Number of ways to select exactly K even numbers from given Array

Given an array arr[] of n integers and an integer K, the task is to find the number of ways to select exactly K even numbers from the given array.

Examples:

Input: arr[] = {1, 2, 3, 4} k = 1
Output: 2
Explanation:
The number of ways in which we can select one even number is 2.

Input: arr[] = {61, 65, 99, 26, 57, 68, 23, 2, 32, 30} k = 2
Output:10
Explanation:
The number of ways in which we can select 2 even number is 10.

Approach: The idea is to apply the rule of combinatorics. For choosing r objects from the given n objects, the total number of ways of choosing is given by ⁿC_r. Below are the steps:

Count the total number of even elements from the given array(say cnt).
Check if the value of K is greater than cnt then the number of ways will be equal to 0.
Otherwise, the answer will be ⁿC_k.

Below is the implementation of the above approach:

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 
    // choosen 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 
    // choosen 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 
    # choosen 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 
    // choosen 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

Output:

Time Complexity: O(N)
Auxiliary Space: O(N)

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My Personal Notes

C++

Java

Python3

C#

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