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Find the date after next half year from a given date

  • Last Updated : 07 Jun, 2021

Given a positive integer D and a string M representing the day and the month of a leap year, the task is to find the date after the next half year.

Examples:

Input: D = 15, M = “January”
Output: 16 July
Explanation: The date from the 15th of January to the next half year is 16th of July.

Input: D = 10, M = “October”
Output: 10 April

Approach: Since a leap year contains 366 days, the given problem can be solved by finding the data after incrementing the current date by 183 days. Follow the steps to solve the problem:

  • Store the number of days for each month in that array, say days[].
  • Initialize a variable, say curMonth as M, to store the index of the current month.
  • Initialize a variable, say curDate as D, to store the current date.
  • Initialize a variable, say count as 183, representing the count of days to increment.
  • Iterate until the value of count is positive and perform the following steps:
    • If the value of count is positive and curDate is at most number of days in the curMonth then decrement the value of count by 1 and increment the value of curDate by 1.
    • If the value of count is 0 then break out of the loop.
    • Update the value of curMonth by (curMonth + 1)%12.
  • After completing the above steps, print the date corresponding to curDate and curMonth as the result.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <iostream>
using namespace std;
 
// Function to find the date
// after the next half-year
void getDate(int d, string m) {
 
    // Stores the number of days in the
    // months of a leap year
    int days[] = {31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
 
    // List of months
    string month[] = {"January", "February",
             "March", "April",
             "May", "June",
             "July", "August",
             "September", "October",
             "November", "December"};
 
    // Days in half of a year
    int cnt = 183;
 
    // Index of current month
    int cur_month;
      for(int i = 0; i < 12; i++)
      if(m == month[i])
          cur_month = i;
 
    // Starting day
    int cur_date = d;
 
    while(1) {
 
        while(cnt > 0 && cur_date <= days[cur_month]) {
 
            // Decrement the value of
            // cnt by 1
            cnt -= 1;
 
            // Increment cur_date
            cur_date += 1;
        }
 
        // If cnt is equal to 0, then
        // break out of the loop
        if(cnt == 0)
            break;
 
        // Update cur_month
        cur_month = (cur_month + 1) % 12;
 
        // Update cur_date
        cur_date = 1;
    }
 
    // Print the resultant date
    cout << cur_date << " " << month[cur_month] << endl;
}
 
// Driver Code
int main() {
 
    int D = 15;
    string M = "January";
 
    // Function Call
    getDate(D, M);
       
    return 0;
}
 
// This code is contributed by Dharanendra L V.

Java




// Java program for the above approach
class GFG{
     
// Function to find the date
// after the next half-year
public static void getDate(int d, String m)
{
     
    // Stores the number of days in the
    // months of a leap year
    int[] days = { 31, 29, 31, 30, 31, 30,
                   31, 31, 30, 31, 30, 31 };
 
    // List of months
    String[] month = { "January", "February", "March",
                       "April", "May", "June", "July",
                       "August", "September", "October",
                       "November", "December" };
 
    // Days in half of a year
    int cnt = 183;
 
    // Index of current month
    int cur_month = 0;
    for(int i = 0; i < 12; i++)
        if (m == month[i])
            cur_month = i;
 
    // Starting day
    int cur_date = d;
 
    while (true)
    {
        while (cnt > 0 && cur_date <= days[cur_month])
        {
             
            // Decrement the value of
            // cnt by 1
            cnt -= 1;
 
            // Increment cur_date
            cur_date += 1;
        }
 
        // If cnt is equal to 0, then
        // break out of the loop
        if (cnt == 0)
            break;
 
        // Update cur_month
        cur_month = (cur_month + 1) % 12;
 
        // Update cur_date
        cur_date = 1;
    }
 
    // Print the resultant date
    System.out.println(cur_date + " " +
                       month[cur_month]);
}
 
// Driver Code
public static void main(String args[])
{
    int D = 15;
    String M = "January";
 
    // Function Call
    getDate(D, M);
}
}
 
// This code is contributed by SoumikMondal

Python3




# Python program for the above approach
 
# Function to find the date
# after the next half-year
def getDate(d, m):
 
    # Stores the number of days in the
    # months of a leap year
    days = [31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
 
    # List of months
    month = ['January', 'February',
             'March', 'April',
             'May', 'June',
             'July', 'August',
             'September', 'October',
             'November', 'December']
 
    # Days in half of a year
    cnt = 183
 
    # Index of current month
    cur_month = month.index(m)
 
    # Starting day
    cur_date = d
 
    while(1):
 
        while(cnt > 0 and cur_date <= days[cur_month]):
 
            # Decrement the value of
            # cnt by 1
            cnt -= 1
 
            # Increment cur_date
            cur_date += 1
 
        # If cnt is equal to 0, then
        # break out of the loop
        if(cnt == 0):
            break
 
        # Update cur_month
        cur_month = (cur_month + 1) % 12
 
        # Update cur_date
        cur_date = 1
 
    # Print the resultant date
    print(cur_date, month[cur_month])
 
 
# Driver Code
D = 15
M = "January"
 
# Function Call
getDate(D, M)

C#




// C# program for the above approach
using System;
 
class GFG{
     
// Function to find the date
// after the next half-year
static void getDate(int d, string m)
{
     
    // Stores the number of days in the
    // months of a leap year
    int[] days = { 31, 29, 31, 30, 31, 30,
                   31, 31, 30, 31, 30, 31 };
 
    // List of months
    string[] month = { "January", "February", "March",
                       "April", "May", "June", "July",
                       "August", "September", "October",
                       "November", "December" };
 
    // Days in half of a year
    int cnt = 183;
 
    // Index of current month
    int cur_month = 0;
    for(int i = 0; i < 12; i++)
        if (m == month[i])
            cur_month = i;
 
    // Starting day
    int cur_date = d;
 
    while (true)
    {
        while (cnt > 0 && cur_date <= days[cur_month])
        {
             
            // Decrement the value of
            // cnt by 1
            cnt -= 1;
 
            // Increment cur_date
            cur_date += 1;
        }
 
        // If cnt is equal to 0, then
        // break out of the loop
        if (cnt == 0)
            break;
 
        // Update cur_month
        cur_month = (cur_month + 1) % 12;
 
        // Update cur_date
        cur_date = 1;
    }
 
    // Print the resultant date
    Console.WriteLine(cur_date + " " +
                      month[cur_month]);
}
 
// Driver Code
public static void Main()
{
    int D = 15;
    string M = "January";
 
    // Function Call
    getDate(D, M);
}
}
 
// This code is contributed by ukasp

Javascript




<script>
 
// Javascript program for the above approach
 
// Function to find the date
// after the next half-year
function getDate(d, m)
{
     
    // Stores the number of days in the
    // months of a leap year
    let days = [ 31, 29, 31, 30, 31, 30,
                 31, 31, 30, 31, 30, 31 ];
  
    // List of months
    let month = [ "January", "February", "March",
                  "April", "May", "June", "July",
                  "August", "September", "October",
                  "November", "December" ];
  
    // Days in half of a year
    let cnt = 183;
  
    // Index of current month
    let cur_month = 0;
    for(let i = 0; i < 12; i++)
        if (m == month[i])
            cur_month = i;
  
    // Starting day
    let cur_date = d;
  
    while (true)
    {
        while (cnt > 0 && cur_date <= days[cur_month])
        {
             
            // Decrement the value of
            // cnt by 1
            cnt -= 1;
  
            // Increment cur_date
            cur_date += 1;
        }
  
        // If cnt is equal to 0, then
        // break out of the loop
        if (cnt == 0)
            break;
  
        // Update cur_month
        cur_month = (cur_month + 1) % 12;
  
        // Update cur_date
        cur_date = 1;
    }
  
    // Print the resultant date
    document.write(cur_date + " " +
                   month[cur_month]);
}
 
// Driver Code
let D = 15;
let M = "January";
 
// Function Call
getDate(D, M);
 
// This code is contributed by susmitakundugoaldanga
 
</script>
Output: 
16 July

 

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

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