comments | difficulty | edit_url | tags | ||||||
---|---|---|---|---|---|---|---|---|---|
true |
Medium |
|
Given an array of meeting time intervals intervals
where intervals[i] = [starti, endi]
, return the minimum number of conference rooms required.
Example 1:
Input: intervals = [[0,30],[5,10],[15,20]] Output: 2
Example 2:
Input: intervals = [[7,10],[2,4]] Output: 1
Constraints:
1 <= intervals.length <= 104
0 <= starti < endi <= 106
class Solution:
def minMeetingRooms(self, intervals: List[List[int]]) -> int:
delta = [0] * 1000010
for start, end in intervals:
delta[start] += 1
delta[end] -= 1
return max(accumulate(delta))
class Solution {
public int minMeetingRooms(int[][] intervals) {
int n = 1000010;
int[] delta = new int[n];
for (int[] e : intervals) {
++delta[e[0]];
--delta[e[1]];
}
int res = delta[0];
for (int i = 1; i < n; ++i) {
delta[i] += delta[i - 1];
res = Math.max(res, delta[i]);
}
return res;
}
}
class Solution {
public:
int minMeetingRooms(vector<vector<int>>& intervals) {
int n = 1000010;
vector<int> delta(n);
for (auto e : intervals) {
++delta[e[0]];
--delta[e[1]];
}
for (int i = 0; i < n - 1; ++i) {
delta[i + 1] += delta[i];
}
return *max_element(delta.begin(), delta.end());
}
};
func minMeetingRooms(intervals [][]int) int {
n := 1000010
delta := make([]int, n)
for _, e := range intervals {
delta[e[0]]++
delta[e[1]]--
}
for i := 1; i < n; i++ {
delta[i] += delta[i-1]
}
return slices.Max(delta)
}
use std::{cmp::Reverse, collections::BinaryHeap};
impl Solution {
#[allow(dead_code)]
pub fn min_meeting_rooms(intervals: Vec<Vec<i32>>) -> i32 {
// The min heap that stores the earliest ending time among all meeting rooms
let mut pq = BinaryHeap::new();
let mut intervals = intervals;
let n = intervals.len();
// Let's first sort the intervals vector
intervals.sort_by(|lhs, rhs| lhs[0].cmp(&rhs[0]));
// Push the first end time to the heap
pq.push(Reverse(intervals[0][1]));
// Traverse the intervals vector
for i in 1..n {
// Get the current top element from the heap
if let Some(Reverse(end_time)) = pq.pop() {
if end_time <= intervals[i][0] {
// If the end time is early than the current begin time
let new_end_time = intervals[i][1];
pq.push(Reverse(new_end_time));
} else {
// Otherwise, push the end time back and we also need a new room
pq.push(Reverse(end_time));
pq.push(Reverse(intervals[i][1]));
}
}
}
pq.len() as i32
}
}