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Amazon·Software Engineer·Onsite - Coding / Algorithms·Intermediate

Intermediate
Jun 2026

Summary

Amazon coding round, one algorithm problem that looks deceptively clean but has some annoying edge cases once you start thinking about cumulative adjustments.

Questions Asked (1)

Q1

You have an integer array. In a single operation, you can pick any contiguous subarray and increment every element in it by 1. What is the minimum number of operations required to make the array non-decreasing?

Algorithms & Data Structures
Author's notes

Took me a while to see that this reduces to summing up the drops between adjacent elements after accounting for whatever increases you've already applied going left to right.

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AI HintsAI Generated

Suggested Approach

Reframe the problem by considering the differences between adjacent elements. The minimum number of operations equals the sum of positive differences between consecutive elements, as each operation can fix a deficit by incrementing a suffix. Explain this insight and provide a simple O(n) algorithm.

Pro tip: Mention that this is equivalent to the total 'upward slope' needed, and note that the operation is like adding 1 to a suffix, so each positive difference requires at least that many operations. This shows deep understanding and efficiency.

1. Understand the operation

Recognize that incrementing a contiguous subarray by 1 can be used to raise elements to meet non-decreasing constraints. The key is to think about how to fix violations efficiently.

2. Identify violations

A violation occurs when an element is greater than the next element (a[i] > a[i+1]). To fix it, we need to increment a[i+1] and possibly subsequent elements.

3. Derive the formula

The minimum operations equal the sum of positive differences between consecutive elements: sum(max(0, a[i] - a[i+1])) for i from 0 to n-2. This is because each operation can reduce one positive difference by 1.

4. Validate with examples

Test with simple arrays like [3,2,1] (answer 3) and [1,2,3] (answer 0) to confirm the formula. Explain why it works: each operation increments a suffix, effectively reducing the deficit at the start of the suffix.

5. Analyze complexity

State that the algorithm runs in O(n) time and O(1) space, as it only requires a single pass through the array.

Key Points to Mention

  • The operation is equivalent to adding 1 to a suffix of the array.
  • The minimum number of operations is the sum of positive differences between adjacent elements.
  • Each operation can fix at most one unit of a positive difference.
  • The formula is sum(max(0, a[i] - a[i+1])) for i from 0 to n-2.
  • Time complexity is O(n) and space complexity is O(1).
  • Edge cases: already non-decreasing array (0 operations), strictly decreasing array (sum of all differences).

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