1700. Number of Students Unable to Eat Lunch

Problem

The school cafeteria offers circular and square sandwiches at lunch break, referred to by numbers 0 and 1 respectively. All students stand in a queue. Each student either prefers square or circular sandwiches.

The number of sandwiches in the cafeteria is equal to the number of students. The sandwiches are placed in a stack. At each step:

• If the student at the front of the queue prefers the sandwich on the top of the stack, they will take it and leave the queue.
• Otherwise, they will leave it and go to the queue's end.

This continues until none of the queue students want to take the top sandwich and are thus unable to eat.

You are given two integer arrays students and sandwiches where sandwiches[i] is the type of the ith sandwich in the stack (i = 0 is the top of the stack) and students[j] is the preference of the jth student in the initial queue (j = 0 is the front of the queue). Return the number of students that are unable to eat.

Example 1:

Input: students = [1,1,0,0], sandwiches = [0,1,0,1]
Output: 0 
Explanation:
- Front student leaves the top sandwich and returns to the end of the line making students = [1,0,0,1].
- Front student leaves the top sandwich and returns to the end of the line making students = [0,0,1,1].
- Front student takes the top sandwich and leaves the line making students = [0,1,1] and sandwiches = [1,0,1].
- Front student leaves the top sandwich and returns to the end of the line making students = [1,1,0].
- Front student takes the top sandwich and leaves the line making students = [1,0] and sandwiches = [0,1].
- Front student leaves the top sandwich and returns to the end of the line making students = [0,1].
- Front student takes the top sandwich and leaves the line making students = [1] and sandwiches = [1].
- Front student takes the top sandwich and leaves the line making students = [] and sandwiches = [].
Hence all students are able to eat.

Example 2:

Input: students = [1,1,1,0,0,1], sandwiches = [1,0,0,0,1,1]
Output: 3

Constraints:

• 1 <= students.length, sandwiches.length <= 100
• students.length == sandwiches.length
• sandwiches[i] is 0 or 1.
• students[i] is 0 or 1.

Problem Summary

The school cafeteria provides circular and square sandwiches, represented by the numbers 0 and 1 respectively. All students stand in a queue, and each student either likes circular sandwiches or square sandwiches. The number of sandwiches in the cafeteria is the same as the number of students. All sandwiches are placed in a stack. In each round:

• If the student at the front of the queue likes the sandwich on top of the stack, they will take it and leave the queue.
• Otherwise, this student will give up this sandwich and return to the end of the queue. This process continues until all students in the queue do not like the sandwich on top of the stack.

You are given two integer arrays students and sandwiches, where sandwiches[i] is the type of the ith sandwich in the stack (i = 0 is the top of the stack), and students[j] is the preference of the jth student in the initial queue (j = 0 is the starting position of the queue). Please return the number of students who are unable to eat lunch.

Solution Approach

• Easy problem. According to the problem statement, no matter how students take turns getting sandwiches, as long as there are enough, they can always get one after multiple rounds. The problem can be equivalently seen as students coming to the front of the queue one by one to get sandwiches. The 2 types of sandwiches have already been placed there. In the end, students who cannot get a sandwich are all unable to do so because there are not enough sandwiches of their preferred type. Following this idea, first count the total number of the 2 types of sandwiches, then subtract the students' total demand for sandwiches; the remaining students are exactly those who cannot be satisfied.

Code

package leetcode

func countStudents(students []int, sandwiches []int) int {
	tmp, n, i := [2]int{}, len(students), 0
	for _, v := range students {
		tmp[v]++
	}
	for i < n && tmp[sandwiches[i]] > 0 {
		tmp[sandwiches[i]]--
		i++
	}
	return n - i
}