Count Days Spent Together

Count Days Spent Together - Problem

Math String Easy

Alice and Bob are traveling to Rome for separate business meetings.

You are given 4 strings arriveAlice, leaveAlice, arriveBob, and leaveBob. Alice will be in the city from the dates arriveAlice to leaveAlice (inclusive), while Bob will be in the city from the dates arriveBob to leaveBob (inclusive). Each will be a 5-character string in the format "MM-DD", corresponding to the month and day of the date.

Return the total number of days that Alice and Bob are in Rome together.

You can assume that all dates occur in the same calendar year, which is not a leap year. Note that the number of days per month can be represented as: [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31].

Input & Output

Example 1 — Partial Overlap

$ Input: arriveAlice = "08-15", leaveAlice = "08-18", arriveBob = "08-16", leaveBob = "08-19"

› Output: 3

💡 Note: Alice is in Rome from Aug 15-18, Bob from Aug 16-19. They overlap on Aug 16, 17, and 18 = 3 days

Example 2 — No Overlap

$ Input: arriveAlice = "10-01", leaveAlice = "10-03", arriveBob = "11-01", leaveBob = "12-03"

› Output: 0

💡 Note: Alice leaves on Oct 3, Bob arrives on Nov 1. No overlap period, so 0 days together

Example 3 — Complete Overlap

$ Input: arriveAlice = "09-01", leaveAlice = "10-19", arriveBob = "09-04", leaveBob = "09-09"

› Output: 6

💡 Note: Bob's entire stay (Sep 4-9) is within Alice's stay (Sep 1 - Oct 19), so 6 days together

Constraints

All dates given in "MM-DD" format
All dates occur in the same non-leap year
1 ≤ MM ≤ 12
1 ≤ DD ≤ days in that month

Visualization

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Asked in

G Google 15 a Amazon 12 M Microsoft 8

The key insight is to convert MM-DD dates to day numbers (1-365), then calculate the overlap using the formula: max(0, min(end1, end2) - max(start1, start2) + 1). Best approach is the direct overlap calculation with Time: O(1), Space: O(1).

Common Approaches

✓ Brute Force Day by Day

⏱️ Time: O(1) Space: O(1)

Convert all dates to day numbers (1-365), then iterate through each day to check if both people are in Rome on that day. Count all such days.

Direct Overlap Calculation

⏱️ Time: O(1) Space: O(1)

Convert dates to day numbers, then use mathematical formula to find overlap: max(0, min(end1, end2) - max(start1, start2) + 1). This avoids checking every day individually.

Brute Force Day by Day — Algorithm Steps

Convert all MM-DD dates to day numbers (1-365)
For each day from 1 to 365, check if both Alice and Bob are present
Count the days where both are present

Visualization

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Step-by-Step Walkthrough

Convert Dates

Convert MM-DD format to day numbers (1-365)

Check Each Day

Iterate through days 1-365 and count overlaps

Count Overlaps

Return total days both are present

Code -

solution.c — C

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

int dateToDay(char* dateStr) {
    int daysInMonth[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    int month = (dateStr[0] - '0') * 10 + (dateStr[1] - '0');
    int day = (dateStr[3] - '0') * 10 + (dateStr[4] - '0');
    
    int dayNumber = day;
    for (int i = 0; i < month - 1; i++) {
        dayNumber += daysInMonth[i];
    }
    return dayNumber;
}

int solution(char* arriveAlice, char* leaveAlice, char* arriveBob, char* leaveBob) {
    int aliceStart = dateToDay(arriveAlice);
    int aliceEnd = dateToDay(leaveAlice);
    int bobStart = dateToDay(arriveBob);
    int bobEnd = dateToDay(leaveBob);
    
    int count = 0;
    for (int day = 1; day <= 365; day++) {
        if (aliceStart <= day && day <= aliceEnd && bobStart <= day && day <= bobEnd) {
            count++;
        }
    }
    
    return count;
}

int main() {
    char arriveAlice[10], leaveAlice[10], arriveBob[10], leaveBob[10];
    
    fgets(arriveAlice, sizeof(arriveAlice), stdin);
    arriveAlice[strcspn(arriveAlice, "\n")] = 0;
    
    fgets(leaveAlice, sizeof(leaveAlice), stdin);
    leaveAlice[strcspn(leaveAlice, "\n")] = 0;
    
    fgets(arriveBob, sizeof(arriveBob), stdin);
    arriveBob[strcspn(arriveBob, "\n")] = 0;
    
    fgets(leaveBob, sizeof(leaveBob), stdin);
    leaveBob[strcspn(leaveBob, "\n")] = 0;
    
    int result = solution(arriveAlice, leaveAlice, arriveBob, leaveBob);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(1)

Always iterates through 365 days regardless of input

✓ Linear Growth

Space Complexity

O(1)

Only uses a few variables to track dates and count

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force Day by Day — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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