Connect N Ropes at Minimum Cost

Connect N Ropes at Minimum Cost is a classic greedy + heap problem. It teaches a powerful instinct: when a cost compounds, attack the smallest pieces first, and let a min-heap keep handing them to you.

The slow way first

You might join ropes in input order, or try every possible order of joins. Trying all orders is exponential — far too slow. Even a fixed order can be wasteful: if you join the two longest ropes early, their large combined length gets re-added in every later join.

The question to ask: which join should I do next so I pay as little as possible over the whole process? The lengths you combine earliest get counted again in every subsequent join, so the cheapest plan keeps the small lengths combining first.

The idea: always join the two shortest

Greedily join the two shortest ropes available, add their sum to the running cost, and treat the result as a new rope. Repeat until one rope is left. A rope joined early contributes to many later sums, so you want the small lengths to be the ones that pile up — not the large ones.

A min-heap is the perfect tool: it always lets you pop the two smallest in O(log n), and you push the new combined rope back in O(log n).

The key insight: this is exactly the structure of a Huffman tree. Combining the two smallest at every step provably minimizes the total weighted cost.

Walk through it

Step through the animation. The cells are the ropes inside the min-heap. Each round, the two smallest cells light up, their sum is added to total, and a new combined cell appears while the originals disappear. Watch total climb 0 → 3 → 11 → 27 as the heap shrinks to a single rope.

Pseudocode

build a min-heap from the rope lengths
total = 0
while more than one rope remains:
    a = pop smallest
    b = pop next smallest
    total += a + b            # cost of joining them
    push (a + b) back         # the new combined rope
return total

The Python solution

import heapq
 
def connect_ropes(ropes):
    heapq.heapify(ropes)
    total = 0
    while len(ropes) > 1:
        a = heapq.heappop(ropes)
        b = heapq.heappop(ropes)
        total += a + b
        heapq.heappush(ropes, a + b)
    return total

• heapq.heapify(ropes) turns the list into a min-heap in place in O(n).
• The loop runs while at least two ropes remain — we still have a join to do.
• heappop twice gives the two shortest ropes, a and b.
• total += a + b charges the cost of joining them.
• heappush(ropes, a + b) puts the combined rope back so it can be joined again later.
• When one rope is left, every rope is connected — return the accumulated total.

Complexity

There are n − 1 joins, each costing O(log n) for the heap operations, giving O(n log n). The heap itself is the O(n) extra space.

When this pattern shows up

Practice