Leetcode #552: Student Attendance Record II
In this guide, we solve Leetcode #552 Student Attendance Record II 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
An attendance record for a student can be represented as a string where each character signifies whether the student was absent, late, or present on that day. The record only contains the following three characters: 'A': Absent.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Dynamic Programming
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: n = 2
Output: 8
Explanation: There are 8 records with length 2 that are eligible for an award:
"PP", "AP", "PA", "LP", "PL", "AL", "LA", "LL"
Only "AA" is not eligible because there are 2 absences (there need to be fewer than 2).
Python Solution
class Solution:
def checkRecord(self, n: int) -> int:
def dfs(i, j, k):
if i >= n:
return 1
ans = 0
if j == 0:
ans += dfs(i + 1, j + 1, 0)
if k < 2:
ans += dfs(i + 1, j, k + 1)
ans += dfs(i + 1, j, 0)
return ans % mod
mod = 10**9 + 7
ans = dfs(0, 0, 0)
dfs.cache_clear()
return ans
Complexity
The time complexity is O(n·m) (typical). The space complexity is O(n·m) or optimized.
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.