What is the time complexity of the optimal min-heap based algorithm to find the minimum cost to connect n sticks, and why?

medium🪤 Complexity Trap Q13 of Q15

Greedy Algorithms - Minimum Cost to Connect Sticks

AO(n log n) because each of the n-1 merges involves two heap pops and one push, each O(log n)

BO(n log k) where k is the maximum stick length, due to heapifying by stick size

CO(n) because heap operations are constant time on average

DO(n^2) because each merge requires scanning all sticks

Step-by-Step Solution

Solution:

Step 1: Identify number of operations
There are n sticks, so n-1 merges are needed.
Step 2: Analyze heap operations per merge
Each merge requires two pops and one push on a min-heap, each operation O(log n).
Final Answer:
Option A -> Option A
Quick Check:
(n-1) merges x 3 heap ops x O(log n) -> O(n log n) [OK]

Quick Trick: Heap operations cost O(log n) each, repeated n times [OK]

Common Mistakes:

MISTAKES

Trap Explanation:

PITFALL

Heap operations are logarithmic in heap size, not constant; confusing this leads to underestimating complexity.

Interviewer Note:

CONTEXT

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