My first instinct was to just simulate it, which works but obviously blows up on time complexity for large inputs.
First, clarify the problem: we need to find the minimum number of left-to-right passes to collect numbers in increasing order from 1 to n. Then, observe that a new pass is required exactly when the next number to collect appears before the current number in the array. Use an array to store the position of each number, then iterate from 1 to n, counting how many times the position decreases.
Pro tip: Mention that this is equivalent to counting the number of 'descents' in the sequence of positions when numbers are ordered by value. This insight shows you can reduce the problem to a simple linear scan, demonstrating strong algorithmic thinking.
Confirm that we start with a fresh pass each time we reach the end without having collected all numbers, and that within a pass we collect numbers in increasing order as they appear.
Realize that a new pass is needed exactly when the next number to collect (i+1) appears before the current number (i) in the array. Thus, the number of passes equals 1 plus the number of such inversions in the order of positions.
Create an array pos of size n+1 where pos[value] = index in the original array. Initialize passes = 1. Iterate i from 1 to n-1: if pos[i+1] < pos[i], increment passes. Return passes.
The algorithm uses O(n) extra space for the pos array and runs in O(n) time: one pass to build pos, one pass to count descents. This meets the O(n) requirement.
Walk through a small example, e.g., [3,1,2] (n=3). pos[1]=1, pos[2]=2, pos[3]=0. pos[2] > pos[1]? 2>1 no; pos[3] < pos[2]? 0<2 yes, so passes=2. Verify by simulation: Pass 1: collect 1, then 2? Actually, scan: 3 (skip), 1 (collect), 2 (collect) -> collected 1,2. Pass 2: collect 3. Total 2 passes.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.