Doctor Appointment Slot Assignment

There are N patients, numbered from 0 to N - 1, who want to visit the doctor. The doctor has S possible appointment slots, numbered from 1 to S.
Each patient has two preferred slots. Patient K would like to visit the doctor during either slot A[K] or slot B[K].
The doctor can treat only one patient during each slot.
Return whether it is possible to assign every patient to one of their preferred slots so that at most one patient is assigned to each slot.

Function
solution(A: int[], B: int[], S: int) → boolean

Examples
Example 1
A = [1,1,3]
B = [2,2,1]
S = 3
return = true
One valid assignment is [1,2,3], where the K-th element is the slot assigned to patient K. Another valid assignment is [2,1,3].
Example 2
A = [3,2,3,1]
B = [1,3,1,2]
S = 3
return = false
There are four patients but only three slots, so at least two patients would have to share a slot.
Example 3
A = [2,5,6,5]
B = [5,4,2,2]
S = 8
return = true
One valid assignment is [5,4,6,2].
Example 4
A = [1,2,1,6,8,7,8]
B = [2,3,4,7,7,8,7]
S = 10
return = false
It is not possible to assign all patients to preferred slots while keeping at most one patient in each slot.

Constraints
N is an integer within the range [1, 100000].
S is an integer within the range [2, 100000].
Each element of arrays A and B is an integer within the range [1, S].
No patient has two preferences for the same slot, i.e. A[i] != B[i].

Each patient is an edge between slot A[K] and slot B[K]. Assigning a patient to a slot means orienting that edge toward one endpoint, and each node can receive at most one edge. So for every connected component, you need edges <= nodes. A tree component (edges = nodes - 1) works. A component with exactly one cycle works (edges = nodes). Anything with more edges than nodes fails. Use union-find: track node count and edge count per component, then check each one. The common pitfall is trying bipartite matching or backtracking, which is too slow at N up to 100000. Another pitfall is forgetting that slots nobody wants are just isolated nodes and don't matter. Example 2 fails quickly since 4 edges sit on 3 nodes. If the reframe slips your mind live, StealthCoder is the hedge that surfaces it while you keep typing.

The honest play: practice the pattern, and have StealthCoder ready for the one you didn't see coming.