README_EN.md

comments	difficulty	edit_url	tags
true	Hard	https://github.com/doocs/leetcode/edit/main/solution/0300-0399/0354.Russian%20Doll%20Envelopes/README_EN.md	Array Binary Search Dynamic Programming Sorting

354. Russian Doll Envelopes

中文文档

Description

You are given a 2D array of integers envelopes where envelopes[i] = [wi, hi] represents the width and the height of an envelope.

One envelope can fit into another if and only if both the width and height of one envelope are greater than the other envelope's width and height.

Return the maximum number of envelopes you can Russian doll (i.e., put one inside the other).

Note: You cannot rotate an envelope.

 

Example 1:

Input: envelopes = [[5,4],[6,4],[6,7],[2,3]]
Output: 3
Explanation: The maximum number of envelopes you can Russian doll is 3 ([2,3] => [5,4] => [6,7]).

Example 2:

Input: envelopes = [[1,1],[1,1],[1,1]]
Output: 1

 

Constraints:

• 1 <= envelopes.length <= 105
• envelopes[i].length == 2
• 1 <= wi, hi <= 105

Solutions

Solution 1

Python3

class Solution:
    def maxEnvelopes(self, envelopes: List[List[int]]) -> int:
        envelopes.sort(key=lambda x: (x[0], -x[1]))
        d = [envelopes[0][1]]
        for _, h in envelopes[1:]:
            if h > d[-1]:
                d.append(h)
            else:
                idx = bisect_left(d, h)
                if idx == len(d):
                    idx = 0
                d[idx] = h
        return len(d)

Java

class Solution {
    public int maxEnvelopes(int[][] envelopes) {
        Arrays.sort(envelopes, (a, b) -> { return a[0] == b[0] ? b[1] - a[1] : a[0] - b[0]; });
        int n = envelopes.length;
        int[] d = new int[n + 1];
        d[1] = envelopes[0][1];
        int size = 1;
        for (int i = 1; i < n; ++i) {
            int x = envelopes[i][1];
            if (x > d[size]) {
                d[++size] = x;
            } else {
                int left = 1, right = size;
                while (left < right) {
                    int mid = (left + right) >> 1;
                    if (d[mid] >= x) {
                        right = mid;
                    } else {
                        left = mid + 1;
                    }
                }
                int p = d[left] >= x ? left : 1;
                d[p] = x;
            }
        }
        return size;
    }
}

C++

class Solution {
public:
    int maxEnvelopes(vector<vector<int>>& envelopes) {
        sort(envelopes.begin(), envelopes.end(), [](const auto& e1, const auto& e2) {
            return e1[0] < e2[0] || (e1[0] == e2[0] && e1[1] > e2[1]);
        });
        int n = envelopes.size();
        vector<int> d{envelopes[0][1]};
        for (int i = 1; i < n; ++i) {
            int x = envelopes[i][1];
            if (x > d[d.size() - 1])
                d.push_back(x);
            else {
                int idx = lower_bound(d.begin(), d.end(), x) - d.begin();
                if (idx == d.size()) idx = 0;
                d[idx] = x;
            }
        }
        return d.size();
    }
};

Go

func maxEnvelopes(envelopes [][]int) int {
	sort.Slice(envelopes, func(i, j int) bool {
		if envelopes[i][0] != envelopes[j][0] {
			return envelopes[i][0] < envelopes[j][0]
		}
		return envelopes[j][1] < envelopes[i][1]
	})
	n := len(envelopes)
	d := make([]int, n+1)
	d[1] = envelopes[0][1]
	size := 1
	for _, e := range envelopes[1:] {
		x := e[1]
		if x > d[size] {
			size++
			d[size] = x
		} else {
			left, right := 1, size
			for left < right {
				mid := (left + right) >> 1
				if d[mid] >= x {
					right = mid
				} else {
					left = mid + 1
				}
			}
			if d[left] < x {
				left = 1
			}
			d[left] = x
		}
	}
	return size
}