Leetcode #1349: Maximum Students Taking Exam
In this guide, we solve Leetcode #1349 Maximum Students Taking Exam in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Given a m * n matrix seats that represent seats distributions in a classroom. If a seat is broken, it is denoted by '#' character otherwise it is denoted by a '.' character.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Bit Manipulation, Array, Dynamic Programming, Bitmask, Matrix
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
Input: seats = [["#",".","#","#",".","#"],
[".","#","#","#","#","."],
["#",".","#","#",".","#"]]
Output: 4
Explanation: Teacher can place 4 students in available seats so they don't cheat on the exam.
Python Solution
class Solution:
def maxStudents(self, seats: List[List[str]]) -> int:
def f(seat: List[str]) -> int:
mask = 0
for i, c in enumerate(seat):
if c == '.':
mask |= 1 << i
return mask
def dfs(seat: int, i: int) -> int:
ans = 0
for mask in range(1 << n):
if (seat | mask) != seat or (mask & (mask << 1)):
continue
cnt = mask.bit_count()
if i == len(ss) - 1:
ans = max(ans, cnt)
else:
nxt = ss[i + 1]
nxt &= ~(mask << 1)
nxt &= ~(mask >> 1)
ans = max(ans, cnt + dfs(nxt, i + 1))
return ans
n = len(seats[0])
ss = [f(s) for s in seats]
return dfs(ss[0], 0)
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.