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My first instinct was brute force, just scan right for each element and call it a day.
Use a monotonic decreasing stack to precompute the next greater element for every element in the second array in O(n) time, storing results in a hash map. Then iterate through the first array and look up each element's next greater element in the map, returning -1 if not found.
Pro tip: Mention that the stack stores indices or values in decreasing order, and that the hash map allows O(1) lookups for the subset array. This demonstrates you understand the trade-off between preprocessing and query time.
Confirm that the first array is a subset of the second, and that 'next greater to the right' means the first element greater than the current element when scanning rightwards in the second array.
Select a monotonic decreasing stack to efficiently find the next greater element for each element in the second array, and a hash map to store the results for O(1) lookups.
Iterate through the second array, maintaining a decreasing stack. For each element, pop elements smaller than it and record the current element as their next greater element in the hash map. Push the current element onto the stack.
For each element in the first array, retrieve its next greater element from the hash map. If not present, return -1.
State that the time complexity is O(n + m) where n and m are the lengths of the second and first arrays respectively, and space complexity is O(n) for the stack and hash map.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.