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Module Build Order

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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 release engineer compiles a project whose moduleCount modules are labelled 0 through moduleCount - 1. Each module is compiled exactly once. Every entry [x, y] of dependencies says that module y has to finish compiling before module x may start.

Plenty of sequences can satisfy the same dependency set, so the build tool fixes one: whenever more than one module is unblocked, it always takes the lowest-numbered candidate. That rule makes the answer the lexicographically smallest valid sequence. Return that sequence.

If the dependencies leave no way to compile every module, return an empty array.

Examples

Example 1

Input: moduleCount = 3, dependencies = [[1, 0]]
Output: [0, 1, 2]

Module 0 gates module 1 while module 2 is free from the start, and always grabbing the lowest unblocked label produces 0, 1, 2.

Example 2

Input: moduleCount = 5, dependencies = [[2, 0], [3, 2], [1, 3], [4, 1]]
Output: [0, 2, 3, 1, 4]

The entries chain the modules single file, so only one sequence exists no matter which candidate rule is applied.

Example 3

Input: moduleCount = 2, dependencies = [[0, 1], [1, 0]]
Output: []

Each of the two modules waits on the other, so neither can ever start and the array comes back empty.

Constraints

1 <= moduleCount <= 2000
0 <= dependencies.length <= moduleCount * (moduleCount - 1)
dependencies[i].length == 2
0 <= x, y < moduleCount
x != y
No pair [x, y] is repeated
Among all valid sequences, return the lexicographically smallest

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 build_order(moduleCount: int, dependencies: list[list[int]]) -> list[int]:

Java

public int[] buildOrder(int moduleCount, int[][] dependencies)

The intended approach

Time O(V log V + E)Space O(V + E)

Count how many unfinished dependencies each module still waits on and keep every module at zero in a min-heap of unblocked candidates. Repeatedly pop the smallest label, emit it, and decrement its dependents, pushing any that reach zero. Popping the smallest at every step is exactly the lexicographic rule; if fewer than moduleCount labels are emitted, a cycle blocked the rest.

Pythonentry point: build_order

def build_order(moduleCount, dependencies):
    import heapq
    unlocks = [[] for _ in range(moduleCount)]
    waiting = [0] * moduleCount
    for later, earlier in dependencies:
        unlocks[earlier].append(later)
        waiting[later] += 1
    ready = [m for m in range(moduleCount) if waiting[m] == 0]
    heapq.heapify(ready)
    order = []
    while ready:
        module = heapq.heappop(ready)
        order.append(module)
        for nxt in unlocks[module]:
            waiting[nxt] -= 1
            if waiting[nxt] == 0:
                heapq.heappush(ready, nxt)
    return order if len(order) == moduleCount else []

The obvious version

Walk every permutation of the labels in lexicographic order and return the first one that places each dependency before its dependent. O(n! * E).

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_build_order(moduleCount, dependencies):
    from itertools import permutations
    for candidate in permutations(range(moduleCount)):
        slot = {}
        for i, module in enumerate(candidate):
            slot[module] = i
        if all(slot[earlier] < slot[later] for later, earlier in dependencies):
            return list(candidate)
    return []

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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