How do you solve Maximum Population Year on LeetCode?

Maximum Population Year (LeetCode #1854) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Population changes can be tracked using a counting array.

What is the brute force solution for Maximum Population Year?

The brute force approach for Maximum Population Year is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we check each year from the earliest birth year to the latest death year and count how many people are alive in that year. This is straightforward but can be slow. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Population Year?

Maximum Population Year uses the following concepts: Array, Counting, Prefix Sum. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Maximum Population Year?

Explain your thought process clearly while coding. Consider edge cases, such as when all births and deaths are in the same year. Practice coding the optimal solution to improve speed and confidence.

What are common mistakes when solving Maximum Population Year?

Not accounting for the year of death correctly (off by one error). Failing to initialize or reset variables properly.

#1854

Maximum Population Year

Easy

ArrayCountingPrefix SumHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

Instead of checking each year individually, we can use a counting array to track population changes at birth and death years. This allows us to efficiently compute the population for each year in a single pass.

⚙️

Algorithm

3 steps

1Step 1: Create an array to count population changes for each year from 1950 to 2050.
2Step 2: For each log entry, increment the birth year and decrement the death year in the counting array.
3Step 3: Iterate through the counting array to calculate the running population and track the maximum population and the earliest year.

solution.py14 lines

1def maximumPopulation(logs):
2    population_changes = [0] * 102
3    for birth, death in logs:
4        population_changes[birth - 1950] += 1
5        population_changes[death - 1950] -= 1
6    max_population = 0
7    current_population = 0
8    year_with_max_population = 0
9    for year in range(102):
10        current_population += population_changes[year]
11        if current_population > max_population:
12            max_population = current_population
13            year_with_max_population = year + 1950
14    return year_with_max_population

ℹ

Complexity note: This complexity is O(n) because we only pass through the logs once to set up the population changes and then through the years once to calculate the population, making it efficient.

1Population changes can be tracked using a counting array.
2The earliest year with maximum population can be found efficiently using a single pass.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.