Date after adding given number of days to the given date - GeeksforGeeks

07 May, 2021
Given a date and a positive integer x. The task is to find date after adding x days to the given date
Examples: 
Input : d1 = 14, m1 = 5, y1 = 2017
        Days to be added x = 10
Output : d2 = 24, m2 = 5, y2 = 2017

Input : d1 = 14, m1 = 3, y1 = 2015
        Days to be added x = 466
Output : d2 = 14, m2 = 3, y2 = 2016
 
1) Let given date be d1, m1 and y1. Find offset (number of days spent from beginning to given date) of given year (Refer offsetDays() below) 2) Let offset found in above step be offset1. Find result year y2 and offset of result year offset2 (Refer highlighted code below) 3) Find days and months from offset2 and y2. (Refer revoffsetDays() below).
Below is the implementation of above steps. 
C++
Java
Python3
C#
Javascript
// C++ program to find date after adding// given number of days.#include<bits/stdc++.h>using namespace std; // Return if year is leap year or not.bool isLeap(int y){    if (y%100 != 0 && y%4 == 0 || y %400 == 0)        return true;     return false;} // Given a date, returns number of days elapsed// from the  beginning of the current year (1st// jan).int offsetDays(int d, int m, int y){    int offset = d;     switch (m - 1)    {    case 11:        offset += 30;    case 10:        offset += 31;    case 9:        offset += 30;    case 8:        offset += 31;    case 7:        offset += 31;    case 6:        offset += 30;    case 5:        offset += 31;    case 4:        offset += 30;    case 3:        offset += 31;    case 2:        offset += 28;    case 1:        offset += 31;    }     if (isLeap(y) && m > 2)        offset += 1;     return offset;} // Given a year and days elapsed in it, finds// date by storing results in d and m.void revoffsetDays(int offset, int y, int *d, int *m){    int month[13] = { 0, 31, 28, 31, 30, 31, 30,                      31, 31, 30, 31, 30, 31 };     if (isLeap(y))        month[2] = 29;     int i;    for (i = 1; i <= 12; i++)    {        if (offset <= month[i])            break;        offset = offset - month[i];    }     *d = offset;    *m = i;} // Add x days to the given date.void addDays(int d1, int m1, int y1, int x){    int offset1 = offsetDays(d1, m1, y1);    int remDays = isLeap(y1)?(366-offset1):(365-offset1);     // y2 is going to store result year and    // offset2 is going to store offset days    // in result year.    int y2, offset2;    if (x <= remDays)    {        y2 = y1;        offset2 = offset1 + x;    }     else    {        // x may store thousands of days.        // We find correct year and offset        // in the year.        x -= remDays;        y2 = y1 + 1;        int y2days = isLeap(y2)?366:365;        while (x >= y2days)        {            x -= y2days;            y2++;            y2days = isLeap(y2)?366:365;        }        offset2 = x;    }     // Find values of day and month from    // offset of result year.    int m2, d2;    revoffsetDays(offset2, y2, &d2, &m2);     cout << "d2 = " << d2 << ", m2 = " << m2         << ", y2 = " << y2;} // Driven Programint main(){    int d = 14, m = 3, y = 2015;    int x = 366;     addDays(d, m, y, x);     return 0;}
// Java program to find date after adding// given number of days. class GFG{     // Find values of day and month from// offset of result year.static int m2, d2;     // Return if year is leap year or not.static boolean isLeap(int y){    if (y % 100 != 0 && y % 4 == 0 || y % 400 == 0)        return true;     return false;} // Given a date, returns number of days elapsed// from the beginning of the current year (1st// jan).static int offsetDays(int d, int m, int y){    int offset = d;     if(m - 1 == 11)        offset += 335;    if(m - 1 == 10)        offset += 304;    if(m - 1 == 9)        offset += 273;    if(m - 1 == 8)        offset += 243;    if(m - 1 == 7)        offset += 212;    if(m - 1 == 6)        offset += 181;    if(m - 1 == 5)        offset += 151;    if(m - 1 == 4)        offset += 120;    if(m - 1 == 3)        offset += 90;    if(m - 1 == 2)        offset += 59;    if(m - 1 == 1)        offset += 31;     if (isLeap(y) && m > 2)        offset += 1;     return offset;} // Given a year and days elapsed in it, finds// date by storing results in d and m.static void revoffsetDays(int offset, int y){    int []month={ 0, 31, 28, 31, 30, 31, 30,                    31, 31, 30, 31, 30, 31 };     if (isLeap(y))        month[2] = 29;     int i;    for (i = 1; i <= 12; i++)    {        if (offset <= month[i])            break;        offset = offset - month[i];    }     d2 = offset;    m2 = i;} // Add x days to the given date.static void addDays(int d1, int m1, int y1, int x){    int offset1 = offsetDays(d1, m1, y1);    int remDays = isLeap(y1) ? (366 - offset1) : (365 - offset1);     // y2 is going to store result year and    // offset2 is going to store offset days    // in result year.    int y2, offset2 = 0;    if (x <= remDays)    {        y2 = y1;        offset2 =offset1 + x;    }     else    {        // x may store thousands of days.        // We find correct year and offset        // in the year.        x -= remDays;        y2 = y1 + 1;        int y2days = isLeap(y2) ? 366 : 365;        while (x >= y2days)        {            x -= y2days;            y2++;            y2days = isLeap(y2) ? 366 : 365;        }        offset2 = x;    }    revoffsetDays(offset2, y2);    System.out.println("d2 = " + d2 + ", m2 = " +                            m2 + ", y2 = " + y2);} // Driven Programpublic static void main(String[] args){    int d = 14, m = 3, y = 2015;    int x = 366;    addDays(d, m, y, x);}} // This code is contributed by mits
# Python3 program to find date after adding# given number of days. # Return if year is leap year or not.def isLeap(y):         if (y % 100 != 0 and y % 4 == 0 or y % 400 == 0):        return True    return False # Given a date, returns number of days elapsed# from the beginning of the current year (1st# jan).def offsetDays(d, m, y):         offset = d    switcher = {        10: 30,        9: 31,        8: 30,        7: 31,        6: 31,        5: 30,        4: 31,        3: 30,        2: 31,        1: 28,        0: 31    }    if (isLeap(y) and m > 1):        offset += 1    offset +=switcher.get(m)    return offset  # Given a year and days elapsed in it, finds# date by storing results in d and m.def revoffsetDays(offset, y,d, m):    month = [ 0, 31, 28, 31, 30, 31, 30,31, 31, 30, 31, 30, 31 ]         if (isLeap(y)):        month[2] = 29    for i in range(1, 13):        if (offset <= month[i]):            break        offset = offset - month[i]             d[0] = offset    m[0] = i + 1 # Add x days to the given date.def addDays(d1, m1, y1, x):         offset1 = offsetDays(d1, m1, y1)    if isLeap(y1):        remDays = 366 - offset1    else:        remDays = 365 - offset1             # y2 is going to store result year and    # offset2 is going to store offset days    # in result year.    if (x <= remDays):        y2 = y1        offset2 = offset1 + x    else:                 # x may store thousands of days.        # We find correct year and offset        # in the year.        x -= remDays        y2 = y1 + 1        if isLeap(y2):            y2days = 366        else:            y2days = 365                     while (x >= y2days):            x -= y2days            y2 += 1            if isLeap(y2):                y2days = 366            else:                y2days = 365                 offset2 = x             # Find values of day and month from    # offset of result year.    m2 = [0]    d2 = [0]    revoffsetDays(offset2, y2, d2, m2)    print("d2 = ",*d2,", m2 = ",*m2,", y2 = ",y2,sep="") # Driven Programd = 14m = 3y = 2015x = 366 addDays(d, m, y, x) # This code is contributed by shubhamsingh10
// C# program to find date after adding// given number of days.using System;class GFG{         // Find values of day and month from    // offset of result year.    static int m2, d2;     // Return if year is leap year or not.static bool isLeap(int y){    if (y % 100 != 0 && y % 4 == 0 || y % 400 == 0)        return true;     return false;} // Given a date, returns number of days elapsed// from the beginning of the current year (1st// jan).static int offsetDays(int d, int m, int y){    int offset = d;     if(m - 1 == 11)        offset += 335;    if(m - 1 == 10)        offset += 304;    if(m - 1 == 9)        offset += 273;    if(m - 1 == 8)        offset += 243;    if(m - 1 == 7)        offset += 212;    if(m - 1 == 6)        offset += 181;    if(m - 1 == 5)        offset += 151;    if(m - 1 == 4)        offset += 120;    if(m - 1 == 3)        offset += 90;    if(m - 1 == 2)        offset += 59;    if(m - 1 == 1)        offset += 31;     if (isLeap(y) && m > 2)        offset += 1;     return offset;} // Given a year and days elapsed in it, finds// date by storing results in d and m.static void revoffsetDays(int offset, int y){    int []month = { 0, 31, 28, 31, 30, 31, 30,                    31, 31, 30, 31, 30, 31 };     if (isLeap(y))        month[2] = 29;    int i;    for (i = 1; i <= 12; i++)    {        if (offset <= month[i])            break;        offset = offset - month[i];    }     d2 = offset;    m2 = i;} // Add x days to the given date.static void addDays(int d1, int m1, int y1, int x){    int offset1 = offsetDays(d1, m1, y1);    int remDays = isLeap(y1) ? (366 - offset1) : (365 - offset1);     // y2 is going to store result year and    // offset2 is going to store offset days    // in result year.    int y2, offset2 = 0;    if (x <= remDays)    {        y2 = y1;        offset2 = offset1+x;    }     else    {        // x may store thousands of days.        // We find correct year and offset        // in the year.        x -= remDays;        y2 = y1 + 1;        int y2days = isLeap(y2) ? 366 : 365;        while (x >= y2days)        {            x -= y2days;            y2++;            y2days = isLeap(y2)?366:365;        }        offset2 = x;    }    revoffsetDays(offset2, y2);    Console.WriteLine("d2 = " + d2 + ", m2 = " +                        m2 + ", y2 = " + y2);} // Driven Programstatic void Main(){    int d = 14, m = 3, y = 2015;    int x = 366;    addDays(d, m, y, x);}}// This code is contributed by mits
<script> // Javascript program to find date after// adding given number of days.    // Find values of day and month from// offset of result year.var m2, d2;     // Return if year is leap year or not.function isLeap(y){    if (y % 100 != 0 &&        y % 4 == 0 ||        y % 400 == 0)        return true;     return false;} // Given a date, returns number of// days elapsed from the beginning// of the current year (1st jan).function offsetDays(d , m , y){    var offset = d;     if (m - 1 == 11)        offset += 335;    if (m - 1 == 10)        offset += 304;    if (m - 1 == 9)        offset += 273;    if (m - 1 == 8)        offset += 243;    if (m - 1 == 7)        offset += 212;    if (m - 1 == 6)        offset += 181;    if (m - 1 == 5)        offset += 151;    if (m - 1 == 4)        offset += 120;    if (m - 1 == 3)        offset += 90;    if (m - 1 == 2)        offset += 59;    if (m - 1 == 1)        offset += 31;     if (isLeap(y) && m > 2)        offset += 1;     return offset;} // Given a year and days elapsed in it, finds// date by storing results in d and m.function revoffsetDays(offset, y){    var month = [ 0, 31, 28, 31, 30, 31, 30,                  31, 31, 30, 31, 30, 31 ];     if (isLeap(y))        month[2] = 29;     var i;    for(i = 1; i <= 12; i++)    {        if (offset <= month[i])            break;                     offset = offset - month[i];    }    d2 = offset;    m2 = i;} // Add x days to the given date.function addDays(d1 , m1 , y1 , x){    var offset1 = offsetDays(d1, m1, y1);    var remDays = isLeap(y1) ? (366 - offset1) :                               (365 - offset1);     // y2 is going to store result year and    // offset2 is going to store offset days    // in result year.    var y2, offset2 = 0;    if (x <= remDays)    {        y2 = y1;        offset2 =offset1 + x;    }    else    {                 // x may store thousands of days.        // We find correct year and offset        // in the year.        x -= remDays;        y2 = y1 + 1;        var y2days = isLeap(y2) ? 366 : 365;                 while (x >= y2days)        {            x -= y2days;            y2++;            y2days = isLeap(y2) ? 366 : 365;        }        offset2 = x;    }         revoffsetDays(offset2, y2);    document.write("d2 = " + d2 +                   ", m2 = " + m2 +                   ", y2 = " + y2);} // Driver codevar d = 14, m = 3, y = 2015;var x = 366; addDays(d, m, y, x); // This code is contributed by Amit Katiyar </script>
Output: 
d2 = 14, m2 = 3, y2 = 2016
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