1120MediumLinked ListTwo PointersStack

The Heaviest Coupled Wagons

Tracked in this browser only

Trains the technique from

LeetCode 2130Maximum Twin Sum of a Linked List

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 chain of wagons runs from front to back and holds an even number of them. Counting from zero, wagon i is coupled with wagon n - 1 - i, so the front wagon pairs with the back one, the second with the second from the back, and so on.

Because the harness passes plain JSON, the chain reaches you as chain, listing the loads in order from the front wagon.

A couple's weight is the sum of its two wagons' loads. Return the largest weight any couple carries.

Examples

Example 1

Input: chain = [1, 100, 100, 1]
Output: 200

The two outer wagons couple for a weight of 2 and the two inner ones for 200.

Example 2

Input: chain = [1, 2, 3, 4, 5, 6]
Output: 7

Every couple weighs 7: the first with the last, the second with the fifth and the third with the fourth.

Example 3

Input: chain = [1, 2]
Output: 3

The chain holds a single couple, so its weight is the answer.

Constraints

chain.length is even
2 <= chain.length <= 10^5
1 <= chain[i] <= 10^5

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 pair_sum(chain: list[int]) -> int:

Java

public int pairSum(int[] chain)

The intended approach

Time O(n) over the chain.Space O(1).

Every couple is one wagon from the front half and one from the back half, and the two halves are walked towards each other in step. So set one marker at the front and one at the back, add the pair they sit on, step the front marker forward and the back marker backward, and stop when they meet. That visits every couple exactly once and no couple twice, because the chain holds an even number of wagons and the markers cross without ever landing on the same one. Keeping a running best rather than a list of weights makes the whole thing one pass with nothing stored.

Pythonentry point: pair_sum

def pair_sum(chain):
    lo = 0
    hi = len(chain) - 1
    best = 0
    while lo < hi:
        total = chain[lo] + chain[hi]
        if total > best:
            best = total
        lo += 1
        hi -= 1
    return best

The obvious version

For each wagon in the front half, work out its partner's place and then walk the whole chain from the front picking up just those two loads, as though the chain could only be read in order. Costs the chain length squared.

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_pair_sum(chain):
    n = len(chain)
    best = None
    for i in range(n):
        j = n - 1 - i
        if i >= j:
            continue
        total = 0
        at = 0
        for value in chain:
            if at == i or at == j:
                total += value
            at += 1
        if best is None or total > best:
            best = total
    return best

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