LeetCode #2409 — EASY

Count Days Spent Together

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

The Problem

Problem Statement

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].

Example 1:

Input: arriveAlice = "08-15", leaveAlice = "08-18", arriveBob = "08-16", leaveBob = "08-19"
Output: 3
Explanation: Alice will be in Rome from August 15 to August 18. Bob will be in Rome from August 16 to August 19. They are both in Rome together on August 16th, 17th, and 18th, so the answer is 3.

Example 2:

Input: arriveAlice = "10-01", leaveAlice = "10-31", arriveBob = "11-01", leaveBob = "12-31"
Output: 0
Explanation: There is no day when Alice and Bob are in Rome together, so we return 0.

Constraints:

All dates are provided in the format "MM-DD".
Alice and Bob's arrival dates are earlier than or equal to their leaving dates.
The given dates are valid dates of a non-leap year.

Step 01

Brute Force Baseline

Problem summary: 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].

Baseline thinking

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

Pattern signal: Math

Example 1

"08-15"
"08-18"
"08-16"
"08-19"

Example 2

"10-01"
"10-31"
"11-01"
"12-31"

Core Insight

What unlocks the optimal approach

For a given day, determine if Alice or Bob or both are in Rome.
Brute force all 365 days for both Alice and Bob.

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 #2409: Count Days Spent Together
class Solution {
    private int[] days = new int[] {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

    public int countDaysTogether(
        String arriveAlice, String leaveAlice, String arriveBob, String leaveBob) {
        String a = arriveAlice.compareTo(arriveBob) < 0 ? arriveBob : arriveAlice;
        String b = leaveAlice.compareTo(leaveBob) < 0 ? leaveAlice : leaveBob;
        int x = f(a), y = f(b);
        return Math.max(y - x + 1, 0);
    }

    private int f(String s) {
        int i = Integer.parseInt(s.substring(0, 2)) - 1;
        int res = 0;
        for (int j = 0; j < i; ++j) {
            res += days[j];
        }
        res += Integer.parseInt(s.substring(3));
        return res;
    }
}

// Accepted solution for LeetCode #2409: Count Days Spent Together
func countDaysTogether(arriveAlice string, leaveAlice string, arriveBob string, leaveBob string) int {
	days := []int{31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}
	f := func(s string) int {
		m, _ := strconv.Atoi(s[:2])
		d, _ := strconv.Atoi(s[3:])
		res := 0
		for i := 0; i < m-1; i++ {
			res += days[i]
		}
		res += d
		return res
	}
	a, b := arriveAlice, leaveBob
	if arriveAlice < arriveBob {
		a = arriveBob
	}
	if leaveAlice < leaveBob {
		b = leaveAlice
	}
	x, y := f(a), f(b)
	ans := y - x + 1
	if ans < 0 {
		return 0
	}
	return ans
}

# Accepted solution for LeetCode #2409: Count Days Spent Together
class Solution:
    def countDaysTogether(
        self, arriveAlice: str, leaveAlice: str, arriveBob: str, leaveBob: str
    ) -> int:
        a = max(arriveAlice, arriveBob)
        b = min(leaveAlice, leaveBob)
        days = (31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31)
        x = sum(days[: int(a[:2]) - 1]) + int(a[3:])
        y = sum(days[: int(b[:2]) - 1]) + int(b[3:])
        return max(y - x + 1, 0)

// Accepted solution for LeetCode #2409: Count Days Spent Together
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2409: Count Days Spent Together
// class Solution {
//     private int[] days = new int[] {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
// 
//     public int countDaysTogether(
//         String arriveAlice, String leaveAlice, String arriveBob, String leaveBob) {
//         String a = arriveAlice.compareTo(arriveBob) < 0 ? arriveBob : arriveAlice;
//         String b = leaveAlice.compareTo(leaveBob) < 0 ? leaveAlice : leaveBob;
//         int x = f(a), y = f(b);
//         return Math.max(y - x + 1, 0);
//     }
// 
//     private int f(String s) {
//         int i = Integer.parseInt(s.substring(0, 2)) - 1;
//         int res = 0;
//         for (int j = 0; j < i; ++j) {
//             res += days[j];
//         }
//         res += Integer.parseInt(s.substring(3));
//         return res;
//     }
// }

// Accepted solution for LeetCode #2409: Count Days Spent Together
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #2409: Count Days Spent Together
// class Solution {
//     private int[] days = new int[] {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
// 
//     public int countDaysTogether(
//         String arriveAlice, String leaveAlice, String arriveBob, String leaveBob) {
//         String a = arriveAlice.compareTo(arriveBob) < 0 ? arriveBob : arriveAlice;
//         String b = leaveAlice.compareTo(leaveBob) < 0 ? leaveAlice : leaveBob;
//         int x = f(a), y = f(b);
//         return Math.max(y - x + 1, 0);
//     }
// 
//     private int f(String s) {
//         int i = Integer.parseInt(s.substring(0, 2)) - 1;
//         int res = 0;
//         for (int j = 0; j < i; ++j) {
//             res += days[j];
//         }
//         res += Integer.parseInt(s.substring(3));
//         return res;
//     }
// }

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(C)

Space

O(C)

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.