Most Profitable Path in a Tree
A medium-tier problem at 68% community acceptance, tagged with Array, Tree, Depth-First Search. Reported in interviews at Intuit and 1 others.
Most Profitable Path in a Tree is a medium-difficulty tree problem that Intuit and Snowflake have both asked in their technical rounds. You're given a tree where each node has a profit value, and you need to find the path from root to leaf that maximizes total profit while respecting a time constraint. The trick isn't the tree traversal itself, it's handling the dual optimization: profit and time simultaneously. If you blank on how to thread both constraints through DFS, you lose points fast. This is exactly where StealthCoder runs invisibly during your live OA and surfaces a working solution in seconds.
Companies that ask "Most Profitable Path in a Tree"
Most Profitable Path 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 used it to pass JPMorgan's OA and system design loop.
Get StealthCoderThe problem requires depth-first search to explore all root-to-leaf paths, but the real complexity is tracking profit accumulation while ensuring you don't exceed the time limit as you traverse. Most candidates try a greedy approach first and hit a wall. The correct pattern is to calculate the maximum profit for each valid path by running DFS and maintaining both the current profit sum and elapsed time at each node, then comparing all leaf-node results. The tree structure lets you prune invalid branches early if time runs out, but you still need to explore multiple paths to find the global maximum. If you've drilled Array and Depth-First Search problems before, this problem chains those concepts together in a way that feels familiar until it doesn't. StealthCoder is the hedge for the one moment you can't see how to merge the two constraints cleanly.
Pattern tags
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Make sure you actually pass it.
Most Profitable Path in a Tree 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.
Most Profitable Path in a Tree interview FAQ
Is this problem really asked at Intuit and Snowflake?+
Yes. Both companies appear in recent reports for this problem. It's medium difficulty, so expect it in phone screen or early technical round. At a 67% acceptance rate, it's not a hard filter, but it's also not trivial. Most candidates who see it unprepared struggle with the dual constraint.
What's the core trick I'm missing if I fail this live?+
You're likely trying greedy (pick the highest profit at each step) or forgetting to track time alongside profit during traversal. The trick is running complete DFS on all paths, storing both metrics, then comparing at leaf nodes. It's not fancy, just methodical.
Does this problem test graph knowledge or just trees?+
It's rooted in tree structure, but the logic overlaps with graph DFS patterns. If you're strong on Depth-First Search fundamentals, the tree aspect is straightforward. The challenge is the state management during traversal, not the graph concept itself.
How do I avoid timeout on large inputs?+
Prune aggressively. If elapsed time exceeds the limit mid-traversal, backtrack immediately. Don't explore subtrees you know are invalid. Memoization of subproblems can help if the tree is deep and paths overlap, but standard DFS with early termination usually suffices.
Should I solve this with BFS or DFS?+
DFS is the natural fit. You need to explore complete root-to-leaf paths and compare their profit totals at the end. BFS would require tracking path state differently and is more cumbersome. DFS lets you naturally backtrack and maintain the current sums.
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