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

Solution 1

Python Code
1
2
3
4
5
6
7
8
9

class Solution:
    def findLongestChain(self, pairs: List[List[int]]) -> int:
        pairs.sort()
        dp = [1] * len(pairs)
        for i, (c, _) in enumerate(pairs):
            for j, (_, b) in enumerate(pairs[:i]):
                if b < c:
                    dp[i] = max(dp[i], dp[j] + 1)
        return max(dp)

Java Code
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20

class Solution {
    public int findLongestChain(int[][] pairs) {
        Arrays.sort(pairs, Comparator.comparingInt(a -> a[0]));
        int n = pairs.length;
        int[] dp = new int[n];
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            dp[i] = 1;
            int c = pairs[i][0];
            for (int j = 0; j < i; ++j) {
                int b = pairs[j][1];
                if (b < c) {
                    dp[i] = Math.max(dp[i], dp[j] + 1);
                }
            }
            ans = Math.max(ans, dp[i]);
        }
        return ans;
    }
}

C++ Code
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16

class Solution {
public:
    int findLongestChain(vector<vector<int>>& pairs) {
        sort(pairs.begin(), pairs.end());
        int n = pairs.size();
        vector<int> dp(n, 1);
        for (int i = 0; i < n; ++i) {
            int c = pairs[i][0];
            for (int j = 0; j < i; ++j) {
                int b = pairs[j][1];
                if (b < c) dp[i] = max(dp[i], dp[j] + 1);
            }
        }
        return *max_element(dp.begin(), dp.end());
    }
};

Go Code
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20

func findLongestChain(pairs [][]int) int {
	sort.Slice(pairs, func(i, j int) bool {
		return pairs[i][0] < pairs[j][0]
	})
	n := len(pairs)
	dp := make([]int, n)
	ans := 0
	for i := range pairs {
		dp[i] = 1
		c := pairs[i][0]
		for j := range pairs[:i] {
			b := pairs[j][1]
			if b < c {
				dp[i] = max(dp[i], dp[j]+1)
			}
		}
		ans = max(ans, dp[i])
	}
	return ans
}

TypeScript Code
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13

function findLongestChain(pairs: number[][]): number {
    pairs.sort((a, b) => a[0] - b[0]);
    const n = pairs.length;
    const dp = new Array(n).fill(1);
    for (let i = 0; i < n; i++) {
        for (let j = 0; j < i; j++) {
            if (pairs[i][0] > pairs[j][1]) {
                dp[i] = Math.max(dp[i], dp[j] + 1);
            }
        }
    }
    return dp[n - 1];
}

Rust Code
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15

impl Solution {
    pub fn find_longest_chain(mut pairs: Vec<Vec<i32>>) -> i32 {
        pairs.sort_by(|a, b| a[0].cmp(&b[0]));
        let n = pairs.len();
        let mut dp = vec![1; n];
        for i in 0..n {
            for j in 0..i {
                if pairs[i][0] > pairs[j][1] {
                    dp[i] = dp[i].max(dp[j] + 1);
                }
            }
        }
        dp[n - 1]
    }
}

Solution 2

Python Code
1
2
3
4
5
6
7
8

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

Java Code
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14

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;
    }
}

C++ Code
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16

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;
    }
};

Go Code
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13

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
}

TypeScript Code
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12

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;
}

Rust Code
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16

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
    }
}