0558EasyArrayHash TableTwo PointersBinary SearchSorting

Shared Spare Parts

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Trains the technique from

LeetCode 350Intersection of Two Arrays II

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.

Two service vans each carry a bin of spare parts. nums1 lists the catalogue number of every part in the first bin and nums2 does the same for the second bin; a bin may hold several copies of the same part.

A dispatcher wants the list of parts the two vans have in common, counted with copies: if one van carries three copies of part 8 and the other carries two, then part 8 belongs in the answer twice, because only two matching pairs can be made.

Return that list. Its entries may be given in any order.

Examples

Example 1

Input: nums1 = [3, 8, 3, 0, 8, 8], nums2 = [8, 3, 8]
Output: [3, 8, 8]

The first bin holds three copies of part 8 and the second holds two, so part 8 is reported twice. Part 3 appears twice in the first bin and once in the second, so it is reported once. Part 0 is missing from the second bin.

Example 2

Input: nums1 = [7, 7, 7], nums2 = [7, 7]
Output: [7, 7]

Only two pairs of part 7 can be made, so the answer lists it twice.

Example 3

Input: nums1 = [4], nums2 = [9]
Output: []

The bins share no catalogue number, so the answer is empty.

Constraints

1 <= nums1.length, nums2.length <= 1000
0 <= nums1[i], nums2[i] <= 1000
The result may be returned in any order.

The values you return may be in any order.

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 intersect(nums1: list[int], nums2: list[int]) -> list[int]:

Java

public int[] intersect(int[] nums1, int[] nums2)

The intended approach

Time O(n + m).Space O(min(n, m)) for the tally, beyond the returned list.

Tally the copies of each catalogue number in the shorter bin with a hash map, then walk the longer bin: whenever a part still has a positive tally, emit it and drop the tally by one. Spending the tally is what keeps the number of emitted copies down to the smaller of the two counts.

Pythonentry point: intersect

def intersect(nums1, nums2):
    # O(n + m) time, O(min(n, m)) space: tally the shorter bin, then walk the
    # longer one spending one unit of tally per match.
    if len(nums1) > len(nums2):
        nums1, nums2 = nums2, nums1
    left = {}
    for v in nums1:
        left[v] = left.get(v, 0) + 1
    out = []
    for v in nums2:
        c = left.get(v, 0)
        if c > 0:
            out.append(v)
            left[v] = c - 1
    return out

The obvious version

Copy the second bin into a working list. For each part in the first bin, scan the working list for a matching entry, and on a hit record the part and physically remove that entry so it cannot be matched twice.

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_intersect(nums1, nums2):
    pool = list(nums2)
    out = []
    for v in nums1:
        for j in range(len(pool)):
            if pool[j] == v:
                out.append(v)
                pool.pop(j)
                break
    return out

How this problem was checked

Reference solution executed against 10 test cases — 10 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

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