Meeting Rooms III - Leetcode 2402 - Python
ฝัง
- เผยแพร่เมื่อ 20 ก.ค. 2024
- 🚀 neetcode.io/ - A better way to prepare for Coding Interviews
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⭐ BLIND-75 PLAYLIST: • Two Sum - Leetcode 1 -...
Problem Link: leetcode.com/problems/meeting...
0:00 - Read the problem
0:30 - Drawing Explanation
8:48 - Coding Explanation
leetcode 2402
#neetcode #leetcode #python
Sorry I'm a little late today, but better late than never! 🚀🚀
No problem! Love your leetcode content as always!
Thank you!! This is the video I wait for every day.
Why did you stop posting in your another channel? Why another channel?
I thought my streak is going to end , thanks mate ❤❤
I solved it without help!!! But at what cost😅? I took 3 hours, used a "TreeMap"🙃🙃 and my solution beats only 5% of others. But I am still happy that I am able to solve, something is better than nothing. Came here to learn how to do it the right way.
fwiw my java solution beats 23%, I posted the code as a comment to this video
Bro you are a lifesaver, there was absolutely no way I could have solved this on my own.
Thanks for the video. My level is too low to solve this hard question. I'm just here to learn
You explained things so easily! Thank you, dude.
I love your lc video. I can see you've uploaded a solution, which means I can learn from you when I can't solve it. At the same time I don't see any spoiler, so I can try first.
Great video as always!
This one was rough, thanks man!
Same thought process ! 👌
I figured out what I needed to do but cant't figure out the approach of using two heaps. Good question.
Love this explanation
Thanks!
explain meeting rooms 2 with priority queue as well
Never been here this early lol
Love your content 😄
ty :)
7 days ago, just about time for me!
In the heap, is it possible to store the difference between two meetings (end,start, meeting_idx) ? Re: Naaah ignore
Hi, when this problem will be added to the neecode website list?
Can you use heapq methods like heappop on a list that you never heapified like that? 3:16 If so that's awesome. Does that mean if your list is already a (eg min) heap, those operations just work?
Yeah exactly, heaps are stored as lists in python. Even in languages like Java they are implemented with arrays / ArrayList under the hood
@@NeetCodeIO That's awesome. That's the thing I love about Python. You always discover new ways that something can be done a little lazier, haha.
I always assumed that while they could be accessed like lists, it was either:
* some magic under the hood so you could use the same methods as on a normal list object
* a list with the same data was generated when you needed to interact with that object type, while there is some tree that is the actual heap somewhere else in memory
When rooms are not available, and we look at the used heap, why do we only pop the first value from used. What if there were other values with the same end time, would we not want to pop all the nodes with the same end time, and use the one with the minimum room number?
Ah okay never mind, I guess I realized later that the min heap comparator function will take care of that automatically. Sorry, was using Javascript, and was using a class for the min heap that I wrote myself.
How long did it take for you to come up with a solution for this? I came up with a solution but because of the constraints it failed 10 TCs.
can anyone post the c++ version of code for this?
You solved without hints ?
What do you use to draw on the screen?
paint 3d and a mouse
@@NeetCodeIOare u joking 😅
I don't think he is joking@@yaswanthkosuru
Brilliant!
this solution no longer working for anyone else?
I reach test cases 69/82
I give up 😅😅
I actually don't think this is a leetcode hard, more like a medium. I think there are enough cues in the question that let you know exactly what data structure to use to solve this. It also helps if you've solved meeting rooms II before. But I see a lot of comments saying this is difficult. I love how we all assimilate problems differently and what is hard to some is straightforward to others and vice versa - because best believe there are some problems I could never solve in a 45 minute interview that are so called medium or even easy 😂.
Which is why as much as we all grind leetcode we can't afford to ignore the amount of good fortune that needs to be present for us to land jobs at top firms
The logic building is medium. But implementation is definitely hard. Too many moving parts and variables to keep track of to come up with a full working solution without skipping a meal.
I agree tbh@@zweitekonto9654
Javascript version:
function mostBooked(n, meetings) {
const available = new MinPriorityQueue({
compare: (a, b) => a - b
});
for (let i = 0; i < n; i++) available.enqueue(i);
const used = new MinPriorityQueue({
compare: (a, b) => a[0] - b[0] || a[1] - b[1]
});
const count = new Array(n).fill(0);
meetings.sort((a, b) => a[0] - b[0]);
meetings.forEach(([start, end]) => {
while (used.size() > 0 && start >= used.front()[0]) {
const [, room] = used.dequeue();
available.enqueue(room);
}
if (available.size() > 0) {
const room = available.dequeue();
used.enqueue([end, room]);
count[room] += 1;
} else {
const [earliestEnd, room] = used.dequeue();
used.enqueue([earliestEnd + (end - start), room]);
count[room] += 1;
}
});
return count.indexOf(Math.max(...count));
}
My boilerplatey Java solution, beats 23%
class Solution {
public int mostBooked(int n, int[][] meetings) {
Map map = new HashMap();
Queue queue3 = new PriorityQueue( (a,b) -> a.getValue() == b.getValue() ? a.getKey() - b.getKey() : b.getValue() - a.getValue());
Arrays.sort(meetings,(a,b)->a[0]==b[0]?a[1]-b[1]:a[0]-b[0]);
Queue queue1 = new PriorityQueue( (a,b) -> a[0]==b[0]?a[1]-b[1]: a[0] - b[0] );
Queue queue2 = new PriorityQueue();
for (int i=0;i
the c++ version, with the "long long"s is a nuisance 🥲