0292MediumArrayBacktrackingSorting

Distinct Step Orders

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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 bench replays a batch of calibration steps, one after another. The batch is given as steps, a list of signed millivolt values that may contain repeats.

A read-out is the sequence of numbers the bench prints when it plays every entry of the batch exactly once, honouring repeats: a value appearing twice in steps is printed twice. Two read-outs are the same when they print the same numbers in the same order, so rearranging entries that hold equal values does not produce a new read-out.

Return every distinct read-out. The read-outs may be returned in any order.

Examples

Example 1

Input: steps = [2,1,2]
Output: [[1,2,2],[2,1,2],[2,2,1]]

Each read-out prints both 2s and the single 1, and there are three ways the 1 can sit relative to the pair of 2s.

Example 2

Input: steps = [-1,0,-1]
Output: [[-1,-1,0],[-1,0,-1],[0,-1,-1]]

The batch holds two copies of -1 and one 0, so the distinct read-outs are the three placements of the 0.

Constraints

1 <= steps.length <= 8
-10 <= steps[i] <= 10

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 distinct_step_orders(steps: list[int]) -> list[list[int]]:

Java

public List<List<Integer>> distinctStepOrders(int[] steps)

The intended approach

Time O(n * d) for d distinct read-outsSpace O(n) recursion depth beyond the returned read-outs

Sort the batch so equal values sit together, then build read-outs position by position with a used flag per entry. An entry is refused when the entry immediately before it holds the same value and is still unused, which pins one fixed order on every group of equal values and emits each distinct read-out once.

Pythonentry point: distinct_step_orders

def distinct_step_orders(steps):
    # Sorting puts equal values next to each other.  At each position a value may
    # only be placed if the identical value just before it in the sorted list has
    # already been placed, which fixes one canonical order among equal values and
    # so emits every distinct read-out exactly once.
    # O(n * d) time for d distinct read-outs, O(n) recursion depth plus the
    # O(n * d) returned list.
    values = sorted(steps)
    n = len(values)
    used = [False] * n
    current = []
    out = []

    def walk():
        if len(current) == n:
            out.append(current[:])
            return
        for i in range(n):
            if used[i]:
                continue
            if i > 0 and values[i] == values[i - 1] and not used[i - 1]:
                continue
            used[i] = True
            current.append(values[i])
            walk()
            current.pop()
            used[i] = False

    walk()
    return out

The obvious version

Generate all n! orderings with a library permutation helper and keep the ones not seen before in a set of tuples.

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

from itertools import permutations


def bf_distinct_step_orders(steps):
    seen = set()
    out = []
    for p in permutations(steps):
        if p not in seen:
            seen.add(p)
            out.append(list(p))
    return out

How this problem was checked

Reference solution executed against 9 test cases — 9 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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