LeetCode #1854 — EASY

Maximum Population Year

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

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: 1993
Explanation: The maximum population is 1, and 1993 is the earliest year with this population.

Example 2:

Input: logs = [[1950,1961],[1960,1971],[1970,1981]]
Output: 1960
Explanation: 
The maximum population is 2, and it had happened in years 1960 and 1970.
The earlier year between them is 1960.

Constraints:

1 <= logs.length <= 100
1950 <= birth_i < death_i <= 2050

Step 01

Brute Force Baseline

Problem summary: You are given a 2D integer array logs where each logs[i] = [birthi, deathi] indicates the birth and death years of the ith person. The population of some year x is the number of people alive during that year. The ith person is counted in year x's population if x is in the inclusive range [birthi, deathi - 1]. Note that the person is not counted in the year that they die. Return the earliest year with the maximum population.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[[1993,1999],[2000,2010]]

Example 2

[[1950,1961],[1960,1971],[1970,1981]]

Core Insight

What unlocks the optimal approach

For each year find the number of people whose birth_i ≤ year and death_i > year.
Find the maximum value between all years.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1854: Maximum Population Year
class Solution {
    public int maximumPopulation(int[][] logs) {
        int[] d = new int[101];
        final int offset = 1950;
        for (var log : logs) {
            int a = log[0] - offset;
            int b = log[1] - offset;
            ++d[a];
            --d[b];
        }
        int s = 0, mx = 0;
        int j = 0;
        for (int i = 0; i < d.length; ++i) {
            s += d[i];
            if (mx < s) {
                mx = s;
                j = i;
            }
        }
        return j + offset;
    }
}

// Accepted solution for LeetCode #1854: Maximum Population Year
func maximumPopulation(logs [][]int) int {
	d := [101]int{}
	offset := 1950
	for _, log := range logs {
		a, b := log[0]-offset, log[1]-offset
		d[a]++
		d[b]--
	}
	var s, mx, j int
	for i, x := range d {
		s += x
		if mx < s {
			mx = s
			j = i
		}
	}
	return j + offset
}

# Accepted solution for LeetCode #1854: Maximum Population Year
class Solution:
    def maximumPopulation(self, logs: List[List[int]]) -> int:
        d = [0] * 101
        offset = 1950
        for a, b in logs:
            a, b = a - offset, b - offset
            d[a] += 1
            d[b] -= 1
        s = mx = j = 0
        for i, x in enumerate(d):
            s += x
            if mx < s:
                mx, j = s, i
        return j + offset

// Accepted solution for LeetCode #1854: Maximum Population Year
struct Solution;

impl Solution {
    fn maximum_population(logs: Vec<Vec<i32>>) -> i32 {
        let mut start = std::i32::MAX;
        let mut end = 0;
        for log in &logs {
            let birth = log[0];
            let death = log[1];
            start = start.min(birth);
            end = end.max(death);
        }
        let mut res = 0;
        let mut max = 0;
        for i in start..end {
            let mut population = 0;
            for log in &logs {
                let birth = log[0];
                let death = log[1];
                if (birth..death).contains(&i) {
                    population += 1;
                }
            }
            if max < population {
                res = i;
                max = population;
            }
        }
        res
    }
}

#[test]
fn test() {
    let logs = vec_vec_i32![[1993, 1999], [2000, 2010]];
    let res = 1993;
    assert_eq!(Solution::maximum_population(logs), res);
    let logs = vec_vec_i32![[1950, 1961], [1960, 1971], [1970, 1981]];
    let res = 1960;
    assert_eq!(Solution::maximum_population(logs), res);
}

// Accepted solution for LeetCode #1854: Maximum Population Year
function maximumPopulation(logs: number[][]): number {
    const d: number[] = new Array(101).fill(0);
    const offset = 1950;
    for (const [birth, death] of logs) {
        d[birth - offset]++;
        d[death - offset]--;
    }
    let j = 0;
    for (let i = 0, s = 0, mx = 0; i < d.length; ++i) {
        s += d[i];
        if (mx < s) {
            mx = s;
            j = i;
        }
    }
    return j + offset;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(C)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.