Find the Maximum Sum of Node Values
A hard-tier problem at 70% community acceptance, tagged with Array, Dynamic Programming, Greedy. Reported in interviews at Deutsche Bank and 1 others.
Deutsche Bank and BlackRock both ask this one, and it trips up candidates who don't see the greedy angle fast enough. You're given a tree structure with node values, and you need to maximize the sum by selectively applying XOR operations. The trap is thinking you need dynamic programming to track every possible state. You don't. Most candidates waste 20 minutes building a DP table when the real solution is a sorted greedy pass. If you blank on the insight during your live assessment, StealthCoder surfaces the pattern in seconds while the proctor sees nothing.
Companies that ask "Find the Maximum Sum of Node Values"
Find the Maximum Sum of Node Values 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 engineer who got tired of watching his cohort grind for six months and still get filtered at the OA stage.
Get StealthCoderThe problem hinges on recognizing that XOR operations have a specific payoff structure. Once you sort the gain from each XOR by magnitude, you can greedily pick operations in descending order of benefit, subject to a parity constraint on how many you can actually perform. The bit manipulation component isn't about manipulating bits at query time; it's about understanding how XOR changes a value and how to measure that delta. The tree structure is real but secondary. Common failure: jumping to DP recurrence relations on subtrees. Smarter approach: compute all possible XOR gains, sort them, and use a greedy or single-pass validation to confirm feasibility. This is where Array, Greedy, and Bit Manipulation intersect. StealthCoder handles the trick for you live, but knowing the sort-then-validate pattern keeps you ahead.
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Find the Maximum Sum of Node Values 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 engineer who got tired of watching his cohort grind for six months and still get filtered at the OA stage. Works on HackerRank, CodeSignal, CoderPad, and Karat.
Find the Maximum Sum of Node Values interview FAQ
Is this actually a tree DP problem or can you solve it greedily?+
It looks like tree DP but solves greedily. Once you identify all available XOR operations and their payoffs, sort by gain and apply a parity constraint. The tree structure sets up which operations exist, but doesn't require state recursion. That insight separates fast solvers from those who time out on DP.
What's the trick most people miss?+
That you can precompute and sort all XOR gains upfront, then validate in a single pass. Many candidates try to build a DP state machine considering all subsets of operations simultaneously. Sorting makes the problem tractable in linear time after a single O(n log n) pass.
How hard is this really for Deutsche Bank and BlackRock interviews?+
Hard, but not because of the tree or math. Hard because you need to spot the greedy structure under time pressure. Once you see it, coding is straightforward. The acceptance rate sits above 69%, meaning most who recognize the pattern pass it.
How do bit manipulation and tree topology work together here?+
Bit manipulation tells you how each XOR changes a node's value. The tree tells you which pairs of nodes can be XOR'd. Once you've isolated all valid pairs and their gains, topology is done. The rest is greedy scheduling on the sorted gain list.
Should I study tree DP or greedy algorithms more for this?+
Greedy, strongly. The tree sets up constraints, but the solution is a sorted pass with a parity check. If you've drilled classic tree DP problems like path sum or subtree max, this will feel like a fake-out. The real skill is recognizing when DP is overkill.
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