LeetCode #354 — HARD

Russian Doll Envelopes

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given a 2D array of integers envelopes where envelopes[i] = [w_i, h_i] 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 <= w_i, h_i <= 10⁵

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Binary Search · Dynamic Programming

Example 1

[[5,4],[6,4],[6,7],[2,3]]

Example 2

[[1,1],[1,1],[1,1]]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #354: Russian Doll Envelopes
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;
    }
}

// Accepted solution for LeetCode #354: Russian Doll Envelopes
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
}

# Accepted solution for LeetCode #354: Russian Doll Envelopes
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)

// Accepted solution for LeetCode #354: Russian Doll Envelopes
struct Solution;

use std::cmp::Reverse;

impl Solution {
    fn max_envelopes(mut envelopes: Vec<Vec<i32>>) -> i32 {
        let n = envelopes.len();
        envelopes.sort_unstable_by_key(|v| (v[0], Reverse(v[1])));
        let mut dp = vec![];
        for i in 0..n {
            let height = envelopes[i][1];
            if let Err(p) = dp.binary_search(&height) {
                if p == dp.len() {
                    dp.push(height);
                } else {
                    dp[p] = height;
                }
            }
        }
        dp.len() as i32
    }
}

#[test]
fn test() {
    let envelopes = vec_vec_i32![[5, 4], [6, 4], [6, 7], [2, 3]];
    let res = 3;
    assert_eq!(Solution::max_envelopes(envelopes), res);
    let envelopes = vec_vec_i32![[4, 5], [4, 6], [6, 7], [2, 3], [1, 1]];
    let res = 4;
    assert_eq!(Solution::max_envelopes(envelopes), res);
}

// Accepted solution for LeetCode #354: Russian Doll Envelopes
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #354: Russian Doll Envelopes
// 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;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.