Count Days Spent Together - Practice Coding Problems

Count Days Spent Together - Problem

Math String Easy

Count Days Spent Together - A Date Intersection Problem

Alice and Bob are traveling to Rome for separate business meetings. You need to find how many days they'll be in the city at the same time.

Given four date strings in "MM-DD" format:
• arriveAlice & leaveAlice - Alice's arrival and departure dates
• arriveBob & leaveBob - Bob's arrival and departure dates

Goal: Return the total number of days both Alice and Bob are in Rome together (inclusive of arrival and departure dates).

Important: All dates are in the same non-leap year. Month lengths: [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]

Example: If Alice is there "08-15" to "08-18" and Bob is there "08-16" to "08-19", they overlap on "08-16", "08-17", "08-18" = 3 days.

Input & Output

example_1.py — Basic 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 is there from Aug 16-19. They overlap on Aug 16, 17, and 18 = 3 days total.

example_2.py — No Overlap

$ Input: arriveAlice = "10-01", leaveAlice = "10-31", arriveBob = "11-01", leaveBob = "12-31"

› Output: 0

💡 Note: Alice leaves on Oct 31 and Bob arrives on Nov 1. There are no overlapping days.

example_3.py — 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) falls within Alice's stay (Sep 1 - Oct 19). Bob is there for 6 days total.

Constraints

All dates are valid and in the format MM-DD
All dates occur in the same non-leap year
1 ≤ MM ≤ 12 and 1 ≤ DD ≤ DaysInMonth[MM]
arriveAlice ≤ leaveAlice and arriveBob ≤ leaveBob
Month lengths: [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]

Visualization

Tap to expand

Understanding the Visualization

Parse Dates

Convert MM-DD strings to day numbers (Jan 1 = Day 1, Dec 31 = Day 365)

Map Ranges

Visualize Alice's and Bob's stays as ranges on a timeline

Find Intersection

The overlap starts at the later arrival and ends at the earlier departure

Count Days

Calculate days = end - start + 1, or 0 if no overlap exists

Key Takeaway

🎯 Key Insight: Date overlap is just range intersection math - convert to numbers first, then use max/min operations to find the overlapping period efficiently in O(1) time!

Asked in

G Google 15 a Amazon 12 ⊞ Microsoft 8 Apple 5

The optimal approach converts MM-DD dates to day numbers (1-365) and finds the intersection of two date ranges. The key insight is that overlapping days = max(0, min(end1, end2) - max(start1, start2) + 1). This gives us O(1) time complexity compared to the brute force O(365) approach.

Common Approaches

Approach	Time	Space	Notes
✓ Optimal: Date Range Intersection	O(1)	O(1)	Convert dates to day numbers and find the intersection of two ranges

Optimal: Date Range Intersection — Algorithm Steps

Create a function to convert MM-DD format to day number (1-365)
Convert all four dates to day numbers
Find intersection start: max(arriveAlice, arriveBob)
Find intersection end: min(leaveAlice, leaveBob)
Return max(0, end - start + 1) to handle no overlap case

Visualization

Tap to expand

Step-by-Step Walkthrough

Convert to Numbers

Transform MM-DD dates to day numbers (1-365)

Find Range Start

Intersection starts at max(alice_start, bob_start)

Find Range End

Intersection ends at min(alice_end, bob_end)

Calculate Days

Days = max(0, end - start + 1)

Code -

solution.c — C

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

int daysInMonth[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

int dateToDay(char* date) {
    int month = (date[0] - '0') * 10 + (date[1] - '0');
    int day = (date[3] - '0') * 10 + (date[4] - '0');
    int dayOfYear = 0;
    for (int i = 0; i < month - 1; i++) {
        dayOfYear += daysInMonth[i];
    }
    return dayOfYear + day;
}

int max(int a, int b) { return a > b ? a : b; }
int min(int a, int b) { return a < b ? a : b; }

int countDaysTogether(char* arriveAlice, char* leaveAlice, char* arriveBob, char* leaveBob) {
    // Convert all dates to day numbers
    int aliceArrive = dateToDay(arriveAlice);
    int aliceLeave = dateToDay(leaveAlice);
    int bobArrive = dateToDay(arriveBob);
    int bobLeave = dateToDay(leaveBob);
    
    // Find intersection of the two ranges
    int overlapStart = max(aliceArrive, bobArrive);
    int overlapEnd = min(aliceLeave, bobLeave);
    
    // Return overlapping days (0 if no overlap)
    return max(0, overlapEnd - overlapStart + 1);
}

Time & Space Complexity

Time Complexity

⏱️

O(1)

Only performing constant number of operations - date conversions and min/max calculations

✓ Linear Growth

Space Complexity

O(1)

Using only a fixed number of variables regardless of input size

✓ Linear Space

28.5K Views

Medium Frequency

~15 min Avg. Time

892 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Optimal: Date Range Intersection — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler