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In this approach, we simulate the exact process of distributing sandwiches using a queue-like structure for the students. We repeatedly check if the student at the front of the queue can take the sandwich at the top of the stack. If not, the student is moved to the back of the queue. This continues until we complete a full cycle without any students taking a sandwich, indicating they can't eat.
Time Complexity: O(n^2) in the worst scenario because each student may be checked multiple times.
Space Complexity: O(1) since it only uses a fixed amount of extra space.
1from collections import deque
2
3def count_students(students, sandwiches):
4 queue = deque(students)
5 j = 0
6 cycle_count = 0
7 while queue and cycle_count < len(queue):
8 if queue[0] == sandwiches[j]:
9 queue.popleft()
10 j += 1
11 cycle_count = 0
12 else:
13 queue.append(queue.popleft())
14 cycle_count += 1
15 return len(queue)
16
17students = [1, 1, 0, 0]
18sandwiches = [0, 1, 0, 1]
19print(count_students(students, sandwiches))This Python solution utilizes a deque for efficient pop and append operations to simulate the queue operations. It keeps track of cycle counts to detect when no further matches can be made, breaking when necessary.
This approach uses counting to determine if students can satisfy their preferences before iterating over sandwiches. By counting the number of students preferring each type, we can efficiently determine how many students will be unable to eat if sandwiches of their preference run out before they can eat.
Time Complexity: O(n) because it processes each list one time.
Space Complexity: O(1) since it only uses fixed additional space for counting.
The C implementation counts the preferences for circular and square sandwiches. It iterates over the sandwiches and decreases the count of the suitable preference whenever a sandwich is taken. If no more students prefer the top sandwich, it returns the remaining count.