Meeting Scheduler
MetaThe problem
You are given two lists of availability slots for two people. Each slot is a list of two integers [start, end], representing the inclusive start time and the exclusive end time of that person's available time.
Your task is to find the earliest time slot that is at least **duration **minutes long and is common between both people's availability. If there is no such slot, return an empty list.
Input: slots1 = [[10, 50], [60, 120], [140, 210]], slots2 = [[0, 15], [60, 70]], duration = 8 **Output: **[60, 68] **Explanation: ** The only overlapping slot that is at least 8 minutes long is [60, 68].
Input: slots1 = [[10, 50], [60, 120], [140, 210]], slots2 = [[0, 15], [60, 70]], duration = 12 Output: [] Explanation: Although [60, 70] overlaps, it is only 10 minutes long, which is less than the required 12 minutes.
Input: slots1 = [[10, 20], [30, 40]], slots2 = [[15, 25], [35, 50]], duration = 5
- 1 <= slots1.length, slots2.length <= 104
- slots1[i].length, slots2[i].length == 2
- slots1[i][0] < slots1[i][1]
- slots2[i][0] < slots2[i][1]
- 0 <= slots1[i][j], slots2[i][j] <= 109
- 1 <= duration <= 106
cpp
class Solution {
public:
vector<int> minAvailableDuration(vector<vector<int>>& slots1, vector<vector<int>>& slots2, int duration) {
//Your Code Goes Here
}
};java
class Solution {
public List<Integer> minAvailableDuration(int[][] slots1, int[][] slots2, int duration) {
//Your Code Goes Here
}
}python
class Solution(object):
def minAvailableDuration(self, slots1, slots2, duration):
"""
:type slots1: List[List[int]]
:type slots2: List[List[int]]
:type duration: int
:rtype: List[int]
"""
//Your Code Goes Herejavascript
/**
* @param {number[][]} slots1
* @param {number[][]} slots2
* @param {number} duration
* @return {number[]}
*/
class Solution{
minAvailableDuration(slots1, slots2, duration){
//your code goes here
}
}csharp
public class Solution
{
public IList<int> MinAvailableDuration(int[][] slots1, int[][] slots2, int duration)
{
//Your Code Goes Here
}
}go
func minAvailableDuration(slots1 [][]int, slots2 [][]int, duration int) []int {
// Your code goes here
}Stuck? Show a way to structure it+
- 01Point at the first slot in each list.
- 02Compute overlap start as the maximum start.
- 03Compute overlap end as the minimum end.
- 04Return when the overlap fits duration.
- 05Advance the interval that ends first.
Reference answer
Then expect these follow-ups
How would you support more than two calendars?
Tests: heap merge
How would you find all feasible slots?
Tests: iteration
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