Student Attendance Record I - Practice Coding Problems

Student Attendance Record I - Problem

String Easy

Imagine you're a teacher evaluating students for the perfect attendance award at the end of the school year! 🎓

You have a student's attendance record represented as a string s, where each character tells you what happened on that day:

'A' - Absent (student didn't come to school)
'L' - Late (student arrived after the bell)
'P' - Present (student was on time)

To win the attendance award, a student must meet both criteria:

🚫 No more than 1 absence - They can miss at most 1 day total
⏰ Never late 3+ days in a row - No streaks of 3 or more consecutive late days

Return true if the student qualifies for the award, false otherwise.

Example: "PPALLP" → true (1 absence, max 2 consecutive lates)
"PPALLL" → false (3 consecutive lates = disqualified!)

Input & Output

example_1.py — Basic Valid Case

$ Input: s = "PPALLP"

› Output: true

💡 Note: The student has 1 absence (≤ 1 allowed) and maximum 2 consecutive lates (< 3 allowed). Both conditions satisfied.

example_2.py — Too Many Consecutive Lates

$ Input: s = "PPALLL"

› Output: false

💡 Note: Although only 0 absences, the student has 3 consecutive lates which violates the attendance policy.

example_3.py — Too Many Absences

$ Input: s = "ALAPPLA"

› Output: false

💡 Note: The student has 2 absences, which exceeds the maximum allowed (1). The consecutive late condition is satisfied but overall ineligible.

Constraints

1 ≤ s.length ≤ 1000
s[i] is either 'A', 'L', or 'P'
Case sensitive: Only uppercase letters are valid

Visualization

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Understanding the Visualization

Start the Review

Begin with clean counters: 0 absences, 0 consecutive lates

Daily Assessment

For each day, update your mental notes and check if limits are exceeded

Immediate Decision

Stop as soon as you find a violation - no need to continue checking

Award Verdict

If you finish without violations, the student gets the award!

Key Takeaway

🎯 Key Insight: We can solve both conditions simultaneously in one pass, making immediate decisions without storing historical data!

Asked in

G Google 15 a Amazon 12 f Meta 8 ⊞ Microsoft 6

The optimal solution uses a single-pass approach with two counters: one tracking total absences and another tracking consecutive late days. We iterate through the string once, updating counters and checking violations immediately. This achieves O(n) time complexity with O(1) space, making it both efficient and elegant.

Common Approaches

Approach	Time	Space	Notes
✓ One-Pass Optimal Solution	O(n)	O(1)	Single pass tracking both absence count and consecutive late streak
Brute Force (Multiple Passes)	O(n)	O(1)	Check absence count first, then scan for consecutive late patterns

One-Pass Optimal Solution — Algorithm Steps

Initialize absence counter and consecutive late counter
Iterate through each character in the string
For 'A': increment absence count, check if ≥ 2
For 'L': increment late streak, check if ≥ 3
For 'P': reset late streak to 0
Return true if we complete the loop without violations

Visualization

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Step-by-Step Walkthrough

Initialize Counters

Start with absent_count=0, consecutive_late=0

Process Each Day

Update counters based on attendance status

Check Violations

Return false immediately if any limit exceeded

Award Decision

If we finish the loop, student qualifies!

Code -

solution.c — C

#include <stdio.h>
#include <string.h>
#include <stdbool.h>

bool checkRecord(char* s) {
    int len = strlen(s);
    int absentCount = 0;
    int consecutiveLate = 0;
    
    for (int i = 0; i < len; i++) {
        if (s[i] == 'A') {
            absentCount++;
            if (absentCount >= 2) {
                return false;
            }
            consecutiveLate = 0; // Reset late streak
        } else if (s[i] == 'L') {
            consecutiveLate++;
            if (consecutiveLate >= 3) {
                return false;
            }
        } else { // s[i] == 'P'
            consecutiveLate = 0; // Reset late streak
        }
    }
    
    return true;
}

int main() {
    char s[100];
    scanf("%s", s);
    // Remove quotes if present
    if (s[0] == '"') {
        memmove(s, s+1, strlen(s));
        s[strlen(s)-1] = '\0';
    }
    printf("%s\n", checkRecord(s) ? "true" : "false");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through the string, with potential early termination

✓ Linear Growth

Space Complexity

O(1)

Only using two integer variables for tracking state

✓ Linear Space

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~15 min Avg. Time

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

One-Pass Optimal Solution — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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