Dada una array arr[] de n enteros y un entero K , la tarea es encontrar el número de formas de seleccionar exactamente K números pares de la array dada.
Ejemplos:
Entrada: arr[] = {1, 2, 3, 4} k = 1
Salida: 2
Explicación:
El número de formas en que podemos seleccionar un número par es 2.Entrada: arr[] = {61, 65, 99, 26, 57, 68, 23, 2, 32, 30} k = 2
Salida: 10
Explicación:
El número de formas en que podemos seleccionar 2 números pares es 10.
Planteamiento: La idea es aplicar la regla de la combinatoria. Para elegir r objetos de los n objetos dados , el número total de formas de elegir está dado por n C r . A continuación se muestran los pasos:
- Cuente el número total de elementos pares de la array dada (por ejemplo, cnt ).
- Compruebe si el valor de K es mayor que cnt , entonces el número de formas será igual a 0.
- De lo contrario, la respuesta será n C k .
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
long long f[12];
// Function for calculating factorial
void fact()
{
// Factorial of n defined as:
// n! = n * (n - 1) * ... * 1
f[0] = f[1] = 1;
for (int i = 2; i <= 10; i++)
f[i] = i * 1LL * f[i - 1];
}
// Function to find the number of ways to
// select exactly K even numbers
// from the given array
void solve(int arr[], int n, int k)
{
fact();
// Count even numbers
int even = 0;
for (int i = 0; i < n; i++) {
// Check if the current
// number is even
if (arr[i] % 2 == 0)
even++;
}
// Check if the even numbers to be
// chosen is greater than n. Then,
// there is no way to pick it.
if (k > even)
cout << 0 << endl;
else {
// The number of ways will be nCk
cout << f[even] / (f[k] * f[even - k]);
}
}
// Driver Code
int main()
{
// Given array arr[]
int arr[] = { 1, 2, 3, 4 };
int n = sizeof arr / sizeof arr[0];
// Given count of even elements
int k = 1;
// Function Call
solve(arr, n, k);
return 0;
}
Java
// Java program for the above approach
class GFG{
static int []f = new int[12];
// Function for calculating factorial
static void fact()
{
// Factorial of n defined as:
// n! = n * (n - 1) * ... * 1
f[0] = f[1] = 1;
for(int i = 2; i <= 10; i++)
f[i] = i * 1 * f[i - 1];
}
// Function to find the number of ways to
// select exactly K even numbers
// from the given array
static void solve(int arr[], int n, int k)
{
fact();
// Count even numbers
int even = 0;
for(int i = 0; i < n; i++)
{
// Check if the current
// number is even
if (arr[i] % 2 == 0)
even++;
}
// Check if the even numbers to be
// chosen is greater than n. Then,
// there is no way to pick it.
if (k > even)
System.out.print(0 + "\n");
else
{
// The number of ways will be nCk
System.out.print(f[even] /
(f[k] * f[even - k]));
}
}
// Driver Code
public static void main(String[] args)
{
// Given array arr[]
int arr[] = { 1, 2, 3, 4 };
int n = arr.length;
// Given count of even elements
int k = 1;
// Function call
solve(arr, n, k);
}
}
// This code is contributed by Rajput-Ji
Python3
# Python3 program for the above approach f = [0] * 12 # Function for calculating factorial def fact(): # Factorial of n defined as: # n! = n * (n - 1) * ... * 1 f[0] = f[1] = 1 for i in range(2, 11): f[i] = i * 1 * f[i - 1] # Function to find the number of ways to # select exactly K even numbers # from the given array def solve(arr, n, k): fact() # Count even numbers even = 0 for i in range(n): # Check if the current # number is even if (arr[i] % 2 == 0): even += 1 # Check if the even numbers to be # chosen is greater than n. Then, # there is no way to pick it. if (k > even): print(0) else: # The number of ways will be nCk print(f[even] // (f[k] * f[even - k])) # Driver Code # Given array arr[] arr = [ 1, 2, 3, 4 ] n = len(arr) # Given count of even elements k = 1 # Function call solve(arr, n, k) # This code is contributed by code_hunt
C#
// C# program for the above approach
using System;
class GFG{
static int []f = new int[12];
// Function for calculating factorial
static void fact()
{
// Factorial of n defined as:
// n! = n * (n - 1) * ... * 1
f[0] = f[1] = 1;
for(int i = 2; i <= 10; i++)
f[i] = i * 1 * f[i - 1];
}
// Function to find the number of ways to
// select exactly K even numbers
// from the given array
static void solve(int []arr, int n, int k)
{
fact();
// Count even numbers
int even = 0;
for(int i = 0; i < n; i++)
{
// Check if the current
// number is even
if (arr[i] % 2 == 0)
even++;
}
// Check if the even numbers to be
// chosen is greater than n. Then,
// there is no way to pick it.
if (k > even)
Console.Write(0 + "\n");
else
{
// The number of ways will be nCk
Console.Write(f[even] /
(f[k] * f[even - k]));
}
}
// Driver Code
public static void Main(String[] args)
{
// Given array []arr
int []arr = { 1, 2, 3, 4 };
int n = arr.Length;
// Given count of even elements
int k = 1;
// Function call
solve(arr, n, k);
}
}
// This code is contributed by sapnasingh4991
Javascript
<script>
// Javascript program for the above approach
var f = Array(12).fill(0);
// Function for calculating factorial
function fact()
{
// Factorial of n defined as:
// n! = n * (n - 1) * ... * 1
f[0] = f[1] = 1;
for (var i = 2; i <= 10; i++)
f[i] = i * 1 * f[i - 1];
}
// Function to find the number of ways to
// select exactly K even numbers
// from the given array
function solve(arr, n, k)
{
fact();
// Count even numbers
var even = 0;
for (var i = 0; i < n; i++) {
// Check if the current
// number is even
if (arr[i] % 2 == 0)
even++;
}
// Check if the even numbers to be
// chosen is greater than n. Then,
// there is no way to pick it.
if (k > even)
document.write( 0 );
else {
// The number of ways will be nCk
document.write( f[even] / (f[k] * f[even - k]));
}
}
// Driver Code
// Given array arr[]
var arr = [ 1, 2, 3, 4 ];
var n = arr.length;
// Given count of even elements
var k = 1;
// Function Call
solve(arr, n, k);
</script>
Producción:
2
Complejidad temporal: O(N)
Espacio auxiliar: O(N)