0016MediumArrayTwo PointersSorting

Charge Neutral Triples

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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 physics teaching kit holds a tray of numbered beads. charges[i] is the electric charge carried by bead i, measured in elementary units, and it may be positive, negative or zero.

A neutral triple is a set of three beads sitting at three different positions in the tray whose charges add to zero.

Report the charge values of every neutral triple the tray allows. Ordering is irrelevant: the three values within a triple may be listed in any order, and the triples themselves may be listed in any order.

A combination of charge values may only be reported once. Beads at different positions can carry equal charges, and many bead sets can therefore land on the same three values, but they all describe one triple in your answer. A single bead may not fill two of the three slots of a triple.

Examples

Example 1

Input: charges = [-3, 1, 2, -1, 4, -1]
Output: [[-3, -1, 4], [-3, 1, 2], [-1, -1, 2]]

Three combinations settle to zero. The last one draws on both beads carrying -1, which is allowed because they sit at separate positions.

Example 2

Input: charges = [0, 0, 0, 0]
Output: [[0, 0, 0]]

Four beads can be picked three at a time in four ways, yet every pick lands on the same three charge values, so the answer holds one triple.

Example 3

Input: charges = [5, -2, 4]
Output: []

The only available set of three beads carries a total of 7, so nothing is reported.

Constraints

3 <= charges.length <= 3000
-10^5 <= charges[i] <= 10^5

The groups you return, and the values inside each group, 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 charge_neutral_triples(charges: list[int]) -> list[list[int]]:

Java

public List<List<Integer>> chargeNeutralTriples(int[] charges)

The intended approach

Time O(n^2)Space O(n)

Sort the charges so equal values sit next to each other and the totals move predictably. Fix the smallest bead of a triple at index `i`, then close in on the remaining two with a pointer at each end of the tail: a total below zero means the low pointer must rise, a total above zero means the high pointer must fall. Repeats are suppressed by skipping any fixed value equal to the previous one and, after recording a hit, by walking both pointers past their duplicates, which is what keeps every value combination reported exactly once.

Pythonentry point: charge_neutral_triples

def charge_neutral_triples(charges):
    vals = sorted(charges)
    n = len(vals)
    found = []
    for i in range(n - 2):
        if vals[i] > 0:
            break
        if i > 0 and vals[i] == vals[i - 1]:
            continue
        low, high = i + 1, n - 1
        while low < high:
            total = vals[i] + vals[low] + vals[high]
            if total < 0:
                low += 1
            elif total > 0:
                high -= 1
            else:
                found.append([vals[i], vals[low], vals[high]])
                low += 1
                high -= 1
                while low < high and vals[low] == vals[low - 1]:
                    low += 1
                while low < high and vals[high] == vals[high + 1]:
                    high -= 1
    return found

The obvious version

Try every set of three positions with three nested loops and collect the sorted value tuples in a set, which removes repeats, O(n^3).

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_neutral_triples(charges):
    n = len(charges)
    seen = set()
    for i in range(n):
        for j in range(i + 1, n):
            for k in range(j + 1, n):
                if charges[i] + charges[j] + charges[k] == 0:
                    seen.add(tuple(sorted((charges[i], charges[j], charges[k]))))
    return [list(t) for t 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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