Minimum Cost of Ropes

Q: What is the time complexity of Minimum Cost of Ropes?

O(n log n) time and O(n) space with a min-heap: n-1 joins, each three heap operations of O(log n) . Scanning the whole pile for the two shortest ropes instead makes it O(n²) .

EasyHeap Greedy

+10

You have some ropes, and ropes[i] is the length of rope i. You can join any two ropes into one rope whose length is the sum of the two, and that join costs the same sum. You keep joining until a single rope is left.

Write a function named minCostRopes that returns the smallest total cost of joining all the ropes into one, or 0 when there is only one rope to begin with.

Function

minCostRopes(ropes: integer-array) → integer

ropesinteger-array: the length of each rope
Returnsinteger: the smallest total cost of joining every rope into one

Constraints

1 ≤ ropes.length ≤ 10⁴
1 ≤ ropes[i] ≤ 100
The answer fits in a 32-bit signed integer.

Examples

Input: ropes = [5, 2, 8, 3]
Output: 33
Explanation: Join 2 and 3 for 5, the two ropes of length 5 for 10, then 8 and 10 for 18. The total is 5 + 10 + 18 = 33.

Input: ropes = [1, 1, 1, 1]
Output: 8
Explanation: Join the 1s in pairs for 2 each, then the two ropes of length 2 for 4: 8 in total. Adding each 1 to one growing rope would cost 2 + 3 + 4 = 9.

Follow-up

Suppose a join may tie up to three ropes and costs their total length. How does the greedy rule change when the number of ropes does not work out evenly?

Solution

Every join pays the full length of both pieces, so a rope joined early is paid again in every later join it is part of. Keep long ropes out of the early joins: join the two shortest ropes every round. A min-heap finds those two in O(log n).

Scan for the two shortest every round

Correct, but does not finish on the largest tests

Intuition

Each round, scan the pile for the shortest rope and take it out, scan again for the next shortest, and put their join back.

Why the two shortest: draw the joins as a tree with the ropes as leaves. A rope's length is paid once for every join above it, which is its depth. The deepest level has two leaves side by side, and moving the two shortest ropes there never raises the cost. So some cheapest plan joins the two shortest first, and the argument repeats on the smaller pile.

There are n-1 rounds of two full scans: O(n²), about 10^8 comparisons for 10^4 ropes.

Algorithm

Copy the ropes into a working list, pile, and start total at 0.
Repeat until one rope is left: find the shortest rope and remove it by moving the last rope into its slot.
Do the same again to get the next shortest rope.
Add the sum of the two to total and append a rope of that length to the pile.
Return total.

def minCostRopes(ropes):
    pile = list(ropes)

    def take_shortest():
        k = 0
        for i in range(1, len(pile)):
            if pile[i] < pile[k]:
                k = i
        pile[k], pile[-1] = pile[-1], pile[k]  # move it to the end, then remove it
        return pile.pop()

    total = 0
    while len(pile) > 1:
        joined = take_shortest() + take_shortest()
        total += joined
        pile.append(joined)
    return total

Min-heap

Intuition

A min-heap keeps the shortest rope at its root, and taking the root out or inserting a rope costs O(log n). A round is two extractions, which hand you the two shortest ropes, and one insertion of the joined rope.

n ropes need n-1 joins of three heap operations each, so the time is O(n log n): about 4 × 10^5 steps for 10^4 ropes. The heap never holds more than n ropes.

Algorithm

Put every rope length in a min-heap and set total = 0.
Repeat until one rope is left: pop the two shortest.
Add their sum to total and push the sum back into the heap.
Return total, which stays 0 when there was only one rope.

import heapq


def minCostRopes(ropes):
    heap = list(ropes)
    heapq.heapify(heap)  # a min-heap: heap[0] is the shortest rope
    total = 0
    while len(heap) > 1:
        joined = heapq.heappop(heap) + heapq.heappop(heap)
        total += joined
        heapq.heappush(heap, joined)
    return total

Pitfalls and edge cases

Sorting once and adding each rope to one running rope. The running rope can grow longer than ropes you have not used yet: for [2, 3, 4, 4] it costs 5 + 9 + 13 = 27, while joining the two shortest costs 5 + 8 + 13 = 26.
Dropping the joined rope. Its length goes back into the pile and is paid again in a later join.
Returning the length of the final rope. For [5, 2, 8, 3] the last rope is 18 long, but the cost is 33.
Using a max-heap by accident. C++ priority_queue and Rust's BinaryHeap hand out the largest value unless you reverse the order.

Frequently asked questions3

What is the time complexity of Minimum Cost of Ropes?

O(n log n) time and O(n) space with a min-heap: n-1 joins, each three heap operations of O(log n). Scanning the whole pile for the two shortest ropes instead makes it O(n²).

Is Minimum Cost of Ropes the same as Huffman coding?

Yes, it is the same algorithm. Huffman coding builds a prefix code by merging the two least frequent symbols again and again, and it minimizes the sum of each frequency times its depth in the tree. Here the ropes play the symbols and their lengths play the frequencies, so the total cost is that same sum.

Can Minimum Cost of Ropes be solved without a heap?

Yes, with two queues. Sort the ropes into one queue and push each joined rope onto the back of a second. Joined ropes come out in non-decreasing order, so the shortest rope left is at the front of one queue. Each round is then O(1), and with lengths up to 100 a counting sort makes the run O(n).

def minCostRopes(ropes):
    # Write code here

Test #1

Test #2

Test #3

Argument

ropes = [5, 2, 8, 3]

Expected Output