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If the candy distribution problem is modified so that children stand in a circle (first and last are neighbors), how does this affect the approach to find the minimum candies needed?

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Greedy Algorithms - Candy Distribution

If the candy distribution problem is modified so that children stand in a circle (first and last are neighbors), how does this affect the approach to find the minimum candies needed?

ASorting the ratings array solves the problem directly.

BThe same two-pass greedy algorithm applies without any changes.

CThe problem becomes more complex; a single two-pass linear scan is insufficient due to circular dependency.

DAssigning one candy to each child is always optimal in a circle.

Step-by-Step Solution

Solution:

Step 1: Understand circular constraint
First and last children are neighbors, creating a cycle.
Step 2: Impact on algorithm
Two-pass linear scan assumes linear order; circular dependency breaks this assumption.
Step 3: Resulting complexity
Requires more complex methods like breaking the circle or using dynamic programming.
Final Answer:
Option C -> Option C
Quick Check:
Circular neighbor relations require special handling [OK]

Quick Trick: Circle breaks linear assumptions; need advanced approach [OK]

Common Mistakes:

MISTAKES

Assuming linear two-pass works unchanged
Thinking sorting solves neighbor constraints
Believing one candy per child suffices always

Trap Explanation:

PITFALL

Ignoring circular neighbor relation leads to incorrect assumptions

Interviewer Note:

CONTEXT

Tests understanding of problem variations and algorithm limitations

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