Clarify that the words are distinct and both appear, then propose a single-pass solution that tracks the most recent index of each word and updates the minimum distance whenever either word is encountered. This achieves O(n) time and O(1) space, which is optimal since every element may need to be examined.
Pro tip: Mention that if the array is static and many queries will be made, you could preprocess positions of each word and use binary search to answer each query in O(log n) time, but for a single query the linear scan is optimal.
Confirm that the two words are distinct, both appear in the array, and that distance is measured as the absolute difference of indices. Ask if the array can be modified or if multiple queries are expected.
Mention that a naive approach would compare every occurrence of word1 with every occurrence of word2, which is O(n^2) in the worst case. This shows you understand the baseline but also its inefficiency.
Iterate through the array once, keeping track of the most recent index of word1 and word2. When you see either word, update its last seen index and if the other word has been seen, compute the distance and update the minimum.
State that the algorithm runs in O(n) time and uses O(1) extra space, which is optimal because you must look at each element at least once in the worst case.
Discuss what happens if one word appears many times, if the words are the same (though problem says distinct), or if the array is empty. Also mention the preprocessing approach for multiple queries.
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