← Walmart Labs Interview Insights
This one took me a minute to even understand what they were asking.
Sort both arrays and use a greedy two-pointer strategy: for each element in sorted A, find the smallest element in sorted B that is strictly greater, sum it, and remove it. This maximizes the sum by ensuring each winning B is as small as possible, preserving larger elements for harder-to-beat A's.
Pro tip: After solving, discuss how the greedy choice is optimal via an exchange argument, and mention that the problem is equivalent to maximizing the sum of a maximum matching in a bipartite graph where edges exist if B[i] > A[i].
Restate the goal: permute B to maximize the sum of B[i] where B[i] > A[i]. Clarify that only strictly greater elements contribute, and each B element can be used at most once.
Sort A and B in ascending order. This simplifies the process of finding the smallest B that beats each A.
Use two pointers: iterate through sorted A, and for each A[i], advance a pointer in B until finding the smallest B[j] > A[i]. If found, add B[j] to sum and mark it used; otherwise, skip A[i].
Explain that the greedy choice is safe: using the smallest possible B for each A leaves larger B's available for other A's, which can only increase or maintain the sum.
Sorting takes O(n log n), and the two-pointer pass takes O(n), so overall O(n log n) time and O(1) extra space (if sorting in place).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.