Use a hash map to store movie lengths and their indices while iterating through the array. For each movie, check if the complement (flight duration minus current movie length) exists in the map; if so, return the pair of indices. This yields O(n) time and O(n) space.
Pro tip: Clarify edge cases upfront, such as whether a movie can be paired with itself (i != j) and if the flight duration can be zero or negative. Also, mention that you'd handle duplicates by storing the first occurrence or using a set of seen complements.
Confirm that i != j, that movie lengths are positive integers, and that the flight duration is an integer. Ask about handling multiple valid pairs and whether to return any pair or all pairs.
Select a hash map (dictionary) to achieve O(n) time complexity by enabling constant-time lookups for complements.
Traverse the array once. For each movie length, compute the complement and check if it exists in the hash map. If found, return the stored index and the current index.
Ensure that the same index is not used twice. If the complement equals the current movie length, verify that the stored index is different from the current index.
If no pair is found, return an empty result. Analyze time and space complexity: O(n) time and O(n) space, which is optimal for this problem.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.