Count Days Without Meetings

Count Days Without Meetings - Problem

Array Sorting Medium

You are given a positive integer days representing the total number of days an employee is available for work (starting from day 1). You are also given a 2D array meetings of size n where meetings[i] = [start_i, end_i] represents the starting and ending days of meeting i (inclusive).

Return the count of days when the employee is available for work but no meetings are scheduled.

Note: The meetings may overlap.

Input & Output

Example 1 — Basic Case with Overlapping

$ Input: days = 10, meetings = [[5,7],[1,3],[9,10]]

› Output: 2

💡 Note: Meeting days are: 1,2,3 (first meeting), 5,6,7 (second meeting), 9,10 (third meeting). Free days are: 4,8. Total free days = 2.

Example 2 — No Meetings

$ Input: days = 4, meetings = []

› Output: 4

💡 Note: No meetings scheduled, so all 4 days (1,2,3,4) are free for work.

Example 3 — All Days Busy

$ Input: days = 3, meetings = [[2,4]]

› Output: 1

💡 Note: Meeting covers days 2,3. Only day 1 is free (day 4 is beyond the available days). Free days = 1.

Constraints

1 ≤ days ≤ 10⁹
0 ≤ meetings.length ≤ 1000
1 ≤ start_i ≤ end_i ≤ days

Visualization

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Asked in

G Google 23 M Microsoft 18 a Amazon 15

The key insight is to handle overlapping meetings efficiently. For small days, use boolean array marking O(days). For large days, sort and merge intervals first O(n log n). Best approach depends on constraints: Sort and Merge is optimal for large days. Time: O(n log n), Space: O(n).

Common Approaches

✓ Binary Search

⏱️ Time: N/A Space: N/A

Boolean Array Marking

⏱️ Time: O(days + sum of meeting lengths) Space: O(days)

Create a boolean array of size days+1. Mark all days covered by meetings as true. Then count the number of false entries to get free days.

Brute Force - Check Each Day

⏱️ Time: O(days × meetings) Space: O(1)

Iterate through each day and for each day, check all meetings to see if any meeting covers that day. Count days that are not covered by any meeting.

Sort and Merge Intervals

⏱️ Time: O(n log n) Space: O(n)

First sort all meetings by start time. Then merge overlapping meetings into non-overlapping intervals. Finally, count the days not covered by any interval.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

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

static bool occupied[100005];

int countFreeDays(int days, int meetings[][2], int meetingCount) {
    // Initialize all days as free
    for (int i = 1; i <= days; i++) {
        occupied[i] = false;
    }
    
    // Mark occupied days
    for (int i = 0; i < meetingCount; i++) {
        int start = meetings[i][0];
        int end = meetings[i][1];
        for (int day = start; day <= end && day <= days; day++) {
            occupied[day] = true;
        }
    }
    
    // Count free days
    int count = 0;
    for (int i = 1; i <= days; i++) {
        if (!occupied[i]) {
            count++;
        }
    }
    
    return count;
}

int main() {
    char line[10000];
    
    // Read days
    fgets(line, sizeof(line), stdin);
    int days = atoi(line);
    
    // Read meetings array
    fgets(line, sizeof(line), stdin);
    
    static int meetings[1000][2];
    int meetingCount = 0;
    
    // Parse meetings array
    char* p = line;
    
    // Skip whitespace and find '['
    while (*p && *p != '[') p++;
    
    if (*p == '[') {
        p++;
        
        // Check if empty array
        while (*p == ' ') p++;
        if (*p == ']') {
            meetingCount = 0;
        } else {
            // Parse meetings
            while (*p && *p != ']') {
                // Skip whitespace and commas
                while (*p == ' ' || *p == ',') p++;
                
                if (*p == '[') {
                    p++;
                    
                    // Parse start
                    while (*p == ' ') p++;
                    meetings[meetingCount][0] = (int)strtol(p, &p, 10);
                    
                    // Skip comma
                    while (*p == ' ' || *p == ',') p++;
                    
                    // Parse end
                    meetings[meetingCount][1] = (int)strtol(p, &p, 10);
                    
                    // Skip to ']'
                    while (*p && *p != ']') p++;
                    if (*p == ']') p++;
                    
                    meetingCount++;
                }
            }
        }
    }
    
    printf("%d\n", countFreeDays(days, meetings, meetingCount));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

✓ Linear Space

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