Student Attendance Record I

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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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

✓ Brute Force - Multiple Passes

⏱️ Time: O(n) Space: O(1)

First pass through the string to count total 'A' characters. Second pass to check for any substring of 3 or more consecutive 'L' characters. Return false if either condition fails.

Single Pass - Optimal

⏱️ Time: O(n) Space: O(1)

Iterate through the string once, maintaining counters for total absences and current consecutive lates. Return false immediately when either condition is violated.

Brute Force - Multiple Passes — Algorithm Steps

Pass 1: Count all 'A' characters in the string
Pass 2: Check for any occurrence of 'LLL' (3 consecutive lates)
Return false if absences ≥ 2 or consecutive lates ≥ 3

Visualization

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Step-by-Step Walkthrough

Pass 1: Count Absences

Count all 'A' characters

Pass 2: Find Consecutive Lates

Look for 'LLL' pattern

Decision

Award if both conditions pass

Code -

solution.c — C

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

bool solution(char* s) {
    int len = strlen(s);
    
    // Count total absences
    int absentCount = 0;
    for (int i = 0; i < len; i++) {
        if (s[i] == 'A') absentCount++;
    }
    if (absentCount >= 2) return false;
    
    // Check for 3 consecutive lates
    if (strstr(s, "LLL") != NULL) return false;
    
    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)

Two passes through string of length n

✓ Linear Growth

Space Complexity

O(1)

Only using counter variables

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Multiple Passes — Algorithm Steps

Visualization

Code -

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