What is the time complexity of Queue Reconstruction by Height?

The optimal solution has a time complexity of O(n^2).

What data structures are used to solve Queue Reconstruction by Height?

This problem uses Array, Binary Indexed Tree, Segment Tree, Sorting. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Queue Reconstruction by Height?

Insight 1: Sorting the people by height in descending order (and then by k in ascending order for same heights) is crucial. This allows us to insert people into the queue without affecting the counts of people already placed. Insight 2: Using an insertion-based approach, where we insert people at the index specified by their 'k' value, effectively places them in the correct position relative to taller or equally tall people.

Which companies ask LeetCode 406. Queue Reconstruction by Height in interviews?

This problem is frequently asked at PhonePe.

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Queue Reconstruction by Height - LeetCode 406 Solution

Queue Reconstruction by Height - Complete Solution Guide

Queue Reconstruction by Height is LeetCode problem 406, a Medium 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 an array of people, people , which are the attributes of some people in a queue (not necessarily in order). Each people[i] = [h i , k i ] represents the i th person of height h i with exactly k i other people in front who have a height greater than or equal to h i . Reconstruct and return the queue that is represented by the input array people . The returned queue should be formatted as an array queue , where queue[j] = [h j , k j ] is the attributes of the j th person in the queue

Detailed Explanation

The problem asks us to reconstruct a queue of people given their height and the number of people in front of them with a height greater than or equal to their own. The input is an array `people`, where each element `people[i] = [h_i, k_i]` represents a person with height `h_i` and `k_i` people ahead of them who are at least as tall. The output should be the reconstructed queue as an array of `[height, k]` pairs.

Solution Approach

The provided solution uses a sorting-based approach followed by inserting each person into the result list (or array) at a specific index. First, the input array is sorted in descending order of height and, in case of a tie, in ascending order of the number of people ahead of them (k). After sorting, the algorithm iterates through the sorted array and inserts each person into the result array at the index specified by the person's 'k' value. This placement ensures that 'k' people taller or equal in height are in front of them in the constructed queue.

Step-by-Step Algorithm

Step 1: Sort the input array `people`. The sorting criteria are first by height in descending order (taller people come first) and then by the number of people in front (k) in ascending order (lower k comes first for same height).
Step 2: Create an empty result list `result` to store the reconstructed queue.
Step 3: Iterate through the sorted `people` array. For each person `p = [h, k]`, insert them into the `result` list at index `k`.
Step 4: After iterating through all people, the `result` list now contains the reconstructed queue. Convert the result list back to an array if required by the problem constraints.

Key Insights

Insight 1: Sorting the people by height in descending order (and then by k in ascending order for same heights) is crucial. This allows us to insert people into the queue without affecting the counts of people already placed.
Insight 2: Using an insertion-based approach, where we insert people at the index specified by their 'k' value, effectively places them in the correct position relative to taller or equally tall people.
Insight 3: The insertion step is the core of the algorithm. It leverages the sorted order to maintain the 'k' constraint during the reconstruction process.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n)

Topics

This problem involves: Array, Binary Indexed Tree, Segment Tree, Sorting.

Companies

Asked at: PhonePe.

Queue Reconstruction by Height | LeetCode 406

Medium

Array Binary Indexed Tree Segment Tree Sorting

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

Solution Code

Available in:

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

class Solution {
    public int[][] reconstructQueue(int[][] people) {
        Arrays.sort(people, (a, b) -> {
            if (b[0] != a[0]) {
                return b[0] - a[0];
            } else {
                return a[1] - b[1];
            }
        });

        List<int[]> result = new ArrayList<>();
        for (int[] p : people) {
            result.add(p[1], p);
        }

        return result.toArray(new int[people.length][2]);
    }
}

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Problem Info

Description

Problem 406: Queue Reconstruction by Height - Detailed explanation and solution approaches.

Topics

Array Binary Indexed Tree Segment Tree Sorting

Companies

PhonePe

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:406

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

4 steps

1
Step 1: Sort the input array `people`. The sorting criteria are first by height in descending order (taller people come first) and then by the number of people in front (k) in ascending order (lower k comes first for same height).
2
Step 2: Create an empty result list `result` to store the reconstructed queue.
3
Step 3: Iterate through the sorted `people` array. For each person `p = [h, k]`, insert them into the `result` list at index `k`.
4
Step 4: After iterating through all people, the `result` list now contains the reconstructed queue. Convert the result list back to an array if required by the problem constraints.

💡

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(n^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The dominant operation is the insertion into the `result` list/array. Inserting an element at a specific index in a list takes O(n) time in the worst case because it might require shifting the existing elements. Since we perform this insertion 'n' times, the overall time complexity is O(n*n) = O(n^2). The sorting operation contributes O(n log n), but it's dominated by the insertion loop. The space complexity is O(n) because the `result` list/array stores 'n' elements in the reconstructed queue.

📖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

Segment Tree or Binary Indexed Tree (BIT)

Use a segment tree or BIT to efficiently find the correct insertion index for each person. The segment tree would track the number of available slots. This allows for insertion in O(log n) time for each person.

Time: O(n log n)Space: O(n)

Trade-offs: Segment tree or BIT would improve time complexity to O(n log n), but the implementation is more complex. Useful when n is very large.

✨

Key Insights

1Insight 1: Sorting the people by height in descending order (and then by k in ascending order for same heights) is crucial. This allows us to insert people into the queue without affecting the counts of people already placed.
2Insight 2: Using an insertion-based approach, where we insert people at the index specified by their 'k' value, effectively places them in the correct position relative to taller or equally tall people.
3Insight 3: The insertion step is the core of the algorithm. It leverages the sorted order to maintain the 'k' constraint during the reconstruction process.

📚

Key Terms Explained

For Students

Array

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

Sorting

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

Edge Cases

Edge case 1: Empty input array. The code handles this implicitly as the loops will not execute, and an empty list will be returned.
Edge case 2: Input array with duplicate heights. Sorting by k in ascending order when heights are equal ensures correct placement relative to people of the same height.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting correctly. Sorting only by height or in the wrong order will lead to incorrect reconstructions.

✗Common mistake 2: Inserting the person at the wrong index. Failing to consider the relative positions based on 'k' will result in an invalid queue.

✅

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: Scheduling tasks with dependencies. Tasks can be viewed as people, height as priority, and k as number of dependencies. Queue reconstruction helps determine optimal task execution order.

🏦Real-world scenario 2: Placing advertisements in a web page layout based on their priority and desired position relative to other ads.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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: July 17, 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

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 14

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 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.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

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 "Queue Reconstruction by Height"?

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

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