Triples with Bitwise AND Equal To Zero
A hard-tier problem at 59% community acceptance, tagged with Array, Hash Table, Bit Manipulation. Reported in interviews at Flipkart and 0 others.
You're staring at an array and the problem asks for triplets where the bitwise AND equals zero. It sounds simple until you realize the brute force is O(n^3) and timeouts are guaranteed. Flipkart has asked this. The trick isn't obvious: you need to recognize that AND can only decrease (or stay the same) as you combine more numbers, and that the number of distinct AND values across any subarray is logarithmic. Once that pattern clicks, the solution shifts from brute force to something tractable. If you hit this live and freeze on the bit manipulation angle, StealthCoder surfaces the working approach in seconds, invisible to the proctor.
Companies that ask "Triples with Bitwise AND Equal To Zero"
Triples with Bitwise AND Equal To Zero 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. Made by an Amazon engineer who watched the leaked-problem repo become an industry secret. He decided you should have it too.
Get StealthCoderThe core insight is that bitwise AND between numbers always decreases or stays the same as you add more operands. This means for any starting index, the set of distinct AND values you can form with subsequent elements is small, roughly O(log max_value). The solution strategy: iterate through the array, and for each position, maintain a hash table of distinct AND values formed with all previous elements. For each new element, compute its AND with every distinct AND already seen, count triplets, and add the new AND values to your set. This reduces what looks like O(n^3) down to O(n^2 log max_value), which passes. The pitfall most candidates hit is trying to track all triplets explicitly instead of using the AND property to prune the search space. Bit Manipulation and Hash Table work together here, not separately.
Pattern tags
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Triples with Bitwise AND Equal To Zero 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. Made by an Amazon engineer who watched the leaked-problem repo become an industry secret. He decided you should have it too. Works on HackerRank, CodeSignal, CoderPad, and Karat.
Triples with Bitwise AND Equal To Zero interview FAQ
Is this really asked at top companies or is it just Flipkart?+
It's in Flipkart's rotation and shows up as a Hard problem across major platforms. The acceptance rate is around 59%, so it's tough but solvable. It tests both bit manipulation depth and algorithm optimization, which is why it stays in the funnel.
What's the trick I'm missing if I only think O(n^3)?+
The trick is the logarithmic bound on distinct AND values. AND can only decrease, so iterating from each position backwards, you hit at most log(max_value) unique AND results, not n results. That's what kills the naive approach and opens the door to a hash table solution.
Do I need to know advanced bit manipulation or is standard AND enough?+
Standard AND is enough, but you need to internalize that AND has this monotonic property. You don't need XOR, bit counting, or Gray codes. The hard part is seeing the logarithmic bound, not the bit operations themselves.
How does this relate to the other Array and Hash Table problems I've drilled?+
It combines both topics: Array iteration and Hash Table for fast lookup and aggregate tracking. Unlike many array-hash problems though, the real optimization comes from understanding the bit property, not the data structure alone.
Is this problem still asked or is it older?+
It's active. The acceptance rate at 59% and the Hard tag mean it's recent enough to stay relevant. If Flipkart is still hiring engineers for roles that need algorithmic depth, this pattern is still in rotation.
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