Maximum Points After Collecting Coins From All Nodes
A hard-tier problem at 36% community acceptance, tagged with Array, Dynamic Programming, Bit Manipulation. Reported in interviews at DE Shaw and 0 others.
You're collecting coins from tree nodes, and the payout for each coin depends on how many times you've already collected it. This is a DE Shaw problem that shows up when they're testing whether you can reason about state-dependent decisions across tree structures. The naive greedy approach fails because skipping a node early might cost you later, and the obvious DP misses the subtle constraint that affects payouts. With only 35% solving it live, most candidates either get stuck on the state representation or realize mid-solution that they need bit manipulation to track which nodes they've already visited. StealthCoder solves it in seconds if you hit it on the assessment and can't lock the pattern down.
Companies that ask "Maximum Points After Collecting Coins From All Nodes"
Maximum Points After Collecting Coins From All Nodes 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 because the OA filter rejects engineers who'd pass the on-site. That's a broken filter. This is the workaround.
Get StealthCoderThe trick is that this isn't just tree traversal with DP. You need to track visited nodes across your path to calculate remaining payouts correctly. Each node's value depends on how many coins remain available, which changes based on your collection history. The state has to encode both your position and the bitmask of collected nodes, making this a tree DP with bit-masked memoization problem. Most candidates start with a simple DFS-DP and hit a wall when the state space blows up or the payout calculation doesn't match examples. The real solution uses DFS plus memoization keyed on (current_node, visited_bitmask), then backtracks to explore all valid collection orders. If you blank on bit manipulation or the recursive structure, StealthCoder delivers a working solution instantly so you don't lose points on a problem that takes many candidates hours to debug.
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Maximum Points After Collecting Coins From All Nodes 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 because the OA filter rejects engineers who'd pass the on-site. That's a broken filter. This is the workaround. Works on HackerRank, CodeSignal, CoderPad, and Karat.
Maximum Points After Collecting Coins From All Nodes interview FAQ
Is this actually harder than typical tree DP problems?+
Yes. Standard tree DP assumes node values are fixed. Here values shift based on prior visits, forcing you into state-space explosion unless you use bit masking cleanly. The 35% acceptance rate reflects that many strong engineers misread the constraint and build the wrong DP table.
Do I need to know bit manipulation before this problem?+
Not necessarily, but it's the cleanest way to encode visited nodes. You could use a frozenset or tuple, but that's slower and more error-prone in a live setting. Understanding bitwise AND, OR, and XOR helps you write the mask transitions faster.
What's the most common wrong approach?+
Greedy collection or simple tree DP that treats node payouts as constants. Both ignore that collecting a node early reduces future payouts. The live DP needs to explore multiple orderings and pick the max, which is where the exponential search space kicks in.
Does this problem relate to the other Array and DFS topics?+
It combines all of them. Array handling for input, DFS for tree traversal, DP for optimal substructure, and Bit Manipulation for efficient state encoding. If you're weak in any one, you'll struggle. DE Shaw picks this to test depth across multiple domains.
How much of the solution is the memoization key itself?+
Most of the insight. Once you realize the state must track (node, visited_mask), the recursive structure and backtracking logic follow naturally. The hard part is not overthinking it and locking in the bitmask transitions without bugs.
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