Formatted question description: https://leetcode.ca/all/1854.html
1854. Maximum Population Year
Level
Easy
Description
You are given a 2D integer array logs
where each logs[i] = [birth_i, death_i]
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 [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
Solution
Use a list to store each person’s first and last year. That is, for each log
in logs
, add log[0]
and (log[1]  1)
to the list. Then store the list according to absolute values in ascending order and according to actual values in descending order. Then loop over the list to count the maximum population and store the corresponding year.

class Solution { public int maximumPopulation(int[][] logs) { List<Integer> birthDeathList = new ArrayList<Integer>(); for (int[] log : logs) { birthDeathList.add(log[0]); birthDeathList.add((log[1]  1)); } Collections.sort(birthDeathList, new Comparator<Integer>() { public int compare(Integer num1, Integer num2) { if (Math.abs(num1) != Math.abs(num2)) return Math.abs(num1)  Math.abs(num2); else return num2  num1; } }); int maxYear = 0; int maxAlive = 0; int curAlive = 0; int size = birthDeathList.size(); for (int i = 0; i < size; i++) { int year = birthDeathList.get(i); if (year > 0) { curAlive++; if (curAlive > maxAlive) { maxYear = year; maxAlive = curAlive; } } else curAlive; } return maxYear; } }

// OJ: https://leetcode.com/problems/maximumpopulationyear/ // Time: O(N^2) // Space: O(1) class Solution { public: int maximumPopulation(vector<vector<int>>& A) { sort(begin(A), end(A)); int ans = 0, mx = 0, N = A.size(); for (int i = 0; i < N; ++i) { int yr = A[i][0], cnt = 0; for (int j = 0; j < N; ++j) { cnt += A[j][0] <= yr && A[j][1] > yr; } if (cnt > mx) { mx = cnt; ans = yr; } } return ans; } };

# 1854. Maximum Population Year # https://leetcode.com/problems/maximumpopulationyear/ class Solution: def maximumPopulation(self, logs: List[List[int]]) > int: people = [0] * 2051 for b,d in logs: people[b] += 1 people[d] = 1 for i in range(1, 2051): people[i] += people[i  1] res = 0 mmax = float('inf') for i,x in enumerate(people): if x > mmax: res = i mmax = x return res