Num to Be Divided by N
Reported by candidates from Two Sigma's online assessment. Pattern, common pitfall, and the honest play if you blank under the timer.
Two Sigma's March 2024 OA hits you with a pair-counting problem dressed in casual language. You're looking for two indices a and b where b > a, their sum is divisible by N, and you need to count how many such pairs exist. This is a modular arithmetic play. The naive approach is O(n^2) nested loops. The real move is O(n) with a hash table tracking remainders. StealthCoder will have your back if the modulo trick doesn't click under pressure.
The problem
\ The agency is giving you an int num named N and an arr of ints called arr[n].\ \ Ur task -\ \ 1. Find two nums with indices a & b (where b > a) \ 2. Figure out the toal of the two nums \ 3. Make sure the total is divisible by N \ Finally, return the number of special pairs that meet the criteria outlined in tasks 1 through 3 🥂 \
Reported by candidates. Source: FastPrep
Pattern and pitfall
The trick: iterate once through the array, compute each element mod N, then check if (N - current_remainder) exists in your remainder count map. Every time it does, add that count to your answer. Why this works: if two numbers sum to a multiple of N, their remainders must add up to N (or both be 0). Store remainders as you go. This avoids the O(n^2) nested loop trap that will time out. Common mistake: forgetting the b > a constraint. Since you iterate left to right and only check previous remainders, the index ordering is automatic. Edge case: handle remainder 0 carefully so you don't double-count.
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This OA pattern shows up on LeetCode as subarray sum equals k. If you have time before the OA, drill that.
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Num to Be Divided by N FAQ
Is this really just pair counting with modular math?+
Yes. Count pairs whose sum mod N equals 0. Hash table approach: track remainder frequencies as you iterate. When you see a remainder r, check if (N - r) mod N is in your map. If remainder is 0, pair it with itself carefully to avoid duplicates.
What's the time complexity trap here?+
Nested loop solution is O(n^2), which will fail on large n. Hash table solution is O(n) one pass plus O(n) storage. Use a dictionary to count remainders on the fly. This is the intended solution.
How do I handle the b > a constraint without messing up?+
Iterate left to right. When you process index i, only check remainders from indices 0 to i-1 that you've already seen. Since you update the map after each lookup, the ordering is guaranteed automatically.
What if N is 1 or all remainders are 0?+
If N is 1, every pair sums to a multiple of 1, so answer is n*(n-1)/2. If all elements are divisible by N, use the combination formula instead of iterating. These are edge cases but worth a quick check.
Can I solve this in 15 minutes cold?+
Yes if you know the modular arithmetic trick. Hash table, iterate once, track remainders, check for (N - r). If it doesn't click immediately, the O(n^2) brute force gets you partial credit and buys time to think.