Dada la longitud de una array de números enteros N y un número entero K . La tarea es modificar la array de tal manera que la array contenga primero todos los enteros impares del 1 al N en orden ascendente, luego todos los enteros pares del 1 al N en orden ascendente y luego imprima el K -ésimo elemento en la array modificada.
Ejemplos:
Entrada: N = 8, K = 5
Salida: 2
La array será {1, 3, 5, 7, 2, 4, 6, 8}
y el quinto elemento es 2.
Entrada: N = 7, K = 2
Salida : 3
Enfoque ingenuo: un enfoque simple es almacenar primero los números impares, uno por uno hasta N , y luego almacenar los números pares uno por uno hasta N , y luego imprimir el k-ésimo elemento.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the kth element
// in the modified array
int getNumber(int n, int k)
{
int arr[n];
int i = 0;
// First odd number
int odd = 1;
while (odd <= n) {
// Insert the odd number
arr[i++] = odd;
// Next odd number
odd += 2;
}
// First even number
int even = 2;
while (even <= n) {
// Insert the even number
arr[i++] = even;
// Next even number
even += 2;
}
// Return the kth element
return arr[k - 1];
}
// Driver code
int main()
{
int n = 8, k = 5;
cout << getNumber(n, k);
return 0;
}
Java
// Java implementation of the approach
class GFG
{
// Function to return the kth element
// in the modified array
static int getNumber(int n, int k)
{
int []arr = new int[n];
int i = 0;
// First odd number
int odd = 1;
while (odd <= n)
{
// Insert the odd number
arr[i++] = odd;
// Next odd number
odd += 2;
}
// First even number
int even = 2;
while (even <= n)
{
// Insert the even number
arr[i++] = even;
// Next even number
even += 2;
}
// Return the kth element
return arr[k - 1];
}
// Driver code
public static void main(String[] args)
{
int n = 8, k = 5;
System.out.println(getNumber(n, k));
}
}
// This code is contributed by 29AjayKumar
Python3
# Python3 implementation of the approach # Function to return the kth element # in the modified array def getNumber(n, k): arr = [0] * n; i = 0; # First odd number odd = 1; while (odd <= n): # Insert the odd number arr[i] = odd; i += 1; # Next odd number odd += 2; # First even number even = 2; while (even <= n): # Insert the even number arr[i] = even; i += 1; # Next even number even += 2; # Return the kth element return arr[k - 1]; # Driver code if __name__ == '__main__': n = 8; k = 5; print(getNumber(n, k)); # This code is contributed by Rajput-Ji
C#
// C# implementation of the approach
using System;
class GFG
{
// Function to return the kth element
// in the modified array
static int getNumber(int n, int k)
{
int []arr = new int[n];
int i = 0;
// First odd number
int odd = 1;
while (odd <= n)
{
// Insert the odd number
arr[i++] = odd;
// Next odd number
odd += 2;
}
// First even number
int even = 2;
while (even <= n)
{
// Insert the even number
arr[i++] = even;
// Next even number
even += 2;
}
// Return the kth element
return arr[k - 1];
}
// Driver code
public static void Main(String[] args)
{
int n = 8, k = 5;
Console.WriteLine(getNumber(n, k));
}
}
// This code is contributed by PrinciRaj1992
Javascript
<script>
// C++ implementation of the approach
// Function to return the kth element
// in the modified array
function getNumber(n, k)
{
var arr = Array(n).fill(n);
var i = 0;
// First odd number
var odd = 1;
while (odd <= n) {
// Insert the odd number
arr[i++] = odd;
// Next odd number
odd += 2;
}
// First even number
var even = 2;
while (even <= n) {
// Insert the even number
arr[i++] = even;
// Next even number
even += 2;
}
// Return the kth element
return arr[k - 1];
}
// Driver code
var n = 8, k = 5;
document.write(getNumber(n, k));
</script>
Producción:
2
Complejidad temporal: O(n)
Espacio auxiliar: O(n), ya que se utiliza un espacio adicional de tamaño n
Enfoque eficiente: encuentre el índice donde se almacenará el primer elemento par en la array generada. Ahora, si el valor de k es menor o igual que el índice, entonces el número deseado será k * 2 – 1, de lo contrario, el número deseado será (k – índice) * 2
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the kth element
// in the modified array
int getNumber(int n, int k)
{
int pos;
// Finding the index from where the
// even numbers will be stored
if (n % 2 == 0) {
pos = n / 2;
}
else {
pos = (n / 2) + 1;
}
// Return the kth element
if (k <= pos) {
return (k * 2 - 1);
}
else
return ((k - pos) * 2);
}
// Driver code
int main()
{
int n = 8, k = 5;
cout << getNumber(n, k);
return 0;
}
Java
// Java implementation of the approach
import java.io.*;
class GFG
{
// Function to return the kth element
// in the modified array
static int getNumber(int n, int k)
{
int pos;
// Finding the index from where the
// even numbers will be stored
if ((n % 2) == 0)
{
pos = n / 2;
}
else
{
pos = (n / 2) + 1;
}
// Return the kth element
if (k <= pos)
{
return (k * 2 - 1);
}
else
return ((k - pos) * 2);
}
// Driver code
public static void main (String[] args)
{
int n = 8, k = 5;
System.out.println (getNumber(n, k));
}
}
// This code is contributed by @tushil.
Python3
# Python3 implementation of the approach # Function to return the kth element # in the modified array def getNumber(n, k) : # Finding the index from where the # even numbers will be stored if (n % 2 == 0) : pos = n // 2; else : pos = (n // 2) + 1; # Return the kth element if (k <= pos) : return (k * 2 - 1); else : return ((k - pos) * 2); # Driver code if __name__ == "__main__" : n = 8; k = 5; print(getNumber(n, k)); # This code is contributed by AnkitRai01
C#
// C# implementation of the approach
using System;
class GFG
{
// Function to return the kth element
// in the modified array
static int getNumber(int n, int k)
{
int pos;
// Finding the index from where the
// even numbers will be stored
if ((n % 2) == 0)
{
pos = n / 2;
}
else
{
pos = (n / 2) + 1;
}
// Return the kth element
if (k <= pos)
{
return (k * 2 - 1);
}
else
return ((k - pos) * 2);
}
// Driver code
static public void Main ()
{
int n = 8, k = 5;
Console.Write(getNumber(n, k));
}
}
// This code is contributed by @ajit.
Javascript
<script>
// Javascript implementation of the approach
// Function to return the kth element
// in the modified array
function getNumber(n, k)
{
let pos;
// Finding the index from where the
// even numbers will be stored
if ((n % 2) == 0)
{
pos = parseInt(n / 2, 10);
}
else
{
pos = parseInt(n / 2, 10) + 1;
}
// Return the kth element
if (k <= pos)
{
return (k * 2 - 1);
}
else
return ((k - pos) * 2);
}
let n = 8, k = 5;
document.write(getNumber(n, k));
</script>
Producción:
2
Tiempo Complejidad: O(1)
Espacio Auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por dangenmaster2 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA