0083MediumArrayTwo PointersSorting

Four-Card Bias Groups

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Trains the technique from

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 bench stacks probe cards to cancel out a bias. You are given the integer array offsets, holding one signed bias figure per card in the tray, and an integer target.

Report every group of four cards whose bias figures total target. A group picks four distinct tray positions, so no card may be counted twice inside one group.

Two groups are the same group when they hold the same four figures with the same repeats, whichever tray positions supplied them; report such a group once only. Write the four figures of a group in non-decreasing order. The groups themselves may come back in any order. Return an empty list when no four cards total target.

Examples

Example 1

Input: offsets = [1, 5, 3, 6], target = 15
Output: [[1, 3, 5, 6]]

All four cards must be used, and 1 + 5 + 3 + 6 is 15, so the single group is the whole tray written in non-decreasing order.

Example 2

Input: offsets = [-6, -1, 0, 1, 6, 7], target = 0
Output: [[-6, -1, 0, 7], [-6, -1, 1, 6]]

Two distinct groups of figures reach 0 here; either ordering of the two groups is accepted.

Example 3

Input: offsets = [9, 9, 9], target = 27
Output: []

Only three cards sit in the tray, so no group of four exists and the report is empty.

Constraints

1 <= offsets.length <= 200
-10^9 <= offsets[i] <= 10^9
-10^9 <= target <= 10^9

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 bias_quartets(offsets: list[int], target: int) -> list[list[int]]:

Java

public List<List<Integer>> biasQuartets(int[] offsets, int target)

The intended approach

Time O(n^3)Space O(n)

Sort the figures so equal ones sit together and the tail can be squeezed from both ends. Fix the two lowest members of a group with a pair of loops, then close on the remaining amount with one index walking up from the left and one walking down from the right, which finds every completing pair in linear time. Stepping past runs of equal figures at each of the four positions is what keeps a repeated group from being reported twice.

Pythonentry point: bias_quartets

def bias_quartets(offsets, target):
    values = sorted(offsets)
    n = len(values)
    groups = []
    for a in range(n - 3):
        if a > 0 and values[a] == values[a - 1]:
            continue
        for b in range(a + 1, n - 2):
            if b > a + 1 and values[b] == values[b - 1]:
                continue
            need = target - values[a] - values[b]
            lo, hi = b + 1, n - 1
            while lo < hi:
                pair = values[lo] + values[hi]
                if pair == need:
                    groups.append([values[a], values[b], values[lo], values[hi]])
                    lo += 1
                    hi -= 1
                    while lo < hi and values[lo] == values[lo - 1]:
                        lo += 1
                    while lo < hi and values[hi] == values[hi + 1]:
                        hi -= 1
                elif pair < need:
                    lo += 1
                else:
                    hi -= 1
    return groups

The obvious version

Enumerate all combinations of four tray positions, keep those that total the target, and drop repeats by putting each surviving group of figures into a set as a sorted tuple. O(n^4), no sorting insight at all.

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_bias_quartets(offsets, target):
    from itertools import combinations
    seen = set()
    n = len(offsets)
    for picks in combinations(range(n), 4):
        if offsets[picks[0]] + offsets[picks[1]] + offsets[picks[2]] + offsets[picks[3]] == target:
            seen.add(tuple(sorted(offsets[p] for p in picks)))
    return [list(group) for group in sorted(seen)]

How this problem was checked

Reference solution executed against 11 test cases — 11 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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