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1854 - Maximum Population Year (Easy)

https://leetcode.com/problems/maximum-population-year/

Problem Statement​

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

Approach 1: Line Sweep​

Written by @wingkwong
class Solution {
public:
    int maximumPopulation(vector<vector<int>>& logs) {
        vector<int> cnt(2100);
        for (auto x : logs) {
            // start: + 1
            cnt[x[0]]++;
            // end: - 1
            cnt[x[1]]--;
        }
        // calculate prefix sum
        partial_sum(cnt.begin(), cnt.end(), cnt.begin());
        // check the maximum
        int mx = *max_element(cnt.begin(), cnt.end());
        // find the first year with the max population
        for (int i = 1950; i <= 2050; i++) {
            if (mx == cnt[i]) {
                return i;
            }
        }
        return -1;
    }
};

We can further revise it due to the constraint 1950<=birthi<deathi<=20501950 <= birth_i < death_i <= 2050.

Written by @wingkwong
class Solution {
public:
    int maximumPopulation(vector<vector<int>>& logs) {
        vector<int> cnt(101);
        for (auto x : logs) {
            // start: + 1
            cnt[x[0] - 1950]++;
            // end: - 1
            cnt[x[1] - 1950]--;
        }
        // calculate prefix sum
        partial_sum(cnt.begin(), cnt.end(), cnt.begin());
        // check the maximum
        int mx = *max_element(cnt.begin(), cnt.end());
        // find the first year with the max population
        for (int i = 0; i <= 100; i++) {
            if (mx == cnt[i]) {
                return 1950 + i;
            }
        }
        return -1;
    }
};