0288MediumArrayHeap (Priority Queue)

Smallest Coupled Trims

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Trains the technique from

LeetCode 373Find K Pairs with Smallest Sums

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 test rig couples one damper from the left tray with one damper from the right tray. Tray contents arrive as left and right, each a list of signed trim values in microns sorted in non-decreasing order. Trims may be negative.

A coupling picks one position in left and one position in right, and its coupled trim is the sum of the two values. Positions matter, not values: if a tray holds the same trim at two positions, each position gives its own coupling.

Return the k couplings whose coupled trim is smallest, each written as [left trim, right trim]. Order the answer by coupled trim ascending; where two couplings share a coupled trim, put the smaller left trim first, and if those match too, put the smaller right trim first. It is guaranteed that k does not exceed the number of couplings.

Examples

Example 1

Input: left = [-4,-1,0], right = [-2,3], k = 4
Output: [[-4,-2],[-1,-2],[0,-2],[-4,3]]

The listed couplings have coupled trims -6, -3, -2 and -1, which are the four smallest, and they appear in ascending order of coupled trim.

Example 2

Input: left = [1,3], right = [1,3], k = 3
Output: [[1,1],[1,3],[3,1]]

Coupled trims are 2, 4 and 4. The two couplings that share the trim 4 are ordered by their left trim, so 1 with 3 precedes 3 with 1.

Constraints

1 <= left.length, right.length <= 10^5
-10^9 <= left[i], right[i] <= 10^9
left and right are both sorted in non-decreasing order.
1 <= k <= 10^4
k <= left.length * right.length

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 smallest_coupled_trims(left: list[int], right: list[int], k: int) -> list[list[int]]:

Java

public List<List<Integer>> smallestCoupledTrims(int[] left, int[] right, int k)

The intended approach

Time O(r + k log r) where r = min(k, len(left))Space O(r) beyond the returned list of k couplings

Seed a min-heap with the coupling of each of the first min(k, len(left)) left trims against the smallest right trim, keyed by coupled trim then left trim then right trim. Each pop is the next coupling in the required order; after popping a coupling, push the same left position against the next right position, which keeps exactly one frontier coupling per active left position.

Pythonentry point: smallest_coupled_trims

import heapq


def smallest_coupled_trims(left, right, k):
    # Only the first k entries of `left` can ever contribute, and within one such
    # entry the coupled trim grows as the `right` index grows.  So a heap holding
    # one frontier coupling per active `left` entry always exposes the next
    # coupling in (coupled trim, left trim, right trim) order.
    # O(r + k log r) time with r = min(k, len(left)), O(r) heap space plus the
    # returned list of at most k couplings.
    rows = min(k, len(left))
    heap = [(left[i] + right[0], left[i], right[0], i, 0) for i in range(rows)]
    heapq.heapify(heap)
    out = []
    while heap and len(out) < k:
        _, a, b, i, j = heapq.heappop(heap)
        out.append([a, b])
        if j + 1 < len(right):
            nxt = right[j + 1]
            heapq.heappush(heap, (left[i] + nxt, left[i], nxt, i, j + 1))
    return out

The obvious version

Materialise every coupling as a (coupled trim, left trim, right trim) triple, sort the whole list, and slice off the first k.

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_smallest_coupled_trims(left, right, k):
    pairs = []
    for a in left:
        for b in right:
            pairs.append((a + b, a, b))
    pairs.sort()
    return [[a, b] for _, a, b in pairs[:k]]

How this problem was checked

Reference solution executed against 8 test cases — 8 passed
2 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

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