How do you solve Student Attendance Record I on LeetCode?

Student Attendance Record I (LeetCode #551) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Count occurrences efficiently in a single pass.

What is the brute force solution for Student Attendance Record I?

The brute force approach for Student Attendance Record I is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check each character in the string to count the number of absences and track consecutive lates. This approach is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Student Attendance Record I?

Student Attendance Record I uses the following concepts: String. The recommended patterns to study are: Array, String.

What are the interview tips for Student Attendance Record I?

Explain your thought process clearly while coding. Consider edge cases like all 'P's or all 'A's. Use descriptive variable names to make your code readable.

What are common mistakes when solving Student Attendance Record I?

Not resetting the late counter when encountering 'P'. Overcomplicating the conditions for eligibility.

#551

Student Attendance Record I

Easy

StringArrayString

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

We can achieve the same result with a single pass through the string while keeping track of absences and consecutive lates. This is efficient and straightforward.

⚙️

Algorithm

6 steps

1Step 1: Initialize counters for absences and consecutive lates.
2Step 2: Loop through each character in the string.
3Step 3: Increment the absence counter for 'A' and check if it exceeds 1.
4Step 4: Increment the late counter for 'L' and check if it reaches 3.
5Step 5: Reset the late counter if the character is 'P'.
6Step 6: Return true if no conditions for disqualification are met.

solution.py15 lines

1def checkRecord(s):
2    absences = 0
3    late_count = 0
4    for char in s:
5        if char == 'A':
6            absences += 1
7            if absences > 1:
8                return False
9        if char == 'L':
10            late_count += 1
11            if late_count == 3:
12                return False
13        else:
14            late_count = 0
15    return True

ℹ

Complexity note: The time complexity is O(n) because we only traverse the string once. The space complexity is O(1) since we are using a constant amount of space for counters.

1Count occurrences efficiently in a single pass.
2Use simple counters to track conditions.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.