Day of the Week

Day of the Week - Problem

Math Easy

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.

Input & Output

Example 1 — Basic Case

$ Input: day = 31, month = 12, year = 1999

› Output: Friday

💡 Note: December 31, 1999 was a Friday. We can verify this by counting days from January 1, 1971 or using Zeller's formula.

Example 2 — Leap Year

$ Input: day = 29, month = 2, year = 2000

› Output: Tuesday

💡 Note: February 29, 2000 was a Tuesday. Year 2000 is a leap year (divisible by 400).

Example 3 — Reference Date

$ Input: day = 1, month = 1, year = 1971

› Output: Friday

💡 Note: January 1, 1971 was a Friday as given in the problem statement.

Constraints

1 ≤ day ≤ 31
1 ≤ month ≤ 12
1971 ≤ year ≤ 2100
The given date is guaranteed to be valid

Visualization

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Asked in

a Amazon 25 M Microsoft 20 G Google 15

The key insight is to either count total days from a reference date or use mathematical formulas like Zeller's congruence. Best approach is Zeller's formula for direct calculation. Time: O(1), Space: O(1)

Common Approaches

✓ Day Counting from Reference

⏱️ Time: O(1) Space: O(1)

Calculate the total number of days between January 1, 1971 (Friday) and the target date, then use modulo 7 to find the day of week.

Zeller's Congruence Formula

⏱️ Time: O(1) Space: O(1)

Apply Zeller's congruence algorithm which uses a mathematical formula to compute the day of week directly without counting days.

Day Counting from Reference — Algorithm Steps

Calculate total days from 1971 to target year
Add days for months in target year
Add remaining days
Use modulo 7 with Friday as day 0

Visualization

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Step-by-Step Walkthrough

Reference Point

Start from Jan 1, 1971 (Friday)

Count Days

Add days from years, months, and remaining days

Find Day

Use modulo 7 to get day of week

Code -

solution.c — C

#include <stdio.h>
#include <stdbool.h>

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

char* solution(int day, int month, int year) {
    int daysInMonth[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    int daysInMonthLeap[] = {31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    char* dayNames[] = {"Friday", "Saturday", "Sunday", "Monday", "Tuesday", "Wednesday", "Thursday"};
    
    int totalDays = 0;
    
    // Add days for complete years
    for (int y = 1971; y < year; y++) {
        totalDays += isLeapYear(y) ? 366 : 365;
    }
    
    // Add days for complete months
    int* monthDays = isLeapYear(year) ? daysInMonthLeap : daysInMonth;
    for (int m = 0; m < month - 1; m++) {
        totalDays += monthDays[m];
    }
    
    // Add remaining days
    totalDays += day - 1;
    
    return dayNames[totalDays % 7];
}

int main() {
    int day, month, year;
    scanf("%d", &day);
    scanf("%d", &month);
    scanf("%d", &year);
    
    char* result = solution(day, month, year);
    printf("%s\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(1)

Fixed number of calculations regardless of input size

✓ Linear Growth

Space Complexity

O(1)

Only uses constant extra variables

✓ Linear Space

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Input & Output

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Related Problems

Common Approaches

Day Counting from Reference — Algorithm Steps

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Code -

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