Which modification to the optimal algorithm correctly handles this variant?
hard🎤 Interviewer Follow-up Q15 of Q15
Greedy Algorithms - Candy Distribution
Suppose the problem is modified so that children can have equal ratings but must still have strictly more candies than neighbors with lower ratings. Which modification to the optimal algorithm correctly handles this variant?
AChange comparison from '>' to '>=' in both passes to handle equal ratings
BKeep '>' comparisons but initialize candies with zeros and adjust accordingly
CUse the same two-pass greedy with '>' comparisons but do not update candies if ratings are equal
DReplace two-pass greedy with a sorting-based approach to assign candies in rating order
Step-by-Step Solution
Step 1: Understand new requirement
Children with equal ratings do not require more candies than each other, only strictly higher ratings require more candies.
Step 2: Adjust comparisons in algorithm
Keep '>' comparisons to ensure only strictly higher ratings get more candies; do not update candies when ratings are equal.
Step 3: Avoid changing to '>=' which would incorrectly increase candies for equal ratings
Changing to '>=' breaks minimality and correctness.
Final Answer:
Option C -> Option C
Quick Check:
Strict '>' comparisons preserve problem constraints with equal ratings [OK]
