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

Intermediate

Apr 2026

Summary

Amazon SWE coding round, one algorithmic problem about distributing candies based on ratings. Pretty standard greedy problem but the constraints trip you up if you're not careful.

Questions Asked (1)

Given an array of children's ratings, assign candies such that each child gets at least one and any child with a higher rating than an adjacent neighbor gets more candies than that neighbor. Return the minimum total number of candies required.

Algorithms & Data Structures

Author's notes

Took me a minute to realize a single pass doesn't cut it.

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AI HintsAI Generated

Suggested Approach

Use a two-pass greedy approach: first initialize each child with 1 candy, then scan left-to-right to ensure higher ratings get more candies than left neighbors, and finally scan right-to-left to handle right neighbors. Sum the candies for the minimum total. This runs in O(n) time and O(n) space.

Pro tip: Mention that this is a classic greedy problem where local optimal choices lead to a global optimum, and emphasize that the two passes are independent and necessary to satisfy both neighbor constraints. Also, note that the space can be optimized to O(1) if only the total is needed, but O(n) is acceptable for clarity.

1. Clarify and Restate

Confirm understanding: each child gets at least one candy; if a child's rating is higher than an adjacent child, they must get more candies. The goal is to minimize the total candies.

2. Initialize and Left-to-Right Pass

Create an array of candies, all initialized to 1. Traverse from left to right: if current rating > previous rating, set candies[i] = candies[i-1] + 1.

3. Right-to-Left Pass

Traverse from right to left: if current rating > next rating, set candies[i] = max(candies[i], candies[i+1] + 1) to satisfy the right neighbor constraint without breaking the left one.

4. Sum and Return

Sum all values in the candies array and return the total. This is the minimum total candies required.

5. Analyze Complexity and Edge Cases

State time complexity O(n) and space O(n). Discuss edge cases: empty array, single child, strictly increasing/decreasing ratings, and equal ratings.

Key Points to Mention

Greedy algorithm with two passes to satisfy both left and right neighbor constraints.
Initialization of each child with at least one candy.
Left-to-right pass ensures higher rating than left neighbor gets more candies.
Right-to-left pass ensures higher rating than right neighbor gets more candies, using max to avoid reducing candies from left pass.
Time complexity O(n) and space complexity O(n), with potential O(1) space optimization if only total is needed.
Edge cases: empty array, single element, strictly increasing/decreasing sequences, and equal ratings.

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