Divide Intervals Into Minimum Number of Groups | Simple Intuition | Leetcode 2406 | codestorywithMIK
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This is the 115th Video of our Playlist "Array 1D/2D : Popular Interview Problems" by codestorywithMIK
In this video we will try to solve a good Array Interval based Problem : Divide Intervals Into Minimum Number of Groups | Simple Intuition | Dry Run | Leetcode 2406 | 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 : Divide Intervals Into Minimum Number of Groups | Simple Intuition | Dry Run | Leetcode 2406 | codestorywithMIK
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My solutions on Github(C++ & JAVA) - https://github.com/MAZHARMIK/Interview_DS_Algo/blob/master/Arrays/Intervals_Based_Qn/Divide Intervals Into Minimum Number of Groups.cpp
Leetcode Link : https://leetcode.com/problems/divide-intervals-into-minimum-number-of-groups
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Summary :
Sort the intervals: First, the intervals are sorted based on their start times, ensuring that the intervals are processed in the correct order.
Use a Min-Heap (Priority Queue): A min-heap (priority_queue) is used to track the end times of ongoing groups. The heap helps manage overlapping intervals efficiently.
Process intervals: For each interval:
If the current interval's start time is greater than the smallest end time in the heap (i.e., the earliest ending group), the group can be reused. Therefore, we pop the top of the heap.
Push the current interval’s end time into the heap to represent a new or reused group.
Result: The size of the heap at the end represents the minimum number of groups required, as each group corresponds to an active interval at some point.
Time and Space Complexity:
Time Complexity: O(n log n) due to sorting the intervals and heap operations for each interval.
Space Complexity: O(n) for storing end times in the heap.
✨ Timelines✨
00:00 - Introduction
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