Removing Minimum Number of Magic Beans

Removing Minimum Number of Magic Beans

What is this problem about?

The Removing Minimum Number of Magic Beans interview question gives you an array of non-negative integers representing magic bean counts in bags. You want to make the array such that all non-empty bags (bags with beans > 0) have the same number of beans. To achieve this, you can remove beans (but not add). You must find the minimum total number of beans to remove.

Why is this asked in interviews?

DE Shaw asks this problem because it combines sorting, prefix sums, and enumeration in an elegant way. The key insight is that the target "remaining count" for non-empty bags must be one of the existing values in the array. By enumerating each possible target and computing the removal cost using prefix sums, you achieve an O(n log n) solution. It tests mathematical reasoning about optimization with discrete choices.

Algorithmic pattern used

The pattern is sort, prefix sum, and enumeration. Sort the array. Compute the prefix sum. For each index i, consider setting the target count to nums[i] (all bags at index i and beyond keep their count, reduced to nums[i], and all bags before index i are emptied). Cost = prefix_sum[i] (beans removed from left bags) + (nums[i] * (n - i) subtracted from total of right bags). Use the total sum and prefix sum to compute efficiently.

Specifically, cost for target nums[i] = prefix[i] + (total - prefix[i+1]) - nums[i] * (n - i - 1).

Simplified: prefix[i] + total - prefix[i] - nums[i] * (n - i - 1) — then take the minimum over all i.

Example explanation

Array: [4, 1, 6, 5]. Sorted: [1, 4, 5, 6]. Total = 16.

• Target = 1 (i=0): empty nothing from left, right bags become 1 each. Remove: (4-1)+(5-1)+(6-1)=3+4+5=12. Plus prefix[0]=0. Total removed: 12.
• Target = 4 (i=1): empty bag with 1 (remove 1). Right bags: (5-4)+(6-4)=1+2=3. Total removed: 1+3=4.
• Target = 5 (i=2): empty bags with 1,4 (remove 5). Right: (6-5)=1. Total removed: 5+1=6.
• Target = 6 (i=3): empty bags with 1,4,5 (remove 10). No right bags. Total removed: 10.

Minimum: 4.

Common mistakes candidates make

• Not sorting the array before applying the prefix sum strategy.
• Forgetting to include the option of making all bags empty (equivalent to target = 0, which is covered by choosing to empty everything).
• Off-by-one errors in prefix sum indices.
• Computing the right-side cost without subtracting the target multiplied by the count of right bags.

Interview preparation tip

For the Removing Minimum Number of Magic Beans coding problem, the array and prefix sum interview pattern combined with sorting and enumeration is the key. Practice computing prefix sums on sorted arrays quickly. The insight that the target must equal one of the array values (sorting enables this) is non-obvious — explain it clearly in your interview at DE Shaw. Walk through cost computation formula derivation step by step to demonstrate mathematical rigor.