# Solution For Maximum Students Taking Exam

Problem Description:

Given a class schedule of n lectures, with start and end time of each lecture, and a limit k on the number of exams a student can take, what is the maximum number of students that can take exams?

Solution:

This problem can be solved using binary search. We need to find the maximum number of students that can take exams by minimizing the number.

First, we need to sort the lectures according to their end time. Then, we can define a function count(exams) that will return the maximum number of students that can take exams, given that each student can take at most exams exams.

We can use a sliding window approach to count the number of exams that can be taken by each student. We keep track of the smallest start time of the next lecture that a student can attend, and the largest end time of the current exams that the student is attending. If the next lecture starts after the student’s current exams have ended, the student attends the next lecture and the current exams are finished. If the student has already taken exams exams, we move to the next student.

To implement the count function, we can loop through all the lectures and count the number of exams that can be taken. If the number of exams is less than or equal to k, we return the number of students that can take exams. Otherwise, we continue with the binary search.

The binary search algorithm repeatedly calls the count function with different values of the number of exams, until the maximum number of students is found. The search range is from 1 to the number of lectures, since each student can take at most one exam per lecture.

Time Complexity:

Sorting the lectures takes O(n log n) time. Each count call takes O(n) time. Therefore, the overall time complexity of the binary search is O(n log n * log n).

Space Complexity:

The space complexity is O(n), since we need to store the lectures.

Here’s the implementation of the solution:

“`

class Solution:

def maxStudents(self, seats: List[List[str]]) -> int:

m, n = len(seats), len(seats[0])

```
def count(exams):
res = 0
for i in range(m):
students = 0
start = 0
for j in range(n):
if seats[i][j] == '.':
if j - start >= exams:
students += 1
start = j
elif seats[i][j] == '#':
start = j + 1
if n - start >= exams:
students += 1
res += students
return res
l, r = 1, n
while l <= r:
mid = (l + r) // 2
if count(mid) >= m:
ans = mid
l = mid + 1
else:
r = mid - 1
return ans
```

“`

In this solution, we loop through all the rows and columns of the seats matrix. We keep track of the number of students that can take exams in each row, using the variable students. We also keep track of the smallest start time of the next lecture that a student can attend, using the variable start.

We then calculate the maximum number of students that can take exams using the count function, and update the binary search range accordingly. Finally, we return the maximum number of students that can take exams.

## Step by Step Implementation For Maximum Students Taking Exam

There are a number of ways to solve this problem, but one approach would be as follows: 1) Sort the students by their exam scores in descending order. 2) Iterate through the students, placing them in the first available seat. If there are no available seats, skip that student. 3) Return the number of students who were able to take the exam. Here is some sample code to illustrate this approach: int maxStudents(int[][] seats) { Arrays.sort(seats, new Comparator() { public int compare(int[] a, int[] b) { return b[1] - a[1]; } }); int numStudents = 0; for (int[] student : seats) { // try to find an available seat boolean foundSeat = false; for (int i = 0; i < seats.length; i++) { if (seats[i][0] == 0) { // found an available seat! seats[i][0] = 1; foundSeat = true; break; } } if (foundSeat) { numStudents++; } } return numStudents; }

def maxStudents(self, seats: List[List[str]]) -> int: # we can represent the students as a binary number, where 1 means the student can sit in that seat # and 0 means the student cannot sit in that seat. For example, if seats = [['#','.','#','#','.','#'], # ['.','#','#','#','#','.'], ['#','.','#','#','.','#']], then the binary representation of the students # would be students = [110, 111, 101]. # We can also represent the seats as a binary number, where 1 means the seat is available and 0 means # the seat is unavailable. For example, if seats = [['#','.','#','#','.','#'], ['.','#','#','#','#','.'], # ['#','.','#','#','.','#']], then the binary representation of the seats would be seats = [110110, # 111111, 110101]. # We want to find the maximum number of students that can take the exam such that no two students are # sitting in the same row and no two students are sitting in the same column. # We can use a brute force approach, where we try every possible combination of students and seats # and check if it is valid. If it is valid, we update our maximum number of students. # We can optimize our brute force approach by only considering combinations of students that are valid. # We can do this by keeping track of the number of students that are sitting in each row and column. # If the number of students sitting in a row or column exceeds the number of available seats in that row # or column, then we know that the combination of students is not valid. # We can also optimize our brute force approach by only considering combinations of students that are valid. # We can do this by keeping track of the number of students that are sitting in each row and column. # If the number of students sitting in a row or column exceeds the number of available seats in that row # or column, then we know that the combination of students is not valid. # We can also optimize our brute force approach by only considering combinations of students that are valid. # We can do this by keeping track of the number of students that are sitting in each row and column. # If the number of students sitting in a row or column exceeds the number of available seats in that row # or column, then we know that the combination of students is not valid. # We can also optimize our brute force approach by only considering combinations of students that are valid. # We can do this by keeping track of the number of students that are sitting in each row and column. # If the number of students sitting in a row or column exceeds the number of available seats in that row # or column, then we know that the combination of students is not valid. # We can also optimize our brute force approach by only considering combinations of students that are valid. # We can do this by keeping track of the number of students that are sitting in each row and column. # If the number of students sitting in a row or column exceeds the number of available seats in that row # or column, then we know that the combination of students is not valid. # We can also optimize our brute force approach by only considering combinations of students that are valid. # 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or column, then we know that the combination of students is not valid. # We can also optimize our brute force approach by only considering combinations of students that are valid. # We can do this by keeping track of the number of students that are sitting in each row and column. # If the number of students sitting in a row or column exceeds the number of available seats in that row # or column, then we know that the combination of students is not valid. # We can also optimize our brute force approach by only considering combinations of students that are valid. # We can do this by keeping track of the number of students that are sitting in each row and column. # If the number of students sitting in a row or column exceeds the number of available seats in that row # or column, then we know that the combination of students is not valid. # We can also optimize our brute force approach by only considering combinations of students that are valid. # We can do this by keeping track of the number of students that are sitting in each row and column. # 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/** * @param {number[][]} seats * @return {number} */ var maxStudents = function(seats) { // TODO };

class Solution { public: int maxStudents(vector>& seats) { int n = seats.size(); int m = seats[0].size(); // dp[i][mask] := the maximum number of students we can place // in the first i rows, given that the i-th row has the students // specified by the mask vector > dp(n + 1, vector (1 << m, -1)); dp[0][0] = 0; for (int i = 1; i <= n; i++) { for (int mask = 0; mask < (1 << m); mask++) { // We can't place more than one student in the same column if (__builtin_popcount(mask) > i) continue; // We can't place students in a row if there's already a student // in that row in the same column bool valid = true; for (int j = 0; j < m; j++) { if ((mask & (1 << j)) && seats[i-1][j]) { valid = false; break; } } if (!valid) continue; // We can't place students in a row if there's already a student // in that row in the column to the left for (int j = 1; j < m; j++) { if ((mask & (1 << j)) && (mask & (1 << (j-1)))) { valid = false; break; } } if (!valid) continue; // Try placing students in all the previous rows for (int prev_mask = 0; prev_mask < (1 << m); prev_mask++) { // We can't place students in a row if there's already a student // in that row in the column to the right bool valid = true; for (int j = 0; j < m-1; j++) { if ((prev_mask & (1 << j)) && (mask & (1 << (j+1)))) { valid = false; break; } } if (!valid) continue; // If we can place students in the previous rows, see if we can // place students in the current row if (dp[i-1][prev_mask] != -1) { dp[i][mask] = max(dp[i][mask], dp[i-1][prev_mask] + __builtin_popcount(mask)); } } } } // Return the maximum number of students we can place in all the rows int ans = 0; for (int mask = 0; mask < (1 << m); mask++) { ans = max(ans, dp[n][mask]); } return ans; } };

public class Solution { public int MaxStudents(char[][] seats) { int res = 0; int n = seats.Length; int m = seats[0].Length; // dp[i][mask] - the maximum number of students that can be seated // in the first i rows using the seating configuration represented by mask int[,] dp = new int[n + 1, 1 << m]; for (int i = 1; i <= n; ++i) { for (int mask = 0; mask < (1 << m); ++mask) { // check if the current seating configuration is valid bool valid = true; for (int j = 0; j < m && valid; ++j) { if ((mask & (1 << j)) > 0 && (mask & (1 << (j + 1))) > 0) { valid = false; } } if (!valid) { continue; } // try to seat students in the current row for (int j = 0; j < m; ++j) { if (seats[i - 1][j] == '#') { continue; } if ((mask & (1 << j)) == 0) { dp[i, mask] = Math.Max(dp[i, mask], dp[i - 1, mask | (1 << j)] + 1); } } } } for (int mask = 0; mask < (1 << m); ++mask) { res = Math.Max(res, dp[n, mask]); } return res; } }