What is the time complexity of Number of Days Between Two Dates?

The optimal solution has a time complexity of O(Y).

What data structures are used to solve Number of Days Between Two Dates?

This problem uses Math, String. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Number of Days Between Two Dates?

**Epoch Normalization for Date Arithmetic:** The core insight is to convert each date string into a single, unambiguous integer representing its total day count from a fixed epoch (e.g., January 1st, 1971). This transforms a potentially messy date-difference calculation involving year/month boundaries and variable month lengths into a simple integer subtraction. **Precise Leap Year and Month Length Rules:** Correctly implementing the `is_leap` function (divisible by 4, unless by 100 but not by 400) and `days_in_month` (handling 28/29 for Feb, 30 for April/June/Sept/Nov, 31 for others) is absolutely fundamental. Any misstep here will propagate errors throughout the day count, especially when dates cross year boundaries or involve February.

Which companies ask LeetCode 1360. Number of Days Between Two Dates in interviews?

This problem is frequently asked at Accenture, Optiver.

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Number of Days Between Two Dates - LeetCode 1360 Solution

Number of Days Between Two Dates - Complete Solution Guide

Number of Days Between Two Dates is LeetCode problem 1360, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Write a program to count the number of days between two dates. The two dates are given as strings, their format is YYYY-MM-DD as shown in the examples. Example 1: Input: date1 = "2019-06-29", date2 = "2019-06-30" Output: 1 Example 2: Input: date1 = "2020-01-15", date2 = "2019-12-31" Output: 15 Constraints: The given dates are valid dates between the years 1971 and 2100 .

Detailed Explanation

This problem challenges us to compute the absolute difference in days between two given dates, presented as "YYYY-MM-DD" strings. At first glance, it seems like a straightforward date arithmetic task, but the devil's in the details: correctly accounting for variable month lengths and, crucially, the intricacies of leap years. The dates are guaranteed to be valid and fall within the years 1971 and 2100, which is a helpful constraint but doesn't let us off the hook for implementing proper calendrical logic.

Solution Approach

The most robust and elegant approach to finding the difference between two dates is to normalize both dates to a single, comparable integer representing the total number of days passed since a fixed reference point. In the provided solution, this reference point is implicitly January 1st, 1971, which aligns with the problem's earliest date constraint. By converting both `date1` and `date2` into their respective total day counts from 1971 (let's call them `total_days_1` and `total_days_2`), the complex date subtraction problem simplifies into a straightforward `abs(total_days_1 - total_days_2)` operation.

Step-by-Step Algorithm

Step 1: Parse the input date strings into year, month, and day components.
Step 2: Determine if each year encountered is a leap year using the leap year rule.
Step 3: Calculate the number of days from the reference date to the input date. This involves iterating through years, adding the number of days in each year (365 or 366 based on leap year status), and adding the days in the months up to the target month and day.
Step 4: Calculate the absolute difference between the day counts for the two input dates.
Step 5: Return the absolute difference as the final result.

Key Insights

**Epoch Normalization for Date Arithmetic:** The core insight is to convert each date string into a single, unambiguous integer representing its total day count from a fixed epoch (e.g., January 1st, 1971). This transforms a potentially messy date-difference calculation involving year/month boundaries and variable month lengths into a simple integer subtraction.
**Precise Leap Year and Month Length Rules:** Correctly implementing the `is_leap` function (divisible by 4, unless by 100 but not by 400) and `days_in_month` (handling 28/29 for Feb, 30 for April/June/Sept/Nov, 31 for others) is absolutely fundamental. Any misstep here will propagate errors throughout the day count, especially when dates cross year boundaries or involve February.
**Accumulative Summation for Robust Day Counting:** The `days_from_1971` helper function is built on an accumulative summation strategy. It first sums days for all *entire years* before the target date's year, then sums days for all *entire months* before the target date's month within that year, and finally adds the day component itself. This systematic, additive approach is less prone to off-by-one errors compared to trying to directly calculate differences between date components.

Complexity Analysis

Time Complexity: O(Y)

Space Complexity: O(1)

Topics

This problem involves: Math, String.

Companies

Asked at: Accenture, Optiver.

Number of Days Between Two Dates | LeetCode 1360

Easy

Math String

Time: O(Y)Space: O(1)

Solution Code

Available in:

class Solution {
    public int daysBetweenDates(String date1, String date2) {
        return Math.abs(days(date1) - days(date2));
    }

    private int days(String date) {
        int year = Integer.parseInt(date.substring(0, 4));
        int month = Integer.parseInt(date.substring(5, 7));
        int day = Integer.parseInt(date.substring(8, 10));

        int[] daysInMonth = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
        int totalDays = 0;

        for (int i = 1971; i < year; i++) {
            if (isLeapYear(i)) {
                totalDays += 366;
            } else {
                totalDays += 365;
            }
        }

        for (int i = 1; i < month; i++) {
            totalDays += daysInMonth[i];
        }

        if (isLeapYear(year) && month > 2) {
            totalDays += 1;
        }

        totalDays += day;

        return totalDays;
    }

    private boolean isLeapYear(int year) {
        return (year % 4 == 0 && year % 100 != 0) || (year % 400 == 0);
    }
}

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Problem Info

Description

Problem 1360: Number of Days Between Two Dates - Detailed explanation and solution approaches.

Topics

Math String

Companies

AccentureOptiver

Stats

Difficulty:Easy

Languages:4

Companies:2

Problem #:1360

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

5 steps

1
Step 1: Parse the input date strings into year, month, and day components.
2
Step 2: Determine if each year encountered is a leap year using the leap year rule.
3
Step 3: Calculate the number of days from the reference date to the input date. This involves iterating through years, adding the number of days in each year (365 or 366 based on leap year status), and adding the days in the months up to the target month and day.
4
Step 4: Calculate the absolute difference between the day counts for the two input dates.
5
Step 5: Return the absolute difference as the final result.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(Y)

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The space used is constant because only a few variables are needed to store the year, month, day, and intermediate day counts. The provided solutions do not use any data structures whose size scales with the input. Note that the C++ solution uses a static array in `days_from_epoch`, but this is constant space as it's allocated only once.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Using a date library

Many programming languages offer built-in date and time libraries. These libraries provide functions to calculate the difference between dates directly, often offering better performance and handling of edge cases compared to manual calculations.

Time: O(1) (generally, as the library functions are optimized)Space: O(1)

Trade-offs: Using a date library can be simpler and more efficient than manual calculation. However, it requires familiarity with the library's API and may be language-specific.

✨

Key Insights

1**Epoch Normalization for Date Arithmetic:** The core insight is to convert each date string into a single, unambiguous integer representing its total day count from a fixed epoch (e.g., January 1st, 1971). This transforms a potentially messy date-difference calculation involving year/month boundaries and variable month lengths into a simple integer subtraction.
2**Precise Leap Year and Month Length Rules:** Correctly implementing the `is_leap` function (divisible by 4, unless by 100 but not by 400) and `days_in_month` (handling 28/29 for Feb, 30 for April/June/Sept/Nov, 31 for others) is absolutely fundamental. Any misstep here will propagate errors throughout the day count, especially when dates cross year boundaries or involve February.
3**Accumulative Summation for Robust Day Counting:** The `days_from_1971` helper function is built on an accumulative summation strategy. It first sums days for all *entire years* before the target date's year, then sums days for all *entire months* before the target date's month within that year, and finally adds the day component itself. This systematic, additive approach is less prone to off-by-one errors compared to trying to directly calculate differences between date components.

📚

Key Terms Explained

For Students

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Dates that span multiple years, including leap years. The solutions correctly handle the added day in February during leap years.
Edge case 2: Dates that are close together (e.g., consecutive days). The solution should correctly return 1.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly handling leap years. Failing to account for leap years correctly will lead to inaccurate results.

✗Common mistake 2: Off-by-one errors. Errors can occur when calculating the number of days in months or years, particularly at the boundaries of months or years.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Calculating the duration of a project or event. Knowing the exact number of days helps in scheduling, resource allocation, and cost estimation.

🏦Real-world scenario 2: Financial applications. Interest calculations, loan amortization, and other financial computations often require accurate calculation of the number of days between dates.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Use this problem to demonstrate clean code practices. Variable naming and code organization matter.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

👨‍💻 From Experience

This is one of the first problems I solved when learning Math. It's a great foundation for more complex problems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: September 9, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

❓

Frequently Asked Questions

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Number of Days Between Two Dates"?

For this easy-level Math problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether string techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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