C program to print number of days in a month

Given a number N, the task is to find the number of days corresponding to each month where 1 is January, 2 is February, 3 is March, and so on.

Examples:

Input: N = 12
Output: 31 Days

Input: N = 2
Output: 28/29 Days

Method – 1: using If Else:



  1. Get the input month as a number N.
  2. If N is one of these value 1, 3, 5, 7, 8, 10, 12, then print “31 Days.”.
  3. If N is one of these value 4, 6, 9, 11, then print “30 Days.”.
  4. If N is 2, then print “28/29 Days.”.
  5. Else print “Invalid Month”.

Below is the implementation of the above approach:

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// C program for the above approach
#include <stdio.h>
  
// Function to find the number of Days
// in month input by user
void printNumberOfDays(int N)
{
  
    // Check for 31 Days
    if (N == 1 || N == 3 || N == 5
        || N == 7 || N == 8 || N == 10
        || N == 12) {
        printf("31 Days.");
    }
  
    // Check for 30 Days
    else if (N == 4 || N == 6
             || N == 9 || N == 11) {
        printf("30 Days.");
    }
  
    // Check for 28/29 Days
    else if (N == 2) {
        printf("28/29 Days.");
    }
  
    // Else Invalid Input
    else {
        printf("Invalid Month.");
    }
}
  
// Driver Code
int main()
{
    // Input Month
    int N = 4;
  
    // Function Call
    printNumberOfDays(N);
  
    return 0;
}

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Output:

30 Days.

Time Complexity: O(1)
Auxiliary Space: O(1)

Method – 2: using Switch Statements:

  1. Get the input month as a number N.
  2. Using switch statement when value of N is one of 1, 3, 5, 7, 8, 10, 12, then print “31 Days.” corresponding to switch case.
  3. If N is one of these value 4, 6, 9, 11, then print “30 Days.” corresponding to switch case.
  4. If N is 2, then print “28/29 Days.” corresponding to switch case.
  5. Else the default condition for the switch case will print “Invalid Month”.

Below is the implementation of the above approach:

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// C program for the above approach
#include <stdio.h>
  
// Function to find the number of Days
// in month input by user usingwwww
// switch statement
void printNumberOfDays(int N)
{
  
    switch (N) {
    // Cases for 31 Days
    case 1:
    case 3:
    case 5:
    case 7:
    case 8:
    case 10:
    case 12:
        printf("31 Days.");
        break;
  
    // Cases for 30 Days
    case 4:
    case 6:
    case 9:
    case 11:
        printf("30 Days.");
        break;
  
    // Case for 28/29 Days
    case 2:
        printf("28/29 Days.");
        break;
  
    default:
        printf("Invalid Month.");
        break;
    }
}
  
// Driver Code
int main()
{
    // Input Month
    int N = 4;
  
    // Function Call
    printNumberOfDays(N);
  
    return 0;
}

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Output:

30 Days.

Time Complexity: O(1)
Auxiliary Space: O(1)

Method – 3: using Arrays:



  1. Store the value of days corresponding to each month in an array as:

    arr[12] = { 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 }

  2. Print the corresponding day to each month from the above array.

Below is the implementation of the above approach:

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// C program to find the number of days
// in a month using arrays
#include <stdio.h>
  
// Driver Code
int main()
{
    // Store the day in array arr[]
    int arr[12] = { 31, 28, 31, 30, 31, 30,
                    31, 31, 30, 31, 30, 31 };
  
    // Input Month
    int N = 4;
  
    // Print the number of days in
    // month 4
    printf("%d Days.", arr[N - 1]);
  
    return 0;
}

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Output:

30 Days.

Time Complexity: O(1)
Auxiliary Space: O(1)

Method – 4: using Pointers:

  1. Store the value of days corresponding to each month in an array as:

    arr[12] = { 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 }

  2. Print the corresponding day to each month from the above array using pointers as:

    printf(“%d Days.”, *(arr + (*N – 1)))

Below is the implementation of the above approach:

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// C program to find the number of days
// in a month using pointers
#include <stdio.h>
  
// Function to print number of Days
void printNumberOfDays(int* arr, int* N)
{
    // Print the number of days for Nth
    // month using *(arr+(*N - 1))
    printf("%d Days.", *(arr + (*N - 1)));
}
  
// Driver Code
int main()
{
    // Store the day in array arr[]
    int arr[12] = { 31, 28, 31, 30, 31, 30,
                    31, 31, 30, 31, 30, 31 };
  
    // Input Month
    int N = 4;
  
    // Print the number of days in
    // month 4
    printNumberOfDays(arr, &N);
  
    return 0;
}

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Output:

30 Days.

Time Complexity: O(1)
Auxiliary Space: O(1)

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