My first instinct was to just simulate it, and that works fine for small inputs, but I had a feeling they wanted something better.
First, clarify the problem and walk through a small example to ensure understanding. Then, derive the key insight: the number of passes equals the maximum over each element of the number of times it appears as a 'record minimum' when scanning left-to-right. Finally, present an efficient algorithm to compute this, analyze its complexity, and discuss edge cases.
Pro tip: Demonstrate strong communication by explaining your thought process step-by-step and verifying with examples; this shows Amazon's Leadership Principle of 'Dive Deep' and 'Customer Obsession' by ensuring the solution is correct and efficient.
Confirm the problem details: array is a permutation of 1..n, scanning left-to-right, target starts at 1, increments when found, restarts from beginning if not all found. Ask if there are any constraints or expected input sizes.
Manually simulate the process for small n (e.g., n=3,4) with different permutations to identify the pattern and build intuition for the number of passes.
Observe that the number of passes is determined by the maximum number of times any element is a 'record minimum' when scanning left-to-right. Equivalently, for each element, count how many smaller elements appear after it; the answer is 1 + max of these counts.
Use a Fenwick tree (BIT) or segment tree to compute for each element the number of smaller elements to its right in O(n log n) time. Alternatively, use a stack-based approach to find the longest decreasing subsequence? Actually, the answer is the length of the longest decreasing subsequence? Wait, need to verify: The number of passes equals the maximum number of times an element is a record minimum, which is the length of the longest decreasing subsequence? Let's test: array [3,1,2]. Passes: pass1: find 1 at index2, then 2 at index3, then restart? Actually, target starts at 1. Pass1: scan: 3 (no), 1 (yes, target=2), 2 (yes, target=3), end? Actually, after finding 2, target=3, but 3 is before, so not found. So pass1 ends, found 1 and 2. Pass2: scan: 3 (yes, target=4), done. So 2 passes. Longest decreasing subsequence: [3,1] length 2. So answer = length of longest decreasing subsequence? Check another: [2,3,1]. Pass1: 2 (no), 3 (no), 1 (yes, target=2), end. Pass2: 2 (yes, target=3), 3 (yes, target=4), done. 2 passes. LDS: [2,1] or [3,1] length 2. [1,2,3]: pass1: 1,2,3 all found, 1 pass. LDS length 1. [3,2,1]: pass1: 3,2,1 -> find 1 at end, target=2, end. Pass2: 3,2 -> find 2, target=3, end. Pass3: 3 -> find 3, done. 3 passes. LDS length 3. So answer = length of longest decreasing subsequence. But careful: is it strictly decreasing? Yes, permutation. So answer is LDS length. So algorithm: compute LDS in O(n log n) using patience sorting or DP with binary search.
Time complexity O(n log n), space O(n). Discuss edge cases: already sorted (1 pass), reverse sorted (n passes), n=1. Also mention that the problem can be solved in O(n) with a stack? Actually, LDS can be computed in O(n log n) with binary search. Confirm with interviewer.
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