Dadas dos fechas, encuentre el número total de días entre ellas. El conteo de días debe calcularse en O(1) tiempo y O(1) espacio auxiliar.
Ejemplos:
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
Le recomendamos encarecidamente que haga clic aquí y lo practique antes de pasar a la solución.
Una solución ingenua es comenzar desde dt1 y seguir contando los días hasta llegar a dt2. Esta solución requiere más de O(1) tiempo.
Una solución mejor y más sencilla es contar el número total de días antes de dt1, es decir, el total de días desde 00/00/0000 hasta dt1, luego contar el número total de días antes de dt2. Finalmente devuelva la diferencia entre dos cuentas.
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
A continuación se muestra la implementación de la idea anterior.
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>
Producción
Difference between two dates is 1461
Complejidad de tiempo: O(1)
Espacio Auxiliar: O(1)
Nota: El programa anterior sigue el Calendario Gregoriano desde el principio de los tiempos.
Este artículo es una contribución de Abhay Rathi . Escriba comentarios si encuentra algo incorrecto o si desea compartir más información sobre el tema tratado anteriormente.
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA