← Pinduoduo Interview Insights
My first instinct was brute force, loop through each query element, find its position in nums, then scan right.
First clarify the problem: for each query, find the next greater element to its right in the nums array. Then propose an efficient solution using a monotonic stack to precompute the next greater element for every element in nums, and store the results in a hash map for O(1) query lookups. Analyze the time and space complexity, and discuss edge cases.
Pro tip: Mention that the monotonic stack processes each element once, giving O(n + m) time, which is optimal. Also, note that since all elements are distinct, the next greater element is unique, simplifying the mapping.
Restate the problem to ensure understanding: for each query value, find the first element to its right in nums that is strictly greater, or -1 if none. Confirm that all elements are distinct and that queries refer to values in nums.
Use a monotonic decreasing stack to efficiently compute the next greater element for each element in nums. The stack stores indices of elements in decreasing order of value.
Iterate through nums from left to right. For each element, while the stack is not empty and the current element is greater than the element at the stack's top index, pop and record the current element as the next greater for that index. Push the current index onto the stack.
After processing, create a hash map where keys are the elements of nums and values are their next greater elements (or -1 if none). This allows O(1) lookup for each query.
For each query, look up its next greater element in the hash map and return the result. Discuss time complexity O(n + m) and space complexity O(n), where n is the length of nums and m is the number of queries.
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