What is the time complexity of Number of Visible People in a Queue?

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

What data structures are used to solve Number of Visible People in a Queue?

This problem uses Array, Stack, Monotonic Stack. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Number of Visible People in a Queue?

Insight 1: The key observation is that we can efficiently determine the number of visible people using a stack to maintain a decreasing sequence of heights from right to left. Insight 2: The monotonic stack data structure is crucial for this problem. It allows us to maintain the tallest people that a person can potentially see.

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Number of Visible People in a Queue - LeetCode 1944 Solution

Q: How to solve Number of Visible People in a Queue?

The solution employs a monotonic stack to keep track of potential visible people to the right of each person. The algorithm iterates through the `heights` array from right to left. For each person, it removes shorter people from the stack until the stack is empty or it finds a person taller than the current person. The number of popped elements represents the number of visible people who are shorter than the current person. If the stack is not empty after popping shorter people, the top of the stack is also visible (because it's taller and blocks the view of anyone behind it).

Q: Which companies ask LeetCode 1944. Number of Visible People in a Queue in interviews?

This problem is frequently asked at Citigroup, DoorDash, Expedia, GE Healthcare, Rippling and 1 more companies.

Number of Visible People in a Queue - Complete Solution Guide

Number of Visible People in a Queue is LeetCode problem 1944, 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

There are n people standing in a queue, and they numbered from 0 to n - 1 in left to right order. You are given an array heights of distinct integers where heights[i] represents the height of the i th person. A person can see another person to their right in the queue if everybody in between is shorter than both of them. More formally, the i th person can see the j th person if i < j and min(heights[i], heights[j]) > max(heights[i+1], heights[i+2], ..., heights[j-1]) . Return an array answer of

Detailed Explanation

The problem asks us to determine, for each person in a queue, how many people they can see to their right. A person 'i' can see another person 'j' to their right (i < j) if all the people standing between them (from i+1 to j-1) are shorter than both person 'i' and person 'j'. We need to return an array where each element represents the number of visible people for the corresponding person in the input array.

Solution Approach

The solution employs a monotonic stack to keep track of potential visible people to the right of each person. The algorithm iterates through the `heights` array from right to left. For each person, it removes shorter people from the stack until the stack is empty or it finds a person taller than the current person. The number of popped elements represents the number of visible people who are shorter than the current person. If the stack is not empty after popping shorter people, the top of the stack is also visible (because it's taller and blocks the view of anyone behind it).

Step-by-Step Algorithm

Step 1: Initialize an empty stack and an array `answer` of the same size as `heights` to store the results.
Step 2: Iterate through the `heights` array from right to left (from index n-1 to 0).
Step 3: For each person at index `i`, initialize a counter `visible_count` to 0.
Step 4: While the stack is not empty and the current person's height (`heights[i]`) is greater than the height of the person at the top of the stack, pop the top element from the stack and increment `visible_count`.
Step 5: If the stack is not empty after the while loop (meaning there's a taller person blocking the view), increment `visible_count`.
Step 6: Assign `visible_count` to `answer[i]`.
Step 7: Push the current person's height (`heights[i]`) onto the stack.
Step 8: After iterating through all the people, return the `answer` array.

Key Insights

Insight 1: The key observation is that we can efficiently determine the number of visible people using a stack to maintain a decreasing sequence of heights from right to left.
Insight 2: The monotonic stack data structure is crucial for this problem. It allows us to maintain the tallest people that a person can potentially see.
Insight 3: Iterating from right to left allows us to progressively build the visibility information by considering the heights to the right of each person.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Stack, Monotonic Stack.

Companies

Asked at: Citigroup, DoorDash, Expedia, GE Healthcare, Rippling, ServiceNow.

Number of Visible People in a Queue | LeetCode 1944

Hard

Array Stack Monotonic Stack

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

Solution Code

Available in:

import java.util.Stack;

class Solution {
    public int[] canSeePersonsCount(int[] heights) {
        int n = heights.length;
        int[] answer = new int[n];
        Stack<Integer> stack = new Stack<>();

        for (int i = n - 1; i >= 0; i--) {
            int current_height = heights[i];
            int visible_count = 0;
            
            while (!stack.isEmpty() && current_height > stack.peek()) {
                stack.pop();
                visible_count += 1;
            }
            
            if (!stack.isEmpty()) {
                visible_count += 1;
            }
            
            answer[i] = visible_count;
            
            stack.push(current_height);
            
        }
        return answer;
    }
}

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Problem Info

Description

Problem 1944: Number of Visible People in a Queue - Detailed explanation and solution approaches.

Topics

Array Stack Monotonic Stack

Companies

CitigroupDoorDashExpediaGE HealthcareRipplingServiceNow

Stats

Difficulty:Hard

Languages:4

Companies:6

Problem #:1944

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

8 steps

1
Step 1: Initialize an empty stack and an array `answer` of the same size as `heights` to store the results.
2
Step 2: Iterate through the `heights` array from right to left (from index n-1 to 0).
3
Step 3: For each person at index `i`, initialize a counter `visible_count` to 0.
4
Step 4: While the stack is not empty and the current person's height (`heights[i]`) is greater than the height of the person at the top of the stack, pop the top element from the stack and increment `visible_count`.
5
Step 5: If the stack is not empty after the while loop (meaning there's a taller person blocking the view), increment `visible_count`.
6
Step 6: Assign `visible_count` to `answer[i]`.
7
Step 7: Push the current person's height (`heights[i]`) onto the stack.
8
Step 8: After iterating through all the people, return the `answer` array.

💡

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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is linear because each element is processed at most twice - once when it's pushed onto the stack and potentially once when it's popped. The space complexity arises from the use of a stack to store heights, which in the worst case can hold all the heights.

📖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

Brute Force

For each person, iterate through all the people to their right and check if they are visible based on the given criteria. This involves nested loops.

Time: O(n^2), where n is the length of the `heights` array.Space: O(1), excluding the output array.

Trade-offs: The brute force approach is simple to understand but is not efficient for large input arrays. It should only be used if efficiency isn't a concern or for small inputs.

✨

Key Insights

1Insight 1: The key observation is that we can efficiently determine the number of visible people using a stack to maintain a decreasing sequence of heights from right to left.
2Insight 2: The monotonic stack data structure is crucial for this problem. It allows us to maintain the tallest people that a person can potentially see.
3Insight 3: Iterating from right to left allows us to progressively build the visibility information by considering the heights to the right of each person.

📚

Key Terms Explained

For Students

Array

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

Stack

LIFO structure - Last In, First Out. Like a stack of plates!

Edge Cases

Edge case 1: Empty input array. The solution implicitly handles this by returning an empty array, as the loop won't execute.
Edge case 2: Input array with only one element. The solution returns [0] as expected since a single person can't see anyone to their right.
Edge case 3: Input array with all elements in increasing order. In this case, each person sees all people to their right until encountering the last element, and the last person sees nobody. The stack will contain all elements until the end. The provided solution handles this correctly.
Edge case 4: Input array with all elements in decreasing order. In this case, everyone sees only one person to their right (the next person). The provided solution handles this case.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not iterating from right to left. Iterating from left to right makes it difficult to maintain the visibility information effectively.

✗Common mistake 2: Incorrectly determining visibility. Forgetting to consider the taller person blocking the view after popping shorter elements from the stack results in incorrect visibility counts.

✗Common mistake 3: Using the wrong data structure. Not using a stack, or not understanding its properties for this specific problem, makes the algorithm difficult to design correctly and efficiently.

✅

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: Stock span problem. The number of visible people is analogous to determining the span of a stock price, where we want to find how many days the current day's price is greater than or equal to the previous days' prices.

🏦Real-world scenario 2: Skyline problem. Determining the visible buildings in a city skyline from a certain viewpoint can be solved using similar stack-based techniques.

💡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: October 23, 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 13

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

@DevExpert92Jul 6

❓

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 problems are best solved with a stack?

Stacks excel at: matching problems (parentheses validation), next greater/smaller element, expression evaluation, backtracking, and undo operations. The key pattern is LIFO (Last In, First Out) - when you need to access the most recently added element.

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 "Number of Visible People in a Queue"?

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 stack techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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