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 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
# 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# 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
<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>
Time Complexity: O(1)
Auxiliary Space: O(1)
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