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Two Shift Conflict Split

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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 repair depot has workstations numbered 0 through n - 1. Two workstations that draw from the same power rail cannot be serviced during the same shift. The clash list is given as conflicts, where conflicts[u] holds every workstation that clashes with workstation u.

Clashes are mutual: whenever v occurs in conflicts[u], u occurs in conflicts[v]. No workstation clashes with itself, and no workstation is listed twice inside one entry. Some workstations may have an empty clash list, and the clashes may break the floor into several groups with no link between them.

The depot runs exactly two shifts, morning and evening, and every workstation must be booked into one of them. Return true when a booking exists that keeps each clashing pair in different shifts, and false when no such booking exists.

Examples

Example 1

Input: conflicts = [[2,3],[3,4],[0],[0,1],[1]]
Output: true

Book workstations 0 and 1 in the morning and workstations 2, 3 and 4 in the evening. Each of the four listed clashes has its two ends in different shifts.

Example 2

Input: conflicts = [[1,2,4],[0,2],[0,1],[4],[0,3]]
Output: false

Every way of booking these five workstations into two shifts leaves at least one listed clash with both ends in the same shift.

Example 3

Input: conflicts = [[1],[0],[3,4],[2,4],[2,3]]
Output: false

No booking of all five workstations keeps every listed clash split across the two shifts.

Constraints

conflicts.length == n
1 <= n <= 100
0 <= conflicts[u].length < n
0 <= conflicts[u][i] <= n - 1
conflicts[u] does not contain u.
All the values of conflicts[u] are unique.
If conflicts[u] contains v, then conflicts[v] contains u.

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 can_split_shifts(conflicts: list[list[int]]) -> bool:

Java

public boolean canSplitShifts(int[][] conflicts)

The intended approach

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

Sweep every workstation as a possible starting point so that groups with no link between them are all covered. From each unlabelled start, run a queue-driven walk that stamps the opposite shift label on each newly reached workstation, and report failure the moment a clash connects two workstations already carrying the same label.

Pythonentry point: can_split_shifts

from collections import deque


def can_split_shifts(conflicts):
    # Walk every clash group with a queue, stamping alternating shift labels on
    # each hop.  A clash whose far end already carries the current label is a
    # contradiction.  O(V + E) time, O(V) space for the label array plus queue.
    n = len(conflicts)
    shift = [0] * n              # 0 unassigned, 1 morning, 2 evening
    for root in range(n):
        if shift[root]:
            continue
        shift[root] = 1
        queue = deque([root])
        while queue:
            u = queue.popleft()
            other = 3 - shift[u]
            for v in conflicts[u]:
                if shift[v] == 0:
                    shift[v] = other
                    queue.append(v)
                elif shift[v] != other:
                    return False
    return True

The obvious version

Enumerate all 2^n morning/evening bookings as bit masks and accept the first mask under which no clash has both ends on the same side.

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_can_split_shifts(conflicts):
    n = len(conflicts)
    for mask in range(1 << n):
        ok = True
        for u in range(n):
            for v in conflicts[u]:
                if ((mask >> u) & 1) == ((mask >> v) & 1):
                    ok = False
                    break
            if not ok:
                break
        if ok:
            return True
    return False

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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