Min Ab Difference of Server Loads
Reported by candidates from Google's online assessment. Pattern, common pitfall, and the honest play if you blank under the timer.
Google's 'Min Ab Difference of Server Loads' hit the OA circuit in February 2024, and it's a partition problem dressed up as infrastructure. You've got a list of server loads, and you need to split them into two groups such that the absolute difference between the sums is minimized. This isn't about optimization theory or system design. It's about recognizing that you're solving a variant of the subset-sum partitioning problem, and knowing which approach scales in time. StealthCoder will save you if you freeze on the state transitions.
Pattern and pitfall
The trap is thinking you can greedily balance loads. You can't. The real pattern is dynamic programming. Build a DP table where dp[i] represents whether you can achieve a sum of i using a subset of the loads. Once you've populated that table, find the largest achievable sum that doesn't exceed half the total, then calculate the difference. The common miss: candidates iterate through loads in the wrong order or fail to recognize that you're looking for the optimal partition point, not the optimal assignment. Time complexity is O(n * sum), space is O(sum). If the total sum is large, you'll want to optimize space by using a set instead. StealthCoder handles the indexing and boundary logic so you don't lose points to off-by-one errors mid-OA.
If you see this problem in your OA tomorrow, the play is to recognize the pattern in 30 seconds. StealthCoder buys you that recognition.
You can drill Min Ab Difference of Server Loads cold, or you can hedge it. StealthCoder runs invisibly during screen share and surfaces a working solution in under 2 seconds. The proctor sees the IDE. They don't see what's behind it. Built by an Amazon engineer who passed his OA cold and still thinks the filter is broken.
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Google reuses patterns across OAs. Built by an Amazon engineer who passed his OA cold and still thinks the filter is broken. Works on HackerRank, CodeSignal, CoderPad, and Karat.
Min Ab Difference of Server Loads FAQ
Can I solve this greedily by always adding to the smaller group?+
No. Greedy fails because a large load might force one group to jump ahead permanently. You need DP to explore all valid partitions and find the one with minimal difference. Test it with loads like [1, 5, 11, 5] to see greedy fail.
What if the total sum is huge?+
Use a set instead of a DP array to track achievable sums. Iterate through each load and update the set in-place, building all possible sums incrementally. Space becomes O(number of unique sums) instead of O(total sum).
Do I need to track which loads go into which group?+
No. The problem only asks for the minimum absolute difference, not the actual partition. Build the DP table, find the best sum split point, and return the difference. Ignore the temptation to backtrack.
How tight is the time limit on this one?+
Google typically gives 90 minutes for two problems. This one should take 20-25 minutes once you recognize the DP pattern. If you're stuck on greedy or brute force after 10 minutes, pivot immediately to subset-sum DP.
What's the trick to not TLE on a large input?+
Use the set-based approach instead of a 2D array. Iterate forward through loads, and for each load, add it to every sum already in your set. This keeps memory and runtime proportional to the actual reachable sums, not the theoretical maximum.