Number of Days Between Two Dates - Practice Coding Problems

Number of Days Between Two Dates - Problem

Math String Easy

You're given two dates in string format as YYYY-MM-DD and need to calculate the absolute number of days between them.

This is a fundamental date calculation problem that appears frequently in real-world applications like subscription billing, project management, and age calculations.

Goal: Calculate the difference in days between two valid dates.

Input: Two date strings in format "YYYY-MM-DD"

Output: A single integer representing the absolute number of days between the dates

Example: Between "2020-01-15" and "2019-12-31", there are 15 days.

Input & Output

example_1.py — Basic Date Difference

$ Input: date1 = "2019-06-29", date2 = "2019-06-30"

› Output: 1

💡 Note: The difference between June 29, 2019 and June 30, 2019 is exactly 1 day. Simple consecutive date calculation.

example_2.py — Cross-Year Calculation

$ Input: date1 = "2020-01-15", date2 = "2019-12-31"

› Output: 15

💡 Note: From Dec 31, 2019 to Jan 15, 2020: 15 days total. This tests year boundary crossing and ensures we return absolute difference.

example_3.py — Same Date Edge Case

$ Input: date1 = "2020-02-29", date2 = "2020-02-29"

› Output: 0

💡 Note: When both dates are identical, the difference is 0. This also tests leap year date handling (Feb 29).

Constraints

The given dates are valid dates between the years 1971 and 2100
Dates are given in format YYYY-MM-DD
Both dates will be valid according to the Gregorian calendar
Years from 1971 to 2100 inclusively

Visualization

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Understanding the Visualization

Reference Point

Establish year 1 AD as our reference epoch (day 0)

Convert First Date

Calculate total days from epoch to first date

Convert Second Date

Calculate total days from epoch to second date

Find Difference

Subtract the two values and take absolute difference

Key Takeaway

🎯 Key Insight: Converting dates to a numerical representation (days since epoch) transforms the problem from iterative counting to simple arithmetic subtraction.

Asked in

G Google 23 a Amazon 18 ⊞ Microsoft 15 Apple 12

The optimal approach is to use mathematical calculation by converting each date to days since a reference epoch, then finding the absolute difference. This achieves O(1) time complexity compared to the brute force O(n) iteration method.

Common Approaches

Approach	Time	Space	Notes
✓ Mathematical Days Since Epoch	O(1)	O(1)	Convert each date to days since a reference point and subtract

Mathematical Days Since Epoch — Algorithm Steps

Parse both date strings into year, month, day components
Create a function to calculate days since epoch (year 1) for any date
Handle leap years using mathematical formulas
Convert both dates to days since epoch
Return the absolute difference between the two values

Visualization

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Step-by-Step Walkthrough

Parse Dates

Extract year, month, day components from date strings

Calculate Days from Epoch

Convert each date to total days since year 1

Handle Leap Years

Add extra days for leap years in calculation

Subtract and Return

Find absolute difference between the two values

Code -

solution.c — C

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

int daysFromEpoch(char* date) {
    int year, month, day;
    sscanf(date, "%d-%d-%d", &year, &month, &day);
    
    int days[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    
    int totalDays = 0;
    
    // Count days from years
    for (int y = 1; y < year; y++) {
        if (y % 4 == 0 && (y % 100 != 0 || y % 400 == 0)) {
            totalDays += 366;
        } else {
            totalDays += 365;
        }
    }
    
    // Add days from months in current year
    for (int m = 1; m < month; m++) {
        totalDays += days[m - 1];
        if (m == 2 && year % 4 == 0 && (year % 100 != 0 || year % 400 == 0)) {
            totalDays += 1;
        }
    }
    
    // Add remaining days
    totalDays += day - 1;
    
    return totalDays;
}

int daysBetweenDates(char* date1, char* date2) {
    int diff = daysFromEpoch(date1) - daysFromEpoch(date2);
    return diff < 0 ? -diff : diff;
}

int main() {
    char line[100];
    fgets(line, sizeof(line), stdin);
    
    char date1[11], date2[11];
    sscanf(line, "\"%10s \"%10s", date1, date2);
    
    // Remove quotes
    date1[strlen(date1)-1] = '\0';
    
    printf("%d\n", daysBetweenDates(date1, date2));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(1)

Constant time operations to calculate days from epoch for each date

✓ Linear Growth

Space Complexity

O(1)

Only using variables for calculations, no additional data structures

✓ Linear Space

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Medium Frequency

~15 min Avg. Time

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Mathematical Days Since Epoch — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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