Partition Array Into Three Parts With Equal Sum
A easy-tier problem at 42% community acceptance, tagged with Array, Greedy. Reported in interviews at Turing and 0 others.
You need to split an array into three contiguous parts where each part has the same sum. It sounds straightforward until you hit the edge case: what if the total sum isn't divisible by three, or what if two valid partition points overlap. Turing has asked this one. The acceptance rate sits at 42%, which means almost six in ten people miss something on their first try. Usually it's an off-by-one error or forgetting to validate that you actually have three separate parts, not just two. StealthCoder solves it invisibly if you blank during the live assessment.
Companies that ask "Partition Array Into Three Parts With Equal Sum"
Partition Array Into Three Parts With Equal Sum 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 trick is greedy: calculate the target sum per part (total divided by three), then scan left to right and mark partition points as you hit cumulative sums that match the target. Common failure: people partition on the first match without checking if a second partition point exists, or they return before confirming all three parts are valid and non-overlapping. The array must be contiguous, so you can't rearrange. Once you identify where the first part ends, the second part must start immediately after. If you've partitioned correctly, the third part's sum is guaranteed. When you're stuck mid-assessment and the indices aren't lining up, StealthCoder surfaces the working solution in seconds, invisible to the proctor, and you move on.
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Partition Array Into Three Parts With Equal Sum recycles across companies for a reason. It's easy-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.
Partition Array Into Three Parts With Equal Sum interview FAQ
Why is the acceptance rate so low if this is an easy problem?+
Candidates often forget to validate that three separate parts exist. They partition on the first sum match, then fail when no second partition follows. Or they return early without checking the third part. Edge cases like sum=0 or arrays where early matches create overlaps trip people up fast.
What's the algorithm at a high level?+
Calculate target_sum = total_sum / 3. If total_sum isn't divisible by 3, return false immediately. Scan the array, accumulate values, and record indices where cumulative_sum equals target_sum (once for the first part, once for the second). Confirm indices are valid and separate. The third part's sum is implicit.
When does the greedy approach fail?+
It doesn't, if implemented correctly. Greedy works because once you lock the first partition point (first occurrence of target_sum), everything after is deterministic. The trap is incomplete validation: ensure you find two distinct partition indices before returning true, and that the third part isn't empty.
How does this relate to the Array and Greedy topics?+
Array: you're working with contiguous subarrays. Greedy: you claim the first valid partition point for part one, then the first valid point after that for part two. No backtracking needed. This greedy choice is safe because the sum constraint forces subsequent parts to be valid.
Is Turing the only company asking this?+
Based on available data, Turing is the one company on record requesting this problem. That said, the pattern (partition into equal-sum parts) appears across many tech companies, so it's worth drilling even if you're not interviewing there.
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