2402. Meeting Rooms III | Sorting | Heap(Priority Queue) | Hard 🥶

In this video, I'll talk about how to solve Leetcode 2402. Meeting Rooms III | Sorting | Heap(Priority Queue) | Hard 🥶
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[@dharmrajmagdum9093]

at 30:27 ha ha ha🤣😂🤣😂 , maksad nahi bhulna hai, ie. return id of room with max meeting held. by the way , great explanation as all ways.

[@fraserdab]

This whole week is heaps man

[@onlysparsh]

maksad nahi bhulna hei

[@prakashnandan7337]

🤣😂

[@fraserdab]

I feel like its better to use an array instead of a map or unordered map and find max in ond way or another then return index of first max element

[@mayankrohilla4105]

Appreciate the explanation very much! just a little suggestion: try to make smaller videos

[@ankushmitra6504]

Did you use any extensions for knowing company tag in leetcode?

[@saunaknandi1814]

Bro he have subscription

[@yuvrajveerdhawan8196]

bro please solve the today weekly contest also

[@harshal8781]

Meeting rooms l,ll solve karne ke liye premium lagega 😢

[@agambedi4142]

bro a small request can you please provide JAVA code too? I try firstly by myself but questions like this is very tricky and need some overview of a solution, please provide java code too.

[@saunaknandi1814]

Bro why are you hiding those company names?

[@harshal8781]

publicly aise show karege to leetcode 🤐

[@akashspace4977]

I have written the same code in JAVA, but all the test cases are not passing. Can anyone help? class Pair { int endTime; int id; Pair(int endTime, int id) { this.endTime = endTime; this.id = id; } } class Solution { public int mostBooked(int n, int[][] meetings) { Arrays.sort(meetings, new Comparator() { @Override public int compare(int[] e1, int[] e2) { if (e1[0] != e2[0]) { if (e1[0] < e2[0]) { return -1; } else { return 1; } } else { if (e1[1] < e2[1]) { return -1; } else { return 1; } } } }); TreeMap booked = new TreeMap(); PriorityQueue free = new PriorityQueue(); Comparator c = new Comparator() { @Override public int compare(Pair p1, Pair p2) { if (p1.endTime == p2.endTime) { if (p1.id < p2.id) { return -1; } else { return 1; } } else { if (p1.endTime < p2.endTime) { return -1; } else { return 1; } } } }; PriorityQueue full = new PriorityQueue(c); for (int i = 0; i < n; i++) { free.add(i); } for (int i = 0; i < meetings.length; i++) { int start = meetings[i][0], end = meetings[i][1]; // If there is some room which was occupied but now is free while (!full.isEmpty() && full.peek().endTime maxMeetings) { maxRoomId = room; maxMeetings = booked.get(room); } } return maxRoomId; } }

[@divyanksrivastava7229]

same

[@divyanksrivastava7229]

hey, I tried to take data type as long and passes all test case give it a try class Solution { public int mostBooked(int n, int[][] meetings) { long[] freq = new long[n]; // Change to long array PriorityQueue free = new PriorityQueue(); PriorityQueue pq = new PriorityQueue(Comparator.comparingLong((ArrayList a) -> a.get(1)).thenComparingLong(a -> a.get(0))); for (int i = 0; i < n; i++) { free.add(i); } Arrays.sort(meetings, Comparator.comparingInt(a -> a[0])); for (int[] a : meetings) { int start = a[0]; int end = a[1]; while (!pq.isEmpty() && pq.peek().get(1) size) { size = freq[i]; ans = i; } } return ans; } }