← Pinduoduo Interview Insights
The brute force is obvious and they'll let you say it out loud, but you better pivot fast.
Use a monotonic decreasing stack to compute the next greater element for all elements 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 from the map, returning -1 if not found.
Pro tip: Clarify that the first array is a subset of the second, so every element in the first array exists in the second. This allows you to precompute for the entire second array and then answer queries in O(1) per element.
Restate the problem: for each element in nums1, find the first greater element to its right in nums2. Confirm that nums1 is a subset of nums2 and that elements are distinct.
Select a monotonic stack to efficiently find the next greater element for each element in nums2, and a hash map to store the results for O(1) lookups.
Iterate through nums2 from left to right, maintaining a decreasing stack. For each element, pop smaller elements from the stack and record the current element as their next greater element in the map. Push the current element onto the stack.
Iterate through nums1, retrieve the next greater element from the map for each value, and use -1 if the value is not in the map.
State that the time complexity is O(n + m) and space complexity is O(n), where n and m are the lengths of nums2 and nums1. Discuss edge cases like no greater element or empty arrays.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.