Given two dates, find total number of days between them. The count of days must be calculated in O(1) time and O(1) auxiliary space.
Examples:
Input: dt1 = {10, 2, 2014}
dt2 = {10, 3, 2015}
Output: 393
dt1 represents "10-Feb-2014" and dt2 represents "10-Mar-2015"
The difference is 365 + 28
Input: dt1 = {10, 2, 2000}
dt2 = {10, 3, 2000}
Output: 29
Note that 2000 is a leap year
Input: dt1 = {10, 2, 2000}
dt2 = {10, 2, 2000}
Output: 0
Both dates are same
Input: dt1 = {1, 2, 2000};
dt2 = {1, 2, 2004};
Output: 1461
Number of days is 365*4 + 1
We strongly recommend that you click here and practice it, before moving on to the solution.
One Naive Solution is to start from dt1 and keep counting days till dt2 is reached. This solution requires more than O(1) time.
A Better and Simple solution is to count total number of days before dt1 from i.e., total days from 00/00/0000 to dt1, then count total number of days before dt2. Finally return the difference between two counts.
Let the given two dates be "1-Feb-2000" and "1-Feb-2004"
dt1 = {1, 2, 2000};
dt2 = {1, 2, 2004};
Count number of days before dt1. Let this count be n1.
Every leap year adds one extra day (29 Feb) to total days.
n1 = 2000*365 + 31 + 1 + Number of leap years
Count of leap years for a date 'd/m/y' can be calculated
using the following formula:
Number leap years
= floor(y/4) - floor(y/100) + floor(y/400) if m > 2
= floor((y-1)/4) - floor((y-1)/100) + floor((y-1)/400) if m <= 2
All above divisions must be done using integer arithmetic
So that the remainder is ignored.
For 01/01/2000, leap year count is 1999/4 - 1999/100
+ 1999/400 which is 499 - 19 + 4 = 484
Therefore n1 is 2000*365 + 31 + 1 + 484
Similarly, the count number of days before dt2.
Let this the count be n2.Finally, return n2-n1
Below is the implementation of the above idea.
C++
// C++ program two find number of // days between two given dates #include <iostream> using namespace std; // A date has day 'd', month 'm' and year 'y' struct Date { int d, m, y; }; // To store number of days in // all months from January to Dec. const int monthDays[12] = { 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 }; // This function counts number of // leap years before the given date int countLeapYears(Date d) { int years = d.y; // Check if the current year needs to be // considered for the count of leap years // or not if (d.m <= 2) years--; // An year is a leap year if it // is a multiple of 4, // multiple of 400 and not a // multiple of 100. return years / 4 - years / 100 + years / 400; } // This function returns number of // days between two given dates int getDifference(Date dt1, Date dt2) { // COUNT TOTAL NUMBER OF DAYS // BEFORE FIRST DATE 'dt1' // initialize count using years and day long int n1 = dt1.y * 365 + dt1.d; // Add days for months in given date for (int i = 0; i < dt1.m - 1; i++) n1 += monthDays[i]; // Since every leap year is of 366 days, // Add a day for every leap year n1 += countLeapYears(dt1); // SIMILARLY, COUNT TOTAL NUMBER OF // DAYS BEFORE 'dt2' long int n2 = dt2.y * 365 + dt2.d; for (int i = 0; i < dt2.m - 1; i++) n2 += monthDays[i]; n2 += countLeapYears(dt2); // return difference between two counts return (n2 - n1); } // Driver code int main() { Date dt1 = { 1, 2, 2000 }; Date dt2 = { 1, 2, 2004 }; // Function call cout << "Difference between two dates is " << getDifference(dt1, dt2); return 0; } |
Java
// Java program two find number of // days between two given dates class GFG { // A date has day 'd', month 'm' and year 'y' static class Date { int d, m, y; public Date(int d, int m, int y) { this.d = d; this.m = m; this.y = y; } }; // To store number of days in // all months from January to Dec. static int monthDays[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; // This function counts number of // leap years before the given date static int countLeapYears(Date d) { int years = d.y; // Check if the current year needs to be considered // for the count of leap years or not if (d.m <= 2) { years--; } // An year is a leap year if it is a multiple of 4, // multiple of 400 and not a multiple of 100. return years / 4 - years / 100 + years / 400; } // This function returns number // of days between two given dates static int getDifference(Date dt1, Date dt2) { // COUNT TOTAL NUMBER OF DAYS BEFORE FIRST DATE 'dt1' // initialize count using years and day int n1 = dt1.y * 365 + dt1.d; // Add days for months in given date for (int i = 0; i < dt1.m - 1; i++) { n1 += monthDays[i]; } // Since every leap year is of 366 days, // Add a day for every leap year n1 += countLeapYears(dt1); // SIMILARLY, COUNT TOTAL NUMBER OF DAYS BEFORE 'dt2' int n2 = dt2.y * 365 + dt2.d; for (int i = 0; i < dt2.m - 1; i++) { n2 += monthDays[i]; } n2 += countLeapYears(dt2); // return difference between two counts return (n2 - n1); } // Driver code public static void main(String[] args) { Date dt1 = new Date(1, 2, 2000); Date dt2 = new Date(1, 2, 2004); System.out.println("Difference between two dates is " + getDifference(dt1, dt2)); } } // This code is contributed by 29AjayKumar |
Python3
# Python3 program two find number of # days between two given dates # A date has day 'd', month # 'm' and year 'y' class Date: def __init__(self, d, m, y): self.d = d self.m = m self.y = y # To store number of days in # all months from January to Dec. monthDays = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31] # This function counts number of # leap years before the given date def countLeapYears(d): years = d.y # Check if the current year needs # to be considered for the count # of leap years or not if (d.m <= 2): years -= 1 # An year is a leap year if it is a # multiple of 4, multiple of 400 and # not a multiple of 100. ans = int(years / 4) ans -= int(years / 100) ans += int(years / 400) return ans # This function returns number of # days between two given dates def getDifference(dt1, dt2): # COUNT TOTAL NUMBER OF DAYS # BEFORE FIRST DATE 'dt1' # initialize count using years and day n1 = dt1.y * 365 + dt1.d # Add days for months in given date for i in range(0, dt1.m - 1): n1 += monthDays[i] # Since every leap year is of 366 days, # Add a day for every leap year n1 += countLeapYears(dt1) # SIMILARLY, COUNT TOTAL NUMBER # OF DAYS BEFORE 'dt2' n2 = dt2.y * 365 + dt2.d for i in range(0, dt2.m - 1): n2 += monthDays[i] n2 += countLeapYears(dt2) # return difference between # two counts return (n2 - n1) # Driver Code dt1 = Date(1, 9, 2014) dt2 = Date(3, 9, 2020) # Function call print("Difference between two dates is", getDifference(dt1, dt2)) # This code is contributed by Smitha |
C#
// C# program two find number of // days between two given dates using System; class GFG { // A date has day 'd', month 'm' and year 'y' public class Date { public int d, m, y; public Date(int d, int m, int y) { this.d = d; this.m = m; this.y = y; } }; // To store number of days in // all months from January to Dec. static int[] monthDays = { 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 }; // This function counts number of // leap years before the given date static int countLeapYears(Date d) { int years = d.y; // Check if the current year // needs to be considered // for the count of leap years or not if (d.m <= 2) { years--; } // An year is a leap year if it is // a multiple of 4, multiple of 400 // and not a multiple of 100. return years / 4 - years / 100 + years / 400; } // This function returns number // of days between two given dates static int getDifference(Date dt1, Date dt2) { // COUNT TOTAL NUMBER OF DAYS // BEFORE FIRST DATE 'dt1' // initialize count using years and day int n1 = dt1.y * 365 + dt1.d; // Add days for months in given date for (int i = 0; i < dt1.m - 1; i++) { n1 += monthDays[i]; } // Since every leap year is of 366 days, // Add a day for every leap year n1 += countLeapYears(dt1); // SIMILARLY, COUNT TOTAL // NUMBER OF DAYS BEFORE 'dt2' int n2 = dt2.y * 365 + dt2.d; for (int i = 0; i < dt2.m - 1; i++) { n2 += monthDays[i]; } n2 += countLeapYears(dt2); // return difference between two counts return (n2 - n1); } // Driver code public static void Main(String[] args) { Date dt1 = new Date(1, 2, 2000); Date dt2 = new Date(1, 2, 2004); // Function call Console.WriteLine("Difference between two dates is " + getDifference(dt1, dt2)); } } // This code is contributed by PrinciRaj1992 |
Javascript
<script> // JavaScript program two find number of // days between two given dates // A date has day 'd', month 'm' and year 'y' class Date { constructor(d,m,y) { this.d = d; this.m = m; this.y = y; } } // To store number of days in // all months from January to Dec. let monthDays=[31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]; // This function counts number of // leap years before the given date function countLeapYears(d) { let years = d.y; // Check if the current year needs to be considered // for the count of leap years or not if (d.m <= 2) { years--; } // An year is a leap year if it is a multiple of 4, // multiple of 400 and not a multiple of 100. return Math.floor(years / 4) - Math.floor(years / 100) + Math.floor(years / 400); } // This function returns number // of days between two given dates function getDifference(dt1,dt2) { // COUNT TOTAL NUMBER OF DAYS BEFORE FIRST DATE 'dt1' // initialize count using years and day let n1 = dt1.y * 365 + dt1.d; // Add days for months in given date for (let i = 0; i < dt1.m - 1; i++) { n1 += monthDays[i]; } // Since every leap year is of 366 days, // Add a day for every leap year n1 += countLeapYears(dt1); // SIMILARLY, COUNT TOTAL NUMBER OF DAYS BEFORE 'dt2' let n2 = dt2.y * 365 + dt2.d; for (let i = 0; i < dt2.m - 1; i++) { n2 += monthDays[i]; } n2 += countLeapYears(dt2); // return difference between two counts return (n2 - n1); } // Driver code let dt1 = new Date(1, 2, 2000); let dt2 = new Date(1, 2, 2004); document.write("Difference between two dates is " + getDifference(dt1, dt2)); // This code is contributed by rag2127 </script> |
Output
Difference between two dates is 1461
Time Complexity: O(1)
Auxiliary Space: O(1)
Note: The above program follows the Gregorian Calendar from the beginning of the time.
This article is contributed by Abhay Rathi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
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