Meta·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Jul 2026

Summary

Meta SWE coding round with an array manipulation problem. Nothing too wild but the edge cases in the problem statement took me a minute to parse correctly.

Questions Asked (1)

Given an integer array, find the minimum number of increment-by-1 operations needed to make the array either strictly increasing or strictly decreasing.

Algorithms & Data Structures

Author's notes

The example they give is [2, 3, 3, 2] with answer 1, which makes sense once you see it, but I initially tried to think about both directions simultaneously and got confused.

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AI HintsAI Generated

Suggested Approach

First, clarify that the problem asks for the minimum total increments to transform the array into a strictly monotonic sequence (either increasing or decreasing). Then, for each direction, compute the minimum cost by greedily adjusting each element to be at least (or at most) the previous element plus (or minus) one, and return the smaller of the two costs. Explain that this greedy approach works because any optimal solution must satisfy the monotonic constraints, and the greedy choice minimizes the increments at each step.

Pro tip: Mention that the greedy algorithm is optimal for this problem, but also note that if the operation were 'set to any value' instead of 'increment by 1', the problem would be different. This shows you understand the nuances of the operation.

1. Clarify the problem

Confirm that the array must become strictly increasing or strictly decreasing, and that only increment-by-1 operations are allowed. Ask if the array can be modified in place or if a new array is needed.

2. Define cost for one direction

For strictly increasing, iterate from left to right, keeping track of the previous value. For each element, if it is not greater than the previous, increment it to previous+1 and add the difference to the total cost.

3. Compute cost for both directions

Apply the same logic for strictly decreasing (iterate left to right, ensuring each element is less than the previous by at least 1). Compute the total increments for both increasing and decreasing.

4. Return the minimum

Compare the two costs and return the smaller one. If the array is already monotonic in one direction, the cost will be 0 for that direction.

5. Analyze complexity

State that the algorithm runs in O(n) time and O(1) extra space, as it only requires a single pass for each direction.

Key Points to Mention

Greedy approach: always increment the current element to the minimum required value to satisfy the monotonic condition.
Two separate passes: one for strictly increasing, one for strictly decreasing.
The cost for a direction is the sum of increments needed at each step.
The greedy choice is optimal because any larger increment would only increase the cost without helping future elements.
Edge cases: empty array, single element, already monotonic array.
Time complexity O(n) and space complexity O(1).

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.