Check If Array Pairs Are Divisible by k
A medium-tier problem at 46% community acceptance, tagged with Array, Hash Table, Counting. Reported in interviews at DevRev and 1 others.
You're asked to determine whether an array can be split into pairs where each pair's sum is divisible by k. DevRev and Visa have both asked this. It looks like a pairing problem until you realize the trick: you don't actually form pairs. Instead, you count remainders and match them mathematically. The 46% acceptance rate reflects how easily candidates miss that insight and waste time on greedy or simulation approaches that don't scale. If this problem hits your live assessment and you blank on the remainder-matching pattern, StealthCoder solves it in seconds invisible to the proctor.
Companies that ask "Check If Array Pairs Are Divisible by k"
Check If Array Pairs Are Divisible by 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 by an engineer at a top-10 tech company who can solve these problems cold but didn't want to trust himself in a 90-minute screen share.
Get StealthCoderThe core insight is remainder arithmetic. When you divide each array element by k, only the remainder matters. Two numbers with remainders r1 and r2 sum to a multiple of k if and only if r1 + r2 equals k (or both are 0). Hash a frequency count of remainders, then iterate through them and check if complementary remainders pair up correctly. The edge case that kills most solutions: remainder 0 and remainder k/2 (when k is even) must appear an even number of times, since they pair with themselves. Candidates who try to form actual pairs or use sorting often run into off-by-one errors or O(n log n) solutions when O(n) is available. This is where the Array and Hash Table topics intersect cleanly. StealthCoder is the safety net if you freeze on how to validate the remainder pairing logic under pressure.
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Check If Array Pairs Are Divisible by k recycles across companies for a reason. It's medium-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 at a top-10 tech company who can solve these problems cold but didn't want to trust himself in a 90-minute screen share. Works on HackerRank, CodeSignal, CoderPad, and Karat.
Check If Array Pairs Are Divisible by k interview FAQ
Why doesn't a greedy pairing approach work here?+
Greedy fails because the order of pairing matters, and you can always rearrange to hit any valid pairing if one exists. The problem isn't asking for a specific pairing, just whether one exists. Remainder counting sidesteps the pairing altogether and runs in one pass.
Is this problem still asked at top companies?+
Yes. DevRev and Visa both have recent reports of asking it. The 46% acceptance rate suggests it's not a gimme, so it's still used to filter candidates. It appears often enough that missing the remainder trick in your live OA would be costly.
What's the hardest part to code correctly?+
Handling the self-pairing edge cases. Remainder 0 pairs with itself (must be even count). Remainder k/2 when k is even also pairs with itself (must be even count). Missing either check causes a wrong answer on test cases with duplicates.
How does this relate to the Hash Table topic?+
You build a hash map of remainder frequencies. Then you iterate the map and match each remainder to its complement (k - remainder). The map lookup is O(1), so overall time is O(n). The counting step is the entire solution.
Should I sort the array first?+
No. Sorting adds O(n log n) and doesn't help. The remainder approach is O(n) time and O(k) space since remainders are bounded by k. Sorting is a red herring that wastes time on a live OA.
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