Partition Array Into Two Arrays to Minimize Sum Difference

Partition Array Into Two Arrays to Minimize Sum Difference

What is this problem about?

The Partition Array Into Two Arrays to Minimize Sum Difference problem asks you to split 2n elements into two groups of n to minimize the absolute difference of their sums. This hard coding problem uses meet-in-the-middle with bitmask enumeration and sorted binary search. The array, bitmask, two pointers, binary search, ordered set, bit manipulation, and dynamic programming interview pattern is fully tested.

Why is this asked in interviews?

Texas Instruments, Samsung, Microsoft, Meta, Google, and Bloomberg ask this because brute-force O(2^(2n)) is infeasible but meet-in-the-middle reduces it to O(2^n * n) by splitting the array in half and combining. This technique is a hallmark of hard combinatorial optimization problems.

Algorithmic pattern used

Meet in the middle. Split array into left half (n elements) and right half (n elements). For each subset size k=0..n of the left half, collect all possible subset sums. For each subset size k of the right half, collect all possible subset sums. For each k: combine left sums of size k with right sums of size n-k (total chosen = n elements). Sort right sums, binary search for the complement minimizing difference. Track minimum difference.

Example explanation

arr=[3,9,7,3], n=2. Split: left=[3,9], right=[7,3]. Total=22. Target sum for one group = 11. Left subsets size 0:{0}, size 1:{3,9}, size 2:{12}. Right subsets size 2:{10}, size 1:{7,3}, size 0:{0}. Combine k=0 left(0) + k=2 right(10): group1=10, diff=|2*10-22|=2. k=1 left(3,9) + k=1 right(7,3): 3+7=10(diff=2), 3+3=6(diff=10), 9+7=16(diff=10), 9+3=12(diff=2). Minimum = 2.

Common mistakes candidates make

• Brute-forcing all 2^(2n) subsets (too slow for n=14).
• Not pairing subset sizes correctly (size k from left + size n-k from right = n elements total).
• Not using binary search on sorted right sums.
• Off-by-one in subset size loop.

Interview preparation tip

Meet in the middle is the go-to technique when 2^n is feasible but 2^(2n) is not. Split the input, enumerate half-subsets for each side, sort one side, binary search for the complement. Practice the classic "4-sum" and "partition array" problems using meet-in-the-middle. The key is correctly pairing sizes (k from left, n-k from right) to enforce the equal-size partition constraint.