What is the time complexity of Russian Doll Envelopes?

The optimal solution has a time complexity of O(nlogn).

What data structures are used to solve Russian Doll Envelopes?

This problem uses Array, Binary Search, Dynamic Programming, Sorting. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Russian Doll Envelopes?

Insight 1: Sorting the envelopes is crucial. Sort by width in ascending order. When widths are the same, sort by height in descending order. This ensures that envelopes with equal widths cannot be nested inside each other. Insight 2: After sorting, the problem transforms into finding the Longest Increasing Subsequence (LIS) of the heights. The descending sort on heights for same widths is essential to avoid incorrectly including envelopes with the same width in the LIS.

Which companies ask LeetCode 354. Russian Doll Envelopes in interviews?

This problem is frequently asked at Atlassian, Intuit, Sprinklr.

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Russian Doll Envelopes - LeetCode 354 Solution

Russian Doll Envelopes - Complete Solution Guide

Russian Doll Envelopes is LeetCode problem 354, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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 maxim

Detailed Explanation

The problem asks us to find the maximum number of Russian doll envelopes, where one envelope can fit inside another if and only if both its width and height are strictly greater than the inner envelope's width and height. We are given a 2D array 'envelopes' where each element 'envelopes[i]' is a pair '[width_i, height_i]' representing the dimensions of an envelope. We cannot rotate envelopes. The goal is to return the maximum number of envelopes that can be nested within each other.

Solution Approach

The solution first sorts the envelopes based on width and height as described in the key insights. Then, it iterates through the sorted envelopes and extracts the height of each envelope. For each height, it performs a binary search in the 'sub' array (which initially is empty). The 'sub' array stores the smallest ending element of each increasing subsequence length found so far. If the height is greater than all elements in 'sub', it extends the LIS by adding the height to 'sub'. Otherwise, it replaces the smallest element in 'sub' that is greater than or equal to the current height with the current height. This substitution maintains the property that 'sub' contains the smallest ending elements, helping to build the LIS efficiently. Finally, the length of the 'sub' array is returned, representing the length of the LIS, which is the maximum number of nested envelopes.

Step-by-Step Algorithm

Step 1: Sort the 'envelopes' array based on width in ascending order and height in descending order (for equal widths).
Step 2: Initialize an empty array 'sub' to store the smallest tail elements of increasing subsequences.
Step 3: Iterate through the sorted 'envelopes', extracting the height of each envelope.
Step 4: Perform a binary search on 'sub' to find the insertion point for the current height.
Step 5: If the insertion point is the end of 'sub', append the height to 'sub' (extend the longest increasing subsequence).
Step 6: Otherwise, replace the element at the insertion point with the current height (potentially find a smaller tail for a subsequence of the same length).
Step 7: After iterating through all envelopes, return the length of 'sub', which represents the maximum number of nested envelopes.

Key Insights

Insight 1: Sorting the envelopes is crucial. Sort by width in ascending order. When widths are the same, sort by height in descending order. This ensures that envelopes with equal widths cannot be nested inside each other.
Insight 2: After sorting, the problem transforms into finding the Longest Increasing Subsequence (LIS) of the heights. The descending sort on heights for same widths is essential to avoid incorrectly including envelopes with the same width in the LIS.
Insight 3: Using binary search (or equivalent) optimizes finding the correct position to update the 'sub' array, which stores the smallest end elements of the LIS of different lengths encountered so far. This ensures an O(n log n) time complexity.

Complexity Analysis

Time Complexity: O(nlogn)

Space Complexity: O(n)

Topics

This problem involves: Array, Binary Search, Dynamic Programming, Sorting.

Companies

Asked at: Atlassian, Intuit, Sprinklr.

Russian Doll Envelopes | LeetCode 354

Hard

Array Binary Search Dynamic Programming Sorting

Time: O(nlogn)Space: O(n)

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public int maxEnvelopes(int[][] envelopes) {
        // Sort by width ascending, and for same width, by height descending.
        // This transforms the problem into finding the Longest Increasing Subsequence (LIS)
        // of the heights. The descending sort on heights for equal widths ensures that
        // envelopes with the same width cannot be part of the same increasing subsequence.
        Arrays.sort(envelopes, (a, b) -> {
            if (a[0] != b[0]) {
                return a[0] - b[0];
            } else {
                return b[1] - a[1];
            }
        });
        
        // This list 'sub' will store the tails of increasing subsequences.
        // It is not the LIS itself, but its length will be the length of the LIS.
        int[] sub = new int[envelopes.length];
        int subLength = 0;
        
        for (int[] envelope : envelopes) {
            int h = envelope[1];
            // Find the insertion point for h in sub to maintain sorted order.
            // Binary search finds the index of the first element in sub that is >= h.
            int idx = Arrays.binarySearch(sub, 0, subLength, h);
            if (idx < 0) {
                idx = -idx - 1;
            }
            
            if (idx == subLength) {
                // If h is greater than all elements in sub, it extends the LIS.
                sub[subLength++] = h;
            } else {
                // Replace the element at idx with h. This means we found a
                // potentially better (smaller) tail for an increasing subsequence
                // of the same length (idx + 1).
                sub[idx] = h;
            }
        }
        
        return subLength;
    }
}

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Problem Info

Description

Problem 354: Russian Doll Envelopes - Detailed explanation and solution approaches.

Topics

Array Binary Search Dynamic Programming Sorting

Companies

AtlassianIntuitSprinklr

Stats

Difficulty:Hard

Languages:4

Companies:3

Problem #:354

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Sort the 'envelopes' array based on width in ascending order and height in descending order (for equal widths).
2
Step 2: Initialize an empty array 'sub' to store the smallest tail elements of increasing subsequences.
3
Step 3: Iterate through the sorted 'envelopes', extracting the height of each envelope.
4
Step 4: Perform a binary search on 'sub' to find the insertion point for the current height.
5
Step 5: If the insertion point is the end of 'sub', append the height to 'sub' (extend the longest increasing subsequence).
6
Step 6: Otherwise, replace the element at the insertion point with the current height (potentially find a smaller tail for a subsequence of the same length).
7
Step 7: After iterating through all envelopes, return the length of 'sub', which represents the maximum number of nested envelopes.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(nlogn)

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

Sorting using standard algorithms like merge sort or quicksort is O(n log n). The binary search step within the loop contributes O(log n) for each envelope. The 'sub' array needs to hold at most 'n' elements, resulting in O(n) space.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Dynamic Programming

An alternative approach is to use dynamic programming. Sort the envelopes by width and then find the LIS of heights. The DP array stores the maximum number of envelopes that can be nested ending at each envelope. Time complexity: O(n^2), Space complexity: O(n).

Time: O(n^2)Space: O(n)

Trade-offs: DP is easier to understand but less efficient than the binary search approach for larger input sizes. The binary search approach is preferred due to its O(n log n) time complexity. DP might be suitable for very small input sizes where the difference in efficiency is negligible.

✨

Key Insights

1Insight 1: Sorting the envelopes is crucial. Sort by width in ascending order. When widths are the same, sort by height in descending order. This ensures that envelopes with equal widths cannot be nested inside each other.
2Insight 2: After sorting, the problem transforms into finding the Longest Increasing Subsequence (LIS) of the heights. The descending sort on heights for same widths is essential to avoid incorrectly including envelopes with the same width in the LIS.
3Insight 3: Using binary search (or equivalent) optimizes finding the correct position to update the 'sub' array, which stores the smallest end elements of the LIS of different lengths encountered so far. This ensures an O(n log n) time complexity.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Empty input 'envelopes'. The solutions correctly handle this case by either implicitly or explicitly returning 0 since no nesting is possible. (Python/Java/C++ implicitly handles empty lists/arrays, C returns 0 if envelopesSize is 0).
Edge case 2: Envelopes with the same width and height. Sorting by width ascending and height descending handles this case by ensuring that envelopes with the same width are not considered as nested. The first one encountered will prevent any others of the same width from being included.
Edge case 3: All envelopes having identical dimensions. Sorting places them in order of width ascending, height descending, and the result will be 1 because they cannot nest in each other.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly sorting the envelopes. Failing to sort heights in descending order when widths are equal will lead to including envelopes with the same width in the LIS, resulting in an incorrect answer.

✗Common mistake 2: Not understanding the LIS concept. Directly trying to find nested envelopes without realizing the connection to LIS will make the problem significantly harder. Failing to use binary search will also lead to O(n^2) complexity.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Resource allocation: Finding the optimal set of resources (e.g., virtual machines with different CPU and memory configurations) that can be nested within each other to maximize resource utilization.

🏦Real-world scenario 2: Task scheduling: Determining the maximum number of tasks that can be scheduled on a processor, where each task requires a certain amount of time and memory, and one task can be scheduled after another if it requires less time and memory.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 22, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 14

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I avoid off-by-one errors in binary search?

Use the pattern: left = 0, right = n-1, while (left <= right), mid = left + (right - left) / 2. Decide if you need to find exact match, first occurrence, or last occurrence, and adjust your conditions accordingly. Test with arrays of size 0, 1, and 2.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Russian Doll Envelopes"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether binary search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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