Dada una array arr[][2] que consta de N pares de strings que representan la hora de inicio y finalización (en formato de 12 horas ) y una string P que representa la hora de la reunión, la tarea es encontrar el recuento de intervalos que contiene el tiempo p
Ejemplos:
Entrada: P = “12:01:AM”, arr[][2] = {{“12:00:AM”, “11:55:PM”}, {“12:01:AM”, “11: 50:AM”}, {“12:30:AM”, “12:00:PM”}, {“11:57:AM”, “11:59:PM”}}
Salida: 2
Explicación: La hora P se encuentra en el primer y segundo intervalo de tiempo.Entrada: P = “12:01:AM”, arr[][2] = {{“09:57:AM”, “12:00:PM”} }
Salida: 0
Explicación: Ningún intervalo contiene la hora P.
Enfoque: La idea es convertir primero todas las horas del formato de 12 horas al formato de 24 horas y luego comparar el rango con la Hora P . Siga los pasos a continuación para resolver el problema:
- Primero, convierta todas las horas del formato de 12 horas al formato de 24 horas y luego almacene el valor entero del formato de hora de 24 horas.
- Inicialice una variable, digamos ans, para almacenar el conteo de intervalos en los que se encuentra P.
- Recorra la array arr[] e incremente el conteo de ans en 1 si (P ≥ L && P ≤ R) o (P ≥ R && P ≤ L) .
- Finalmente, después de completar los pasos anteriores, imprima ans .
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to convert a time in 24
// hour format to an equivalent integer
int convert(string str)
{
// Removes ":" at 3rd position
str.replace(2, 1, "");
// Calculate hours
int h1 = (int)str[1] - '0';
int h2 = (int)str[0] - '0';
int hh = (h2 * 10 + h1 % 10);
// Stores the time in 24 hours format
int time = 0;
// If time is in "AM"
if (str[5] == 'A') {
// If hh is equal to 12
if (hh == 12)
time += stoi(str.substr(2, 2));
else {
time += stoi(str.substr(0, 2));
}
}
// If time is in "PM"
else {
// If hh is equal to 12
if (hh == 12) {
time += stoi(str.substr(0, 4));
}
else {
time += stoi(str.substr(0, 4));
time += 1200;
}
}
// Return time
return time;
}
// Function to count number
// of intervals in which p lies
int countOverlap(string arr[][2],
int n, string p)
{
// Stores the count
int ans = 0;
// Stores the integer value of
// 24 hours time format of P
int M = convert(p);
// Traverse the array
for (int i = 0; i < n; i++) {
// Stores the integer value of
// 24 hours time format of arr[i][0]
int L = convert(arr[i][0]);
// Stores the integer value of
// 24 hours time format of arr[i][1]
int R = convert(arr[i][1]);
// If M lies within the [L, R]
if ((L <= M && M <= R)
|| (M >= R && M <= L))
// Increment ans by 1
ans++;
}
// Return ans
return ans;
}
// Driver Code
int main()
{
string arr[][2] = { { "12:00:AM", "11:55:PM" },
{ "12:01:AM", "11:50:AM" },
{ "12:30:AM", "12:00:PM" },
{ "11:57:AM", "11:59:PM" } };
string P = "12:01:PM";
int N = sizeof(arr) / sizeof(arr[0]);
cout << countOverlap(arr, N, P) << endl;
}
Java
// Java implementation of the above approach
import java.io.*;
class GFG
{
// Function to convert a time in 24
// hour format to an equivalent integer
static int convert(String str)
{
// Removes ":" at 3rd position
str = str.substring(0, 2) + str.substring(3);
// Calculate hours
int h1 = (int)str.charAt(1) - '0';
int h2 = (int)str.charAt(0) - '0';
int hh = (h2 * 10 + h1 % 10);
// Stores the time in 24 hours format
int time = 0;
// If time is in "AM"
if (str.charAt(5) == 'A') {
// If hh is equal to 12
if (hh == 12)
time += Integer.parseInt(
str.substring(2, 4));
else {
time += Integer.parseInt(
str.substring(0, 2));
}
}
// If time is in "PM"
else {
// If hh is equal to 12
if (hh == 12) {
time += Integer.parseInt(
str.substring(0, 4));
}
else {
time += Integer.parseInt(
str.substring(0, 4));
time += 1200;
}
}
// Return time
return time;
}
// Function to count number
// of intervals in which p lies
static int countOverlap(String arr[][], int n, String p)
{
// Stores the count
int ans = 0;
// Stores the integer value of
// 24 hours time format of P
int M = convert(p);
// Traverse the array
for (int i = 0; i < n; i++)
{
// Stores the integer value of
// 24 hours time format of arr[i][0]
int L = convert(arr[i][0]);
// Stores the integer value of
// 24 hours time format of arr[i][1]
int R = convert(arr[i][1]);
// If M lies within the [L, R]
if ((L <= M && M <= R) || (M >= R && M <= L))
// Increment ans by 1
ans++;
}
// Return ans
return ans;
}
// Driver Code
public static void main(String[] args)
{
String[][] arr
= new String[][] { { "12:00:AM", "11:55:PM" },
{ "12:01:AM", "11:50:AM" },
{ "12:30:AM", "12:00:PM" },
{ "11:57:AM", "11:59:PM" } };
String P = "12:01:PM";
int N = arr.length;
System.out.println(countOverlap(arr, N, P));
}
}
// This code is contributed by Dharanendra L V
C#
// C# implementation of the above approach
using System;
class GFG{
// Function to convert a time in 24
// hour format to an equivalent integer
static int convert(String str)
{
// Removes ":" at 3rd position
str = str.Substring(0, 2) + str.Substring(3);
// Calculate hours
int h1 = (int)str[1] - '0';
int h2 = (int)str[0] - '0';
int hh = (h2 * 10 + h1 % 10);
// Stores the time in 24 hours format
int time = 0;
// If time is in "AM"
if (str[5] == 'A')
{
// If hh is equal to 12
if (hh == 12)
time += Int32.Parse(
str.Substring(2, 2));
else
{
time += Int32.Parse(
str.Substring(0, 2));
}
}
// If time is in "PM"
else
{
// If hh is equal to 12
if (hh == 12)
{
time += Int32.Parse(
str.Substring(0, 4));
}
else
{
time += Int32.Parse(
str.Substring(0, 4));
time += 1200;
}
}
// Return time
return time;
}
// Function to count number
// of intervals in which p lies
static int countOverlap(String [,]arr, int n,
String p)
{
// Stores the count
int ans = 0;
// Stores the integer value of
// 24 hours time format of P
int M = convert(p);
// Traverse the array
for(int i = 0; i < n; i++)
{
// Stores the integer value of
// 24 hours time format of arr[i,0]
int L = convert(arr[i,0]);
// Stores the integer value of
// 24 hours time format of arr[i,1]
int R = convert(arr[i,1]);
// If M lies within the [L, R]
if ((L <= M && M <= R) ||
(M >= R && M <= L))
// Increment ans by 1
ans++;
}
// Return ans
return ans;
}
// Driver Code
public static void Main(String[] args)
{
String[,] arr = new String[,]{
{ "12:00:AM", "11:55:PM" },
{ "12:01:AM", "11:50:AM" },
{ "12:30:AM", "12:00:PM" },
{ "11:57:AM", "11:59:PM" } };
String P = "12:01:PM";
int N = arr.GetLength(0);
Console.WriteLine(countOverlap(arr, N, P));
}
}
// This code is contributed by 29AjayKumar
Producción:
2
Complejidad temporal: O(N)
Espacio auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por shreyajaiswal3 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA