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-yearvoid 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 Codeint 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 approachclass GFG{ // Function to find the date// after the next half-yearpublic 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 Codepublic 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-yeardef 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 CodeD = 15M = "January"# Function CallgetDate(D, M) |
C#
// C# program for the above approachusing System;class GFG{ // Function to find the date// after the next half-yearstatic 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 Codepublic 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-yearfunction 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 Codelet D = 15;let M = "January";// Function CallgetDate(D, M);// This code is contributed by susmitakundugoaldanga</script> |
Output:
16 July
Time Complexity: O(1)
Auxiliary Space: O(1)
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