Maximum Strong Pair XOR II
A hard-tier problem at 31% community acceptance, tagged with Array, Hash Table, Bit Manipulation. Reported in interviews at ZScaler and 0 others.
Maximum Strong Pair XOR II is a hard problem with a 31% acceptance rate, asked by ZScaler. You're given an array and need to find the maximum XOR of two elements that form a strong pair, meaning their absolute difference stays within a threshold. The trick is that a brute-force check of every pair times out. This problem combines bit manipulation with a constraint that kills naive solutions. If you hit this in a live OA and panic on the optimization, StealthCoder solves it in seconds, invisible to the proctor.
Companies that ask "Maximum Strong Pair XOR II"
Maximum Strong Pair XOR II 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 naive approach checks all pairs and validates the strong pair condition, but that's O(n^2) and fails on large inputs. The real solution uses a Trie to build a max-XOR lookup while a sliding window maintains only valid pairs. You insert elements into the Trie, slide the left pointer to keep the pair constraint satisfied, and query the Trie for the maximum XOR at each step. The insight is that XOR maximization (choosing 1 bits greedily) and the strong pair constraint work together if you sort the array first and slide correctly. Most candidates either miss the Trie optimization or mess up the window bounds. StealthCoder handles the Trie construction, pointer logic, and the bit-by-bit XOR maximization so you don't crater on implementation.
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Maximum Strong Pair XOR II 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.
Maximum Strong Pair XOR II interview FAQ
Why does sorting help here?+
Sorting lets you use a sliding window on the strong pair constraint. If abs(arr[j] - arr[i]) > threshold, then all elements before i are also too far from j. This turns the problem from O(n^2) to O(n log n) with Trie help. Without sorting, you can't efficiently prune pairs.
What's the role of the Trie?+
The Trie stores numbers in binary form. At each window position, you query it greedily, starting from the most significant bit, to find the element already in the window that produces the maximum XOR with the current element. This is O(log max_value) per query, much faster than checking all pairs.
Is this problem still asked at companies like ZScaler?+
Yes, ZScaler has reported it. Hard bit-manipulation problems with sliding-window twists remain common at companies building systems-level software or working with network protocols. Expect it in senior or backend-focused rounds.
What's the most common mistake?+
Forgetting to remove elements from the Trie when the window slides left. If you don't delete stale entries, you'll compute XOR with elements that violate the strong pair constraint, giving wrong answers on edge cases.
How does this relate to the Bit Manipulation topic?+
XOR maximization requires thinking in binary, you want to flip as many high-order bits as possible. The Trie exploits this by branching left (0) or right (1) and always preferring the opposite bit of your current number. Understanding XOR at the bit level is non-negotiable.
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