Find Subarray With Bitwise OR Closest to K
A hard-tier problem at 30% community acceptance, tagged with Array, Binary Search, Bit Manipulation. Reported in interviews at Salesforce and 0 others.
Salesforce asked this hard problem once, and candidates who haven't seen the bitwise OR + binary search combo often blank on it. You're given an array and a target K. The goal is to find a contiguous subarray whose bitwise OR is closest to K. The acceptance rate sits at 30 percent, which means most people either brute-force it or miss the optimization entirely. You can't just try all subarrays naively. If this problem hits your live OA and you freeze on the approach, StealthCoder solves it invisibly and surfaces the working solution in seconds.
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Find Subarray With Bitwise OR Closest to K 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 trap is thinking you need to check every possible subarray. That's O(n squared) or worse. The real insight is that the bitwise OR only grows monotonically as you expand a subarray. Once you fix a starting position, the number of unique OR values is capped at roughly 30 (the bit width), not n. This lets you use binary search or a two-pointer technique to stay efficient. You'll iterate through potential subarrays, compute their OR values, and track which one gets closest to K. The common failure is implementing the brute force and timing out, or missing the monotonic property altogether. Segment Tree is mentioned in the topics but usually overkill here. When you hit this live and the naive solution times out, StealthCoder's answer walks you through the optimized scan with early termination and the correct distance calculation.
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Find Subarray With Bitwise OR Closest to K 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.
Find Subarray With Bitwise OR Closest to K interview FAQ
Is this really a 30 percent problem or am I just bad at bit manipulation?+
The 30 percent acceptance rate reflects the bitwise monotonicity trick. Most candidates who bomb this problem didn't know that OR values can't decrease as you expand a subarray, so they code a brute-force O(n squared) solution. Once you know that property, the problem becomes tractable. Bit manipulation itself isn't the barrier, recognizing the constraint is.
How does Binary Search fit into this problem?+
Binary Search helps optimize the inner loop. For a fixed starting position, you can binary search on the ending position to narrow down which subarrays might have OR values close to K. Alternatively, you iterate carefully and stop early when you've seen all distinct OR values for that start, since OR can't grow anymore.
Do I really need a Segment Tree for this?+
No. Segment Tree is listed in the topics but is almost never the intended solution. The monotonicity property and efficient iteration through unique OR values is enough. Segment Tree adds complexity without real benefit here.
What's the actual runtime I should aim for?+
O(n log n) or O(n * log(max_value)) is typical for the optimized approach. Since OR values can have at most 30 unique states per starting position, you iterate through O(n) starts and O(30) unique ORs, making it roughly O(30n). Segment Tree solutions may also hit O(n log n) but are unnecessary.
Did Salesforce ask this once or multiple times?+
The data shows Salesforce in the list, but only one company reported it. This is a rare, high-difficulty ask. If it shows up in your OA, it's likely a difficult round meant to separate top performers. Expect the problem setter to know the bitwise monotonicity trick.
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