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Solution: Russian Doll Envelopes

seanpgallivan profile image seanpgallivan ・4 min read

This is part of a series of Leetcode solution explanations (index). If you liked this solution or found it useful, please like this post and/or upvote my solution post on Leetcode's forums.


Leetcode Problem #354 (Hard): Russian Doll Envelopes


Description:


(Jump to: Solution Idea || Code: JavaScript | Python | Java | C++)

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 is greater than the width and height of the other envelope.

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

Note: You cannot rotate an envelope.


Examples:

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 <= 5000
  • envelopes[i].length == 2
  • 1 <= wi, hi <= 10^4

Idea:


(Jump to: Problem Description || Code: JavaScript | Python | Java | C++)

The naive approach here would be to try every single permutation of our envelope array (E), but that would be a time complexity of O(N!) which is frankly an incomprehensible number when N goes up to 5000.

As the naive approach would involve repeating many of the same individual comparisons over and over again, we can quickly see that a dynamic programming (DP) solution would be beneficial.

In order for a DP solution to be effective, however, we'd need to find a way to progress from the easiest subsolution and build from there for each successively more complex subsolution. The best way to do this would be to sort E first by width (E[i][0]), and then by height (E[i][1]).

Then we could start with the smallest envelope and work our way up, storing in our DP array (dp) the result of how many smaller envelopes it is possible to fit in the corresponding envelope. That way we could simplify each iteration to checking to see which of the entries in dp corresponding to smaller envelopes is the largest. This would drop the time complexity to O(N^2), which is a definite improvement.

But it should also be apparent that if we were to define a subsequence of E that was the ideal nesting order of envelopes for the solution, then that array would be strictly increasing in both width and height.

If we've already sorted E primarily by width, we should then be able to consider a corresponding array of just the heights and realize that the solution would be defined as the longest increasing subsequence of that.

The only difficulty would be for consecutive envelopes with the same sorted width. To avoid that, we can simply make sure that our sort function sorts height in descending order so that the first envelope encountered for any given width would be the largest one.

At the end of the longest increasing subsequence algorithm, the length of dp is equal to the length of the subsequence. Due to the sort function and the binary searches required for the algorithm, the time complexity now shrinks to O(N log N).


Implementation:

Python has a built-in binary search function, bisect().

Java has a built-in binary search function as well (Arrays.binarySearch()), but in order to use the more performant int[] rather than a List< Integer >, we'll need to specify a max length for dp and then keep track of the current index of the longest subsequence separately in ans.

C++ has a built-in binary search function, lower_bound().


Javascript Code:


(Jump to: Problem Description || Solution Idea)

var maxEnvelopes = function(E) {
    E.sort((a,b) => a[0] === b[0] ? b[1] - a[1] : a[0] - b[0])
    let len = E.length, dp = []
    for (let i = 0; i < len; i++) {
        let height = E[i][1], left = 0, right = dp.length   
        while (left < right) {
            let mid = (left + right) >> 1
            if (dp[mid] < height) left = mid + 1
            else right = mid
        }
        dp[left] = height
    }
    return dp.length
};
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Python Code:


(Jump to: Problem Description || Solution Idea)

class Solution:
    def maxEnvelopes(self, E: List[List[int]]) -> int:
        E.sort(key=lambda x: (x[0], -x[1]))
        dp = []
        for _,height in E:
            left = bisect_left(dp, height)
            if left == len(dp): dp.append(height)
            else: dp[left] = height
        return len(dp)
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Java Code:


(Jump to: Problem Description || Solution Idea)

class Solution {
    public int maxEnvelopes(int[][] E) {
        Arrays.sort(E, (a,b) -> a[0] == b[0] ? b[1] - a[1] : a[0] - b[0]);
        int[] dp = new int[E.length];
        int ans = 0;
        for (int[] env : E) {
            int height = env[1];
            int left = Arrays.binarySearch(dp, 0, ans, height);
            if (left < 0) left = -left - 1;
            if (left == ans) ans++;
            dp[left] = height;
        }
        return ans;
    }
}
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C++ Code:


(Jump to: Problem Description || Solution Idea)

class Solution {
public:
    int maxEnvelopes(vector<vector<int>>& E) {
        sort(E.begin(), E.end(), [](vector<int>& a, vector<int>& b) 
             -> bool {return a[0] == b[0] ? b[1] < a[1] : a[0] < b[0];});
        vector<int> dp;
        for (auto& env : E) {
            int height = env[1];
            int left = lower_bound(dp.begin(), dp.end(), height) - dp.begin();
            if (left == dp.size()) dp.push_back(height);
            dp[left] = height;
        }
        return dp.size();
    }
};
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