# Leetcode – Maximum Population Year Solution

The “Maximum Population Year” problem on Leetcode is an interesting one that involves working with a time-series dataset to identify the year with the highest population. While the problem may appear to be straightforward, its underlying subtleties make it an exciting challenge. This article provides a detailed analysis of the problem, along with multiple Python-based solutions. We will look into the complexities of each approach and how they perform under different scenarios.

## Problem Statement

You are given a 2D integer array logs where each logs[i] = [birthi, deathi] indicates the birth and death years of the i-th person. The year x is considered to be bad if more people die or are born in year x than in any other year.

Return the earliest year that is not considered bad.

### Example

Input: logs = [[1993,1999],[2000,2010]]

Output: 1993

## Approach 1: Brute-force Iteration

A naive approach to this problem is to simply iterate through all possible years between 1950 and 2050, counting the births and deaths for each year and keeping track of the year with the maximum population.

Here’s how it can be implemented in Python:

def maximumPopulation(logs):
year_count = {}
for birth, death in logs:
for year in range(birth, death):
if year not in year_count:
year_count[year] = 0
year_count[year] += 1
max_population = max(year_count.values())
return min(year for year, count in year_count.items() if count == max_population)

### Time Complexity

The time complexity of this approach is O(n×m), where n is the number of logs and m is the maximum number of years a person could live (essentially 100 in this problem’s constraint).

### Space Complexity

The space complexity is O(u), where u is the number of unique years that have population changes.

## Approach 2: Timeline Events and Sorting

Another approach is to mark each birth and death year as an “event” on a timeline and then sort these events. By traversing the sorted timeline, you can keep track of the current population and identify the year with the maximum population.

def maximumPopulation(logs):
timeline = []
for birth, death in logs:
timeline.append((birth, 1))
timeline.append((death, -1))
timeline.sort()
max_population = 0
current_population = 0
result_year = 0
for year, delta in timeline:
current_population += delta
if current_population > max_population:
max_population = current_population
result_year = year
return result_year

### Time Complexity

The time complexity for sorting the timeline is O(n log⁡ n), and the time complexity for the subsequent traversal is O(n), making the total time complexity O(n log⁡ n).

### Space Complexity

The space complexity is O(n) due to the use of the timeline list.

## Approach 3: Using Prefix Sum

Instead of counting births and deaths for each year individually, we could maintain a “prefix sum” array that keeps track of the population change for each year.

def maximumPopulation(logs):
prefix_sum =  * 101  # As 2050 - 1950 = 100
for birth, death in logs:
prefix_sum[birth - 1950] += 1
prefix_sum[death - 1950] -= 1
max_population = 0
current_population = 0
result_year = 0
for i, delta in enumerate(prefix_sum):
current_population += delta
if current_population > max_population:
max_population = current_population
result_year = i + 1950
return result_year

### Time Complexity

The time complexity is O(n), where n is the total number of years to be checked.

### Space Complexity

The space complexity is O(1) since the size of the prefix_sum array is constant (101).

## Unit Testing

Regardless of the approach used, it’s crucial to test the solution to ensure it works for various cases.

def test_maximumPopulation():
assert maximumPopulation([[1993, 1999], [2000, 2010]]) == 1993
assert maximumPopulation([[1950, 1961], [1960, 1971], [1970, 1981]]) == 1960
print("All test cases pass")

test_maximumPopulation()

## Conclusion

The “Maximum Population Year” problem provides an intriguing challenge that combines elements of sorting, array manipulation, and optimization. The problem provides an excellent opportunity to explore various algorithmic techniques. Each approach comes with its trade-offs and offers valuable insights into effective problem-solving. Whether you’re a beginner looking to understand the basics or an experienced programmer aiming for optimization, this problem has something for everyone.