Student Attendance Record I - Practice Coding Problems

Student Attendance Record I - Problem

String Easy

You are given a string s representing an attendance record for a student 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
'L': Late
'P': Present

The student is eligible for an attendance award if they meet both of the following criteria:

The student was absent ('A') for strictly fewer than 2 days total
The student was never late ('L') for 3 or more consecutive days

Return true if the student is eligible for an attendance award, or false otherwise.

Input & Output

Example 1 — Basic Valid Case

$ Input: s = "PPALLP"

› Output: true

💡 Note: The student has 1 absence (< 2) and at most 2 consecutive lates (< 3), so they are eligible for the award.

Example 2 — Too Many Absences

$ Input: s = "PPALLL"

› Output: false

💡 Note: The student has 3 consecutive lates, which violates the attendance policy.

Example 3 — Multiple Absences

$ Input: s = "ALPPLA"

› Output: false

💡 Note: The student has 2 absences, which is not strictly fewer than 2, so they are not eligible.

Constraints

1 ≤ s.length ≤ 1000
s[i] is either 'A', 'L', or 'P'

Visualization

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Understanding the Visualization

Input

Attendance string with A (absent), L (late), P (present)

Check Criteria

Count absences < 2 and no 3+ consecutive lates

Award Decision

Return true if both conditions are satisfied

Key Takeaway

🎯 Key Insight: Track two simple counters while scanning the string once

Asked in

G Google 12 a Amazon 8 M Microsoft 5

The key insight is to track two counters in a single pass: total absences and consecutive lates. Best approach uses single pass with early termination. Time: O(n), Space: O(1)

Common Approaches

Approach	Time	Space	Notes
✓ Single Pass - Optimal	O(n)	O(1)	One pass tracking both absent count and consecutive late days
Brute Force - Multiple Passes	O(n)	O(1)	Two separate passes: one to count absences, another to find consecutive lates

Single Pass - Optimal — Algorithm Steps

Initialize absent_count = 0 and consecutive_lates = 0
For each character, update counters and check conditions
Return false immediately if absences ≥ 2 or consecutive lates ≥ 3

Visualization

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Step-by-Step Walkthrough

Initialize

Start with absent_count=0, consecutive_lates=0

Process Each Character

Update counters based on current character

Early Exit

Return false immediately if any condition fails

Code -

solution.c — C

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

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

int main() {
    char s[1000];
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0; // Remove newline
    
    bool result = solution(s);
    printf(result ? "true\n" : "false\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through string of length n

✓ Linear Growth

Space Complexity

O(1)

Only using two counter variables

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Single Pass - Optimal — Algorithm Steps

Visualization

Code -

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