My first instinct was some kind of greedy single pass and it was wrong.
Use a two-pass greedy algorithm: first traverse left-to-right ensuring each child with a higher rating than the left neighbor gets more candies, then traverse right-to-left ensuring each child with a higher rating than the right neighbor gets more candies. Sum the maximum of the two passes for each child to get the minimum total candies.
Pro tip: Emphasize that the two-pass approach is optimal because it captures both local constraints independently, and mention that a single pass would fail to handle peaks correctly. Also, note that the algorithm runs in O(n) time and O(n) space, which is optimal for this problem.
Restate the problem to ensure understanding: each child gets at least one candy, and children with higher ratings than adjacent neighbors must receive more candies. Ask about edge cases like empty array or single child.
Explain that you will use two arrays (or one array updated twice) to track candies. First pass left-to-right: if current rating > left neighbor, set candies[i] = candies[i-1] + 1, else 1. Second pass right-to-left: if current rating > right neighbor, set candies[i] = max(candies[i], candies[i+1] + 1).
Choose a small example like ratings = [1,0,2] and demonstrate how the two passes yield candies = [2,1,2] with total 5. This shows how peaks are handled correctly.
State that the algorithm runs in O(n) time with two passes and O(n) space for the candies array. Mention that space can be optimized to O(1) if we only need the total sum, but O(n) is standard.
Explain why the two-pass approach guarantees the minimum: the left-to-right pass satisfies all left-neighbor constraints, the right-to-left pass satisfies all right-neighbor constraints, and taking the maximum ensures both are satisfied. Handle edge cases like all equal ratings or strictly increasing/decreasing sequences.
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