1800 - 1899
1854 - Maximum Population Year (Easy)
Problem Link
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 <= 1001950 <= birthi < deathi <= 2050
Approach 1: Line Sweep
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 .
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;
}
};