Connect Sticks - Implementation

Here is the solution:

function connectSticks(sticks)
 heap := min-heap of sticks
 cost := 0
 while heap.size > 1
 a := heap.pop()
 b := heap.pop()
 merged := a + b
 cost := cost + merged
 heap.push(merged)
 return cost

Time: $O(n \log n)$. Space: $O(n)$.

A min-heap efficiently gives you the two smallest sticks. Merge them, add to cost, push the result back.