LeetCode #1185 — EASY

Day of the Week

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

The Problem

Problem Statement

Given a date, return the corresponding day of the week for that date.

The input is given as three integers representing the day, month and year respectively.

Return the answer as one of the following values {"Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"}.

Note: January 1, 1971 was a Friday.

Example 1:

Input: day = 31, month = 8, year = 2019
Output: "Saturday"

Example 2:

Input: day = 18, month = 7, year = 1999
Output: "Sunday"

Example 3:

Input: day = 15, month = 8, year = 1993
Output: "Sunday"

Constraints:

The given dates are valid dates between the years 1971 and 2100.

Step 01

Brute Force Baseline

Problem summary: Given a date, return the corresponding day of the week for that date. The input is given as three integers representing the day, month and year respectively. Return the answer as one of the following values {"Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"}. Note: January 1, 1971 was a Friday.

Baseline thinking

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

Pattern signal: Math

Example 1

31
8
2019

Example 2

18
7
1999

Example 3

15
8
1993

Step 02

Core Insight

What unlocks the optimal approach

Sum up the number of days for the years before the given year.
Handle the case of a leap year.
Find the number of days for each month of the given year.

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 #1185: Day of the Week
import java.util.Calendar;

class Solution {
    private static final String[] WEEK
        = {"Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"};

    public static String dayOfTheWeek(int day, int month, int year) {
        Calendar calendar = Calendar.getInstance();
        calendar.set(year, month - 1, day);
        return WEEK[calendar.get(Calendar.DAY_OF_WEEK) - 1];
    }
}

// Accepted solution for LeetCode #1185: Day of the Week
func dayOfTheWeek(d int, m int, y int) string {
	if m < 3 {
		m += 12
		y -= 1
	}
	c := y / 100
	y %= 100
	w := (c/4 - 2*c + y + y/4 + 13*(m+1)/5 + d - 1) % 7
	weeks := []string{"Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"}
	return weeks[(w+7)%7]
}

# Accepted solution for LeetCode #1185: Day of the Week
class Solution:
    def dayOfTheWeek(self, day: int, month: int, year: int) -> str:
        return datetime.date(year, month, day).strftime('%A')

// Accepted solution for LeetCode #1185: Day of the Week
struct Solution;

impl Solution {
    fn day_of_the_week(day: i32, mut month: i32, mut year: i32) -> String {
        let days = [
            "Sunday",
            "Monday",
            "Tuesday",
            "Wednesday",
            "Thursday",
            "Friday",
            "Saturday",
        ];
        if month < 3 {
            month += 12;
            year -= 1;
        }
        let week =
            (day + ((month + 1) * 26) / 10 + year + year / 4 + 6 * (year / 100) + year / 400 + 5)
                % 7
                + 1;
        days[week as usize % 7].to_string()
    }
}

#[test]
fn test() {
    let day = 31;
    let month = 8;
    let year = 2019;
    let res = "Saturday".to_string();
    assert_eq!(Solution::day_of_the_week(day, month, year), res);
    let day = 18;
    let month = 7;
    let year = 1999;
    let res = "Sunday".to_string();
    assert_eq!(Solution::day_of_the_week(day, month, year), res);
    let day = 15;
    let month = 8;
    let year = 1993;
    let res = "Sunday".to_string();
    assert_eq!(Solution::day_of_the_week(day, month, year), res);
}

// Accepted solution for LeetCode #1185: Day of the Week
function dayOfTheWeek(d: number, m: number, y: number): string {
    if (m < 3) {
        m += 12;
        y -= 1;
    }
    const c: number = (y / 100) | 0;
    y %= 100;
    const w = (((c / 4) | 0) - 2 * c + y + ((y / 4) | 0) + (((13 * (m + 1)) / 5) | 0) + d - 1) % 7;
    const weeks: string[] = [
        'Sunday',
        'Monday',
        'Tuesday',
        'Wednesday',
        'Thursday',
        'Friday',
        'Saturday',
    ];
    return weeks[(w + 7) % 7];
}

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(1)

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.