Took me a beat to see what they actually wanted.
Start by clarifying the problem: we need to sum the differences (previous - current) for each adjacent pair where current < previous, in a single pass. Then walk through the algorithm: initialize total to 0, iterate from index 1 to n-1, compare each element with its predecessor, and if it's smaller, add the drop to the total. Finally, discuss edge cases and complexity.
Pro tip: Mention that this is essentially summing the negative deltas of the array, and that the single-pass constraint means we should avoid storing extra data. Also, proactively discuss how you'd handle edge cases like an empty array or a single element, and note that the solution is O(n) time and O(1) space.
Restate the problem in your own words to ensure you understand: sum all drops where a value is strictly less than the one before it. Ask clarifying questions about input size, data types, and whether the array can be empty.
Explain that you'll iterate through the array once, keeping a running total. For each index i from 1 to n-1, if arr[i] < arr[i-1], add (arr[i-1] - arr[i]) to the total.
Choose a small example, such as [5, 3, 4, 1], and manually compute the total drops to verify your algorithm. This demonstrates your understanding and catches off-by-one errors.
State that the time complexity is O(n) and space complexity is O(1). Discuss edge cases: empty array, single element, strictly increasing array (total 0), and strictly decreasing array (sum of all consecutive differences).
Implement the solution in your preferred language with clear variable names and comments. If time permits, mention alternative approaches (e.g., using zip in Python) but emphasize the single-pass requirement.
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