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646. Maximum Length of Pair Chain

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

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

  • class Solution {
public:
    int findLongestChain(vector<vector<int>>& pairs) {
        sort(pairs.begin(), pairs.end(), [](vector<int>& a, vector<int> b) {
            return a[1] < b[1];
        });
        int ans = 0, cur = INT_MIN;
        for (auto& p : pairs) {
            if (cur < p[0]) {
                cur = p[1];
                ++ans;
            }
        }
        return ans;
    }
};

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


  • func findLongestChain(pairs [][]int) int {
	sort.Slice(pairs, func(i, j int) bool {
		return pairs[i][1] < pairs[j][1]
	})
	ans, cur := 0, math.MinInt32
	for _, p := range pairs {
		if cur < p[0] {
			cur = p[1]
			ans++
		}
	}
	return ans
}

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


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

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