs in ascending order by the second number, and use a variable $\textit{pre}$ to maintain the maximum value of the second number of the selected pairs.

We traverse the sorted pairs. If the first number of the current pair is greater than $\textit{pre}$, we can greedily select the current pair, increment the answer by one, and update $\textit{pre}$ to the second number of the current pair.

After the traversal, we return the answer.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$. Here, $n$ is the number of pairs.

Python3

class Solution:
    def findLongestChain(self, pairs: List[List[int]]) -> int:
        pairs.sort(key=lambda x: x[1])
        ans, pre = 0, -inf
        for a, b in pairs:
            if pre < a:
                ans += 1
                pre = b
        return ans

Java

class Solution {
    public int findLongestChain(int[][] pairs) {
        Arrays.sort(pairs, (a, b) -> Integer.compare(a[1], b[1]));
        int ans = 0, pre = Integer.MIN_VALUE;
        for (var p : pairs) {
            if (pre < p[0]) {
                ++ans;
                pre = p[1];
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int findLongestChain(vector<vector<int>>& pairs) {
        ranges::sort(pairs, {}, [](const auto& p) { return p[1]; });
        int ans = 0, pre = INT_MIN;
        for (const auto& p : pairs) {
            if (pre < p[0]) {
                pre = p[1];
                ++ans;
            }
        }
        return ans;
    }
};

Go

func findLongestChain(pairs [][]int) (ans int) {
	sort.Slice(pairs, func(i, j int) bool { return pairs[i][1] < pairs[j][1] })
	pre := math.MinInt
	for _, p := range pairs {
		if pre < p[0] {
			ans++
			pre = p[1]
		}
	}
	return
}

TypeScript

function findLongestChain(pairs: number[][]): number {
    pairs.sort((a, b) => a[1] - b[1]);
    let [ans, pre] = [0, -Infinity];
    for (const [a, b] of pairs) {
        if (pre < a) {
            ++ans;
            pre = b;
        }
    }
    return ans;
}

Rust

impl Solution {
    pub fn find_longest_chain(mut pairs: Vec<Vec<i32>>) -> i32 {
        pairs.sort_by_key(|pair| pair[1]);
        let mut ans = 0;
        let mut pre = i32::MIN;
        for pair in pairs {
            let (a, b) = (pair[0], pair[1]);
            if pre < a {
                ans += 1;
                pre = b;
            }
        }
        ans
    }
}