1080MediumArrayGreedyHeap (Priority Queue)

Joining Every Rod Into One

Tracked in this browser only

Trains the technique from

LeetCode 1167Minimum Cost to Connect Sticks

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A pile of rods has the lengths rods. Joining two rods costs their two lengths added together and leaves a single rod of that length in their place.

Join rods until only one remains. Return the smallest total cost.

Examples

Example 1

Input: rods = [1, 2, 3, 4]
Output: 19

Joining 1 and 2 costs three, joining that with 3 costs six, and joining that with 4 costs ten, nineteen in all. Any other order charges the short rods more times over.

Example 2

Input: rods = [1, 1]
Output: 2

One join of the two rods, charging their two lengths.

Example 3

Input: rods = [1, 2, 4, 8, 16]
Output: 56

Each join takes the two shortest rods left, so the four joins cost three, seven, fifteen and thirty-one.

Constraints

1 <= rods.length <= 10^4
1 <= rods[i] <= 10^4

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def connect_sticks(rods: list[int]) -> int:

Java

public int connectSticks(int[] rods)

The intended approach

Time O(n log n) for the heap operations.Space O(n) for the heap.

Each join charges the two lengths going into it, so a rod's original length is charged once for every join it later takes part in. That means a rod joined early is charged many times over, so the shortest rods should be joined first and the longest as late as possible. Keeping the pile in a heap makes that easy: take the two shortest rods, charge their total, and put a rod of that total back. Repeat until one rod is left. The heap keeps the two shortest to hand without sorting the whole pile again after each join. The greedy is right because in any cheapest sequence of joins, the two rods charged the most times can be taken to be the two shortest: swapping a shorter rod into a deeper place never costs more. A pile of one rod needs no joins and costs nothing.

Pythonentry point: connect_sticks

def connect_sticks(rods):
    import heapq
    pile = list(rods)
    heapq.heapify(pile)
    total = 0
    while len(pile) > 1:
        one = heapq.heappop(pile)
        two = heapq.heappop(pile)
        total += one + two
        heapq.heappush(pile, one + two)
    return total

The obvious version

Keep the pile as a plain list and, for each join, scan it twice to find the two shortest rods, remove them, charge their total and append a rod of that length. Costs a scan per join rather than a heap.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_connect_sticks(rods):
    pile = list(rods)
    total = 0
    while len(pile) > 1:
        first = 0
        for at in range(len(pile)):
            if pile[at] < pile[first]:
                first = at
        one = pile.pop(first)
        second = 0
        for at in range(len(pile)):
            if pile[at] < pile[second]:
                second = at
        two = pile.pop(second)
        total += one + two
        pile.append(one + two)
    return total

How this problem was checked

Reference solution executed against 15 test cases — 15 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

Create a free account to unlock the solution

See the step-by-step animation, the intuition, and clean code in every language — free, no credit card.

Create a free account Already have an account? Sign in