# Find the date after next half year from a given date

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 JulyExplanation: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**.

- If the value of
- 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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