1854

1854. Maximum Population Year (Easy)

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.

 

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 <= birthi < deathi <= 2050

Companies:
Microsoft

Related Topics:
Array

Solution 1. Brute Force

// OJ: https://leetcode.com/problems/maximum-population-year/
// Author: github.com/lzl124631x
// 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;
    }
};

Solution 2. Line Sweep

// OJ: https://leetcode.com/problems/maximum-population-year/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(Y) where Y is the range of the year
class Solution {
public:
    int maximumPopulation(vector<vector<int>>& A) {
        int diff[101] = {}, mx = 0, ans = 0, cur = 0;
        for (auto &v : A) {
            diff[v[0] - 1950]++;
            diff[v[1] - 1950]--;
        }
        for (int i = 0; i < 101; ++i) {
            cur += diff[i];
            if (cur > mx) {
                mx = cur;
                ans = i;
            }
        }
        return ans + 1950;
    }
};