Date after adding given number of days to the given date

Last Updated : 22 Dec, 2022

Given a date and a positive integer x. The task is to find the date after adding x days to the given date

Examples:

Input : d1 = 14, m1 = 5, y1 = 2017, x = 10
Output : d2 = 24, m2 = 5, y2 = 2017

Input : d1 = 14, m1 = 3, y1 = 2015, x = 366
Output : d2 = 14, m2 = 3, y2 = 2016

Method 1:

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++

// 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 Program
int main()
{
    int d = 14, m = 3, y = 2015;
    int x = 366;

    addDays(d, m, y, x);

    return 0;
}

Java

// 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 Program 
public 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

# 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 Program
d = 14
m = 3
y = 2015
x = 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 Program 
static void Main() 
{ 
    int d = 14, m = 3, y = 2015; 
    int x = 366; 
    addDays(d, m, y, x); 
} 
}
// This code is contributed by mits

JavaScript

<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 code
var 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

Time complexity: O(1) as it is performing constant operations
Auxiliary space: O(1)

Date after adding given number of days to the given date

kartik

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