Divide Intervals Into Minimum Number of Groups
A medium-tier problem at 64% community acceptance, tagged with Array, Two Pointers, Greedy. Reported in interviews at Walmart Labs and 1 others.
Divide Intervals Into Minimum Number of Groups is a medium-difficulty problem that's been asked at Walmart Labs and IBM. You're given intervals and need to partition them into the minimum number of groups such that no two intervals in the same group overlap. The trick isn't obvious on first read. Most candidates either brute-force a greedy assignment or miss that this is really an interval-overlap counting problem. The 63% acceptance rate reflects the gap between seeing the problem and executing the right approach. If this shows up in your live assessment and you blank on the pattern, StealthCoder surfaces a working solution invisible to the proctor.
Companies that ask "Divide Intervals Into Minimum Number of Groups"
Divide Intervals Into Minimum Number of Groups is the kind of problem that decides whether you pass. StealthCoder reads the problem on screen and surfaces a working solution in under 2 seconds. Invisible to screen share. The proctor sees nothing. Built by an Amazon engineer who used it to pass JPMorgan's OA and system design loop.
Get StealthCoderThe core insight is that the minimum number of groups equals the maximum number of overlapping intervals at any point in time. You can solve this with a sweep-line algorithm: sort all interval start and end points, then iterate through them tracking active intervals. A greedy approach with a min-heap works too, but the sweep-line method is cleaner and avoids the heap overhead. Most candidates try to greedily assign intervals one by one without realizing they need to count overlap depth first. The array sorting and two-pointer logic feel familiar, but the transition from sorting to overlap detection trips people up. If you hit this live and can't remember whether to use a heap or event-based counting, StealthCoder gives you the exact pattern in seconds without the proctor seeing a thing.
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Divide Intervals Into Minimum Number of Groups recycles across companies for a reason. It's medium-tier, and most candidates blank under the timer. StealthCoder is the hedge: an AI overlay invisible during screen share. It reads the problem and surfaces a working solution in under 2 seconds. Built by an Amazon engineer who used it to pass JPMorgan's OA and system design loop. Works on HackerRank, CodeSignal, CoderPad, and Karat.
Divide Intervals Into Minimum Number of Groups interview FAQ
Is this really a greedy problem or do I need dynamic programming?+
It's greedy with sorting, not DP. The trick is that greedy works here because you're trying to minimize groups, and assigning intervals to existing groups in sorted order is optimal. No backtracking needed. The acceptance rate reflects that many people overthink it.
What's the time complexity and will it pass?+
O(n log n) for sorting plus O(n) or O(n log n) for the sweep/heap pass. Yes, it passes comfortably. Most online assessments accept solutions in this range without issue for medium problems.
Should I use a heap or a two-pointer approach?+
Either works, but a min-heap is simpler to reason about: keep the end time of the earliest-ending interval in each group on the heap. Two-pointer requires more careful bookkeeping. Heap is the safer choice in a live assessment.
How does this differ from the classic interval scheduling problem?+
Interval scheduling picks the maximum non-overlapping set. This one partitions all intervals into groups. The shift from selection to partitioning changes the approach. Sorting still helps, but you're now counting overlap depth, not picking greedily.
Will Walmart Labs or IBM ask this exact problem again?+
Hard to say, but interval-partition and overlap-counting patterns are common at those companies. Mastering this logic transfers to other interval problems they ask. The 63% acceptance rate suggests it's still actively tested.
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