The Number of the Smallest Unoccupied Chair | Brute Force | Optimal |Leetcode 1942 |codestorywithMIK
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This is the 114th Video of our Playlist "Array 1D/2D : Popular Interview Problems" by codestorywithMIK
In this video we will try to solve a good Array Practice Problem : The Number of the Smallest Unoccupied Chair | Brute Force | Optimal | Full Dry Run | Leetcode 1942 | codestorywithMIK
I will explain the intuition so easily that you will never forget and start seeing this as cakewalk EASYYY.
We will do live coding after explanation and see if we are able to pass all the test cases.
Also, please note that my Github solution link below contains both C++ as well as JAVA code.
Problem Name : The Number of the Smallest Unoccupied Chair | Brute Force | Optimal | Full Dry Run | Leetcode 1942 | codestorywithMIK
Company Tags : Uber, Meta, Microsoft, Yelp, Google, Snapchat, Amazon, Cisco - Qn had small Variations
My solutions on Github(C++ & JAVA) - https://github.com/MAZHARMIK/Interview_DS_Algo/blob/master/Arrays/Intervals_Based_Qn/The Number of the Smallest Unoccupied Chair.cpp
Leetcode Link : https://leetcode.com/problems/the-number-of-the-smallest-unoccupied-chair
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Summary :
Naive O(n^2) Approach:
Time Complexity: O(n^2)
Space Complexity: O(n)
Explanation:
This approach first sorts the times based on the arrival time to maintain the original friend order.
For each friend's arrival, it searches for the first available chair (linear scan through endTimes array).
Once an unoccupied chair is found, it updates the endTimes array with that friend's departure time.
If the arrival time matches the target friend, it returns the chair index.
The time complexity is quadratic because for each friend, it linearly searches through the chair list to find an available chair.
Min-Heap Approach:
Time Complexity: O(nlogn)
Space Complexity: O(n)
Explanation:
This approach uses two min-heaps:
One for tracking occupied chairs with their departure times.
Another for tracking available chairs (unoccupied).
It sorts the times based on the arrival time.
For each friend's arrival, it frees up any chairs that have become available (those with a departure time less than or equal to the current arrival time).
If no chair is available, a new chair is assigned. Otherwise, the least numbered chair from the free heap is used.
When the target friend's arrival time is encountered, it returns the corresponding chair.
This approach improves efficiency by using heaps, reducing the time complexity to O(nlogn) due to the heap operations.
✨ Timelines✨
00:00 - Introduction
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