LeetCode #1360 — EASY

Number of Days Between Two Dates

Build confidence with an intuition-first walkthrough focused on math fundamentals.

The Problem

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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math

Example 1

"2019-06-29"
"2019-06-30"

Example 2

"2020-01-15"
"2019-12-31"

Core Insight

What unlocks the optimal approach

Create a function f(date) that counts the number of days from 1900-01-01 to date. How can we calculate the answer ?
The answer is just |f(date1) - f(date2)|.
How to construct f(date) ?
For each year from 1900 to year - 1 sum up 365 or 366 in case of leap years. Then sum up for each month the number of days, consider the case when the current year is leap, finally sum up the days.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1360: Number of Days Between Two Dates
class Solution {
    public int daysBetweenDates(String date1, String date2) {
        return Math.abs(calcDays(date1) - calcDays(date2));
    }

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

    private int daysInMonth(int year, int month) {
        int[] days = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
        days[1] += isLeapYear(year) ? 1 : 0;
        return days[month - 1];
    }

    private int calcDays(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));
        int days = 0;
        for (int y = 1971; y < year; ++y) {
            days += isLeapYear(y) ? 366 : 365;
        }
        for (int m = 1; m < month; ++m) {
            days += daysInMonth(year, m);
        }
        days += day;
        return days;
    }
}

// Accepted solution for LeetCode #1360: Number of Days Between Two Dates
func daysBetweenDates(date1 string, date2 string) int {
	return abs(calcDays(date1) - calcDays(date2))
}

func isLeapYear(year int) bool {
	return year%4 == 0 && (year%100 != 0 || year%400 == 0)
}

func daysInMonth(year, month int) int {
	days := [12]int{31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}
	if isLeapYear(year) {
		days[1] = 29
	}
	return days[month-1]
}

func calcDays(date string) int {
	year, _ := strconv.Atoi(date[:4])
	month, _ := strconv.Atoi(date[5:7])
	day, _ := strconv.Atoi(date[8:])
	days := 0
	for y := 1971; y < year; y++ {
		days += 365
		if isLeapYear(y) {
			days++
		}
	}
	for m := 1; m < month; m++ {
		days += daysInMonth(year, m)
	}
	days += day
	return days
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

# Accepted solution for LeetCode #1360: Number of Days Between Two Dates
class Solution:
    def daysBetweenDates(self, date1: str, date2: str) -> int:
        def isLeapYear(year: int) -> bool:
            return year % 4 == 0 and (year % 100 != 0 or year % 400 == 0)

        def daysInMonth(year: int, month: int) -> int:
            days = [
                31,
                28 + int(isLeapYear(year)),
                31,
                30,
                31,
                30,
                31,
                31,
                30,
                31,
                30,
                31,
            ]
            return days[month - 1]

        def calcDays(date: str) -> int:
            year, month, day = map(int, date.split("-"))
            days = 0
            for y in range(1971, year):
                days += 365 + int(isLeapYear(y))
            for m in range(1, month):
                days += daysInMonth(year, m)
            days += day
            return days

        return abs(calcDays(date1) - calcDays(date2))

// Accepted solution for LeetCode #1360: Number of Days Between Two Dates
struct Solution;

impl Solution {
    fn days_between_dates(date1: String, date2: String) -> i32 {
        let (y1, m1, d1) = Self::parse(date1);
        let (y2, m2, d2) = Self::parse(date2);
        let mut s1 = 0;
        let mut s2 = 0;
        for i in 1971..y1 {
            s1 += if Self::is_leap(i) { 366 } else { 365 };
        }
        for i in 1971..y2 {
            s2 += if Self::is_leap(i) { 366 } else { 365 };
        }
        s1 += Self::day_of_year(y1, m1, d1);
        s2 += Self::day_of_year(y2, m2, d2);
        (s1 as i32 - s2 as i32).abs()
    }

    fn parse(date: String) -> (usize, usize, usize) {
        let a: Vec<&str> = date.split_terminator('-').collect();
        let year = a[0].parse::<usize>().unwrap();
        let month = a[1].parse::<usize>().unwrap();
        let day = a[2].parse::<usize>().unwrap();
        (year, month, day)
    }

    fn is_leap(year: usize) -> bool {
        year % 4 == 0 && (year % 100 != 0 || year % 400 == 0)
    }

    fn day_of_year(year: usize, month: usize, day: usize) -> usize {
        let mut days: Vec<usize> = vec![31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31];
        let mut sum = 0;
        if year % 4 == 0 && (year % 100 != 0 || year % 400 == 0) {
            days[1] += 1;
        }
        for i in 0..month - 1 {
            sum += days[i];
        }
        sum += day;
        sum
    }
}

#[test]
fn test() {
    let date1 = "2019-06-29".to_string();
    let date2 = "2019-06-30".to_string();
    let res = 1;
    assert_eq!(Solution::days_between_dates(date1, date2), res);
    let date1 = "2020-01-15".to_string();
    let date2 = "2019-12-31".to_string();
    let res = 15;
    assert_eq!(Solution::days_between_dates(date1, date2), res);
}

// Accepted solution for LeetCode #1360: Number of Days Between Two Dates
function daysBetweenDates(date1: string, date2: string): number {
    return Math.abs(calcDays(date1) - calcDays(date2));
}

function isLeapYear(year: number): boolean {
    return year % 4 === 0 && (year % 100 !== 0 || year % 400 === 0);
}

function daysOfMonth(year: number, month: number): number {
    const days = [31, isLeapYear(year) ? 29 : 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31];
    return days[month - 1];
}

function calcDays(date: string): number {
    let days = 0;
    const [year, month, day] = date.split('-').map(Number);
    for (let y = 1971; y < year; ++y) {
        days += isLeapYear(y) ? 366 : 365;
    }
    for (let m = 1; m < month; ++m) {
        days += daysOfMonth(year, m);
    }
    days += day - 1;
    return days;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

ITERATIVE

O(n) time

O(1) space

Simulate the process step by step — multiply n times, check each number up to n, or iterate through all possibilities. Each step is O(1), but doing it n times gives O(n). No extra space needed since we just track running state.

MATH INSIGHT

O(log n) time

O(1) space

Math problems often have a closed-form or O(log n) solution hidden behind an O(n) simulation. Modular arithmetic, fast exponentiation (repeated squaring), GCD (Euclidean algorithm), and number theory properties can dramatically reduce complexity.

Shortcut: Look for mathematical properties that eliminate iteration. Repeated squaring → O(log n). Modular arithmetic avoids overflow.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.