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354. Russian Doll Envelopes

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

Thinking

An envelope nests only if both width and height increase; we want the longest chain. A 2D \(O(n^2)\) LIS fails for \(n\le 10^5\). After sorting by width, LIS on height is the answer; equal widths must not nest.

Sort by width ascending and height descending, then greedy LIS on heights: append if larger than the tail, else binary-replace. Descending heights keep at most one envelope per width.

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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)
                d[idx] = h
        return len(d)

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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;
    }
}

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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();
    }
};

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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
}

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