Skip Lunch Count

Count the number of days an employee skips lunch based on breakfast timing after 11:00 AM

Understand the Problem

Problem Statement

An employee skips lunch on a given day if he has breakfast after 11:00 AM. Given the breakfast times for N days in 24-hour format, determine how many days the employee skipped lunch.

Constraints

1 ≤ N ≤ 100
Breakfast time is given in HH:MM 24-hour format
Time values are valid (HH from 00-23, MM from 00-59)
Each day's breakfast time is separated by a space

Examples

Example 1

Input

5
10:00 11:00 11:01 9:00 11:05

Output

Explanation

The employee had breakfast after 11:00 AM on the 3rd day (11:01) and 5th day (11:05), so he skipped lunch on 2 days. Breakfast at exactly 11:00 AM (2nd day) does not count as after 11:00 AM.

Solution

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

int main() {
    int n;
    scanf("%d", &n);
    
    int skipCount = 0;
    char time[6];
    
    for(int i = 0; i < n; i++) {
        scanf("%s", time);
        
        // Parse hour and minute
        int hour = (time[0] - '0') * 10 + (time[1] - '0');
        int minute = (time[3] - '0') * 10 + (time[4] - '0');
        
        // Check if breakfast was after 11:00 AM
        if(hour > 11 || (hour == 11 && minute > 0)) {
            skipCount++;
        }
    }
    
    printf("%d", skipCount);
    return 0;
}

Time:O(N) - We process each day exactly once

Space:O(1) - We only use a constant amount of extra space for variables

Approach:

1. Read the number of days N
2. For each day, read the breakfast time as a string
3. Parse the hour (first two characters) and minute (characters 4-5) from the time string
4. Check if hour > 11 OR (hour == 11 AND minute > 0)
5. Increment counter for each day breakfast was after 11:00 AM
6. Print the final count

Visual Explanation

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