Given an array arr[] of length N, the task is to count the number of pairs (X, Y) such that XY is even and count the number of pairs such that XY is odd.
Examples:
Input: arr[] = {2, 3, 4, 5}
Output:
6
6
Explanation: (2, 3), (2, 4), (2, 5), (4, 2), (4, 3) and (4, 5) are the pairs with even values
and (3, 2), (3, 4), (3, 5), (5, 2), (5, 3) and (5, 4) are the pairs with odd values.Input: arr[] = {10, 11, 20, 60, 70}
Output:
16
4
Explanation: (10, 11), (10, 20), (10, 60), (10, 70), (20, 10), (20, 11), (20, 60), (20, 70), (60, 10), (60, 11), (60, 20), (60, 70), (70, 10), (70, 11), (70, 20), (70, 60) are the pairs with even values and (11, 10), (11, 20), (11, 60), (11, 70) are the pairs with odd values.
Naive approach: Calculate the powers for every single pair possible and find whether the calculated value is even or odd.
Efficient approach: Count the even and odd elements in the array and then use the concept pow (even, any element except itself) is even and pow (odd, any element except itself) is odd.
So, the number of pairs (X, Y) are,
- pow(X, Y) is even = (count of even number * (n – 1))
- pow(X, Y) is odd = (count of odd number * (n – 1))
Below is the implementation of the above approach:
CPP
// C++ implementation of the approach #include <iostream> using namespace std; // Function to find and print the // required count of pairs void countPairs(int arr[], int n) { // Find the count of even and // odd elements in the array int even = 0, odd = 0; for (int i = 0; i < n; i++) { if (arr[i] % 2 == 0) even++; else odd++; } // Print the required count of pairs cout << (even) * (n - 1) << endl; cout << (odd) * (n - 1) << endl; } // Driver code int main() { int arr[] = { 2, 3, 4, 5 }; int n = sizeof(arr) / sizeof(arr[0]); countPairs(arr, n); return 0; } |
Java
// Java implementation of the approach class GFG { // Function to find and print the // required count of pairs static void countPairs(int arr[], int n) { // Find the count of even and // odd elements in the array int even = 0, odd = 0; for (int i = 0; i < n; i++) { if (arr[i] % 2 == 0) even++; else odd++; } // Print the required count of pairs System.out.println((even) * (n - 1)); System.out.println((odd) * (n - 1)); } // Driver code public static void main(String args[]) { int arr[] = { 2, 3, 4, 5 }; int n = arr.length; countPairs(arr, n); } } // This code is contributd by ANKITUMAR34 |
C#
// C# implementation of the approach using System; class GFG { // Function to find and print the // required count of pairs static void countPairs(int []arr, int n) { // Find the count of even and // odd elements in the array int even = 0, odd = 0; for (int i = 0; i < n; i++) { if (arr[i] % 2 == 0) even++; else odd++; } // Print the required count of pairs Console.WriteLine((even) * (n - 1)); Console.WriteLine((odd) * (n - 1)); } // Driver code public static void Main() { int []arr = { 2, 3, 4, 5 }; int n = arr.Length; countPairs(arr, n); } } // This code is contributd by AnkitRai01 |
Python3
# Python3 implementation of the approach # Function to find and print the # required count of pairs def countPairs(arr, n): # Find the count of even and # odd elements in the array odd = 0 even = 0 for i in range(n): if (arr[i] % 2 == 0): even += 1 else: odd += 1 # Count the number of odd pairs odd_pairs = odd*(n-1) # Count the number of even pairs even_pairs = even*(n-1) print(odd_pairs) print(even_pairs) # Driver code if __name__ == '__main__': arr = [2, 3, 4, 5] n = len(arr) countPairs(arr, n) |
6 6
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