What is the time complexity of Maximum Population Year?

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

What data structures are used to solve Maximum Population Year?

This problem uses Array, Counting, Prefix Sum. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Maximum Population Year?

Insight 1: Using a data structure to efficiently track population per year is crucial. A simple dictionary/hashmap or an array is suitable. Insight 2: The problem can be solved efficiently using a prefix sum approach (or a difference array technique) to avoid iterating over all people for each year. This significantly improves time complexity.

Which companies ask LeetCode 1854. Maximum Population Year in interviews?

This problem is frequently asked at PayPal, Zoho.

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Maximum Population Year - LeetCode 1854 Solution

Maximum Population Year - Complete Solution Guide

Maximum Population Year is LeetCode problem 1854, a Easy 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 integer array logs where each logs[i] = [birth i , death i ] indicates the birth and death years of the i th person. The population of some year x is the number of people alive during that year. The i th person is counted in year x 's population if x is in the inclusive range [birth i , death i - 1] . Note that the person is not counted in the year that they die. Return the earliest year with the maximum population . Example 1: Input: logs = [[1993,1999],[2000,2010]] Output: 1

Detailed Explanation

The problem asks to find the earliest year with the maximum population given a list of birth and death years for individuals. Each person is considered alive in a year from their birth year (inclusive) until their death year (exclusive). The population of a year is the number of people alive during that year. The output should be the earliest year with the highest population count.

Solution Approach

The provided solutions utilize a variation of the difference array approach. Instead of directly counting the population for each year by iterating through all people, they cleverly use a difference array to track population changes. This approach reduces the computation significantly. The solutions iterate once through the logs, updating a difference array that represents the change in population from one year to the next. A final pass calculates the actual population for each year by accumulating the changes in the difference array. The earliest year with the maximum population is then identified.

Step-by-Step Algorithm

Step 1: Initialize a difference array (or a hashmap in the Python solution). The Python solution uses a dictionary, while Java, C++, and C use arrays. The array size is determined by the range of years (101 in this case, representing years 1950-2050).
Step 2: Iterate through the `logs`. For each log `[birth, death]`, increment the population count for the `birth` year and decrement it for the `death` year in the difference array (or hashmap).
Step 3: Create an array to store the actual population for each year by calculating the cumulative sum of the difference array. (This step is implicit in the Java, C++, and C solutions)
Step 4: Iterate through the population array (or the dictionary entries in Python) to find the earliest year with the maximum population.

Key Insights

Insight 1: Using a data structure to efficiently track population per year is crucial. A simple dictionary/hashmap or an array is suitable.
Insight 2: The problem can be solved efficiently using a prefix sum approach (or a difference array technique) to avoid iterating over all people for each year. This significantly improves time complexity.
Insight 3: Handling edge cases where multiple years have the same maximum population is important. The problem specifies that we should return the earliest year with the maximum population.

Complexity Analysis

Time Complexity: O(n*w)

Space Complexity: O(w)

Topics

This problem involves: Array, Counting, Prefix Sum.

Companies

Asked at: PayPal, Zoho.

Maximum Population Year | LeetCode 1854

Easy

Array Counting Prefix Sum

Time: O(n*w)Space: O(w)

Solution Code

Available in:

class Solution {
    public int maximumPopulation(int[][] logs) {
        int[] years = new int[101];
        for (int[] log : logs) {
            years[log[0] - 1950]++;
            years[log[1] - 1950]--;
        }
        int maxPop = 0;
        int maxYear = 1950;
        int currentPop = 0;
        for (int i = 0; i < 101; i++) {
            currentPop += years[i];
            if (currentPop > maxPop) {
                maxPop = currentPop;
                maxYear = 1950 + i;
            }
        }
        return maxYear;
    }
}

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Problem Info

Description

Problem 1854: Maximum Population Year - Detailed explanation and solution approaches.

Topics

Array Counting Prefix Sum

Companies

PayPalZoho

Stats

Difficulty:Easy

Languages:4

Companies:2

Problem #:1854

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: Initialize a difference array (or a hashmap in the Python solution). The Python solution uses a dictionary, while Java, C++, and C use arrays. The array size is determined by the range of years (101 in this case, representing years 1950-2050).
2
Step 2: Iterate through the `logs`. For each log `[birth, death]`, increment the population count for the `birth` year and decrement it for the `death` year in the difference array (or hashmap).
3
Step 3: Create an array to store the actual population for each year by calculating the cumulative sum of the difference array. (This step is implicit in the Java, C++, and C solutions)
4
Step 4: Iterate through the population array (or the dictionary entries in Python) to find the earliest year with the maximum population.

💡

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*w)

💾

Space Complexity

O(w)

📝

Why?

The time complexity is determined by the two loops. The space complexity is determined by the size of the difference array (or hashmap), which is directly proportional to the range of years.

📖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 year, iterate through all logs and count the number of people alive in that year. This involves nested loops.

Time: O(n*w), where n is the number of logs and w is the range of years.Space: O(1)

Trade-offs: Simple to understand but significantly less efficient than the difference array approach, especially for a large number of logs and a wide range of years.

✨

Key Insights

1Insight 1: Using a data structure to efficiently track population per year is crucial. A simple dictionary/hashmap or an array is suitable.
2Insight 2: The problem can be solved efficiently using a prefix sum approach (or a difference array technique) to avoid iterating over all people for each year. This significantly improves time complexity.
3Insight 3: Handling edge cases where multiple years have the same maximum population is important. The problem specifies that we should return the earliest year with the maximum population.

📚

Key Terms Explained

For Students

Array

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

Edge Cases

Edge case 1: Empty input `logs`. The solutions handle this correctly by returning 0 in the case of an empty array, although the problem statement doesn't explicitly define the behavior for an empty input. The Python solution implicitly handles this as the dictionary remains empty.
Edge case 2: Multiple years with the same maximum population. The solutions correctly identify and return the earliest year among those with the maximum population.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly handling the inclusive and exclusive bounds of birth and death years. Failing to account for the fact that a person is not counted in their death year.

✗Common mistake 2: Not handling the case where multiple years have the same maximum population. Returning an incorrect year (not the earliest).

✅

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: Analyzing historical population trends. This algorithm can be used to identify years with peak population in demographic studies.

🏦Real-world scenario 2: Resource allocation based on population density. In urban planning, knowing the years of peak population can aid in determining optimal resource allocation (e.g., schools, hospitals).

💡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

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ 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 25, 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 should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

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 "Maximum Population Year"?

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

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