Cycle Length Queries in a Tree
A hard-tier problem at 58% community acceptance, tagged with Array, Tree, Binary Tree. Reported in interviews at Arcesium and 0 others.
Cycle Length Queries in a Tree is a hard-tier problem that forces you to think about tree traversal and path queries in ways most candidates skip during prep. Arcesium has asked it. The acceptance rate sits around 58 percent, which looks deceptive, that number hides a wall where most people either misunderstand the problem or build the wrong data structure. You'll see this in your OA and blank on the trick. StealthCoder solves it in seconds, invisible to the proctor, if you hit that wall.
Companies that ask "Cycle Length Queries in a Tree"
Cycle Length Queries in a Tree 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 realized the OA tests how well you memorized 200 problems, not how well you code.
Get StealthCoderThe trap is thinking you need to recompute paths for every query. What actually works: you build a parent-pointer tree, find the lowest common ancestor between two nodes, then compute the cycle length in constant time using the depth values you've cached. Most candidates either implement full tree reconstruction per query (timeout) or try to precompute all pairs (memory bomb). The problem tests whether you understand ancestor relationships and can derive cycle length from LCA and depth. Common pitfall: off-by-one errors when summing distances. If you're not immediately clear on the LCA pattern or how to extract cycle length from it, StealthCoder surfaces the working solution during the assessment so you move to the next problem instead of burning 30 minutes.
Pattern tags
You know the problem.
Make sure you actually pass it.
Cycle Length Queries in a Tree recycles across companies for a reason. It's hard-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 realized the OA tests how well you memorized 200 problems, not how well you code. Works on HackerRank, CodeSignal, CoderPad, and Karat.
Cycle Length Queries in a Tree interview FAQ
Is this really asked in live interviews at big companies, or just Arcesium?+
Arcesium is the reported company in this data. Tree problems with ancestor queries show up across multiple firms, but this exact variant is lower frequency. Don't bank on seeing it, but know it's live possibility. The pattern, caching parent pointers and computing LCA, transfers to other hard tree problems.
What's the main trick that separates 'solves' from 'times out'?+
Precomputing parent pointers and depth values for every node, then using those to find LCA in O(log n) or O(height). Without this, you're recomputing ancestors per query. The cycle length is then depth(u) + depth(v) - 2 * depth(lca). Miss that formula and you rebuild the whole tree every query.
How does this connect to the Array and Binary Tree topics listed?+
The array part is storing parent and depth pointers in a flat structure for O(1) lookup. Binary Tree is the underlying structure. You're not building a balanced BST or heap. You're using tree properties to answer path queries fast.
Is 58 percent acceptance rate actually hard, or are candidates just rushing?+
58 percent is the result of candidates who understand LCA and depth caching solving it cleanly, but a lot of people misread the problem or implement the naive approach. It's hard because the insight isn't obvious on first read. It's not a 20-minute problem.
What should I drill before an OA that includes this type of problem?+
Practice building a parent-pointer tree from a given structure, computing depth for all nodes in one pass, and writing an LCA function that uses those cached values. Do that three times. Then do one full Cycle Length problem end-to-end. That's enough coverage for the pattern.
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