Given two arrays, arr[] and brr[] of size N and M respectively, the task is to find the count of pairs (arr[i], brr[j]) such that the product of elements of the pairs is an even number.
Examples:
Input: arr[] = { 1, 2, 3 }, brr[] = { 1, 2 }
Output: 4
Explanation:
Pairs with even product are: { (arr[0], brr[1]), (arr[1], brr[0]), (arr[1], brr[1]), (arr[2], brr[1]) }.
Therefore, the required output is 4.Input: arr[] = { 3, 2, 1, 4, 4}, brr[] = { 1, 4, 2, 3, 1 }
Output: 19
Naive Approach: The simplest approach to solve this problem is to traverse both the arrays and generate all possible pairs (arr[i], brr[j]) from both the arrays. For every pair, (arr[i], brr[j]), check if their product is an even number or not. If found to be true, then increment the count. Finally, print the count obtained.
Time Complexity: O(N * M)
Auxiliary Space: O(1)
Efficient Approach: The above approach can be optimized based on the following properties of product of two numbers:
Odd * Odd = Odd
Even * Odd = Even
Even * Even = Even
Follow the steps below to solve the problem:
- Initialize two variables, say cntOddArr and cntOddBrr, to store the count of odd numbers present in the arrays arr[] and brr[] respectively.
- Traverse the array arr[] and for every array element, check if it is an odd number or not. If found to be true, then update cntOddArr += 1.
- Traverse the array brr[] and for every array element, index, check if is is an odd number or not. If found to be true, then update cntOddBrr += 1.
- Finally, print the value of ((N * M) – cntOddArr * cntOddBrr).
Below is the implementation of the above approach:
C++
// C++ program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Function to count pairs (arr[i], brr[j])// whose product is an even numberint cntPairsInTwoArray(int arr[], int brr[], int N, int M){ // Stores count of odd // numbers in arr[] int cntOddArr = 0; // Stores count of odd // numbers in brr[] int cntOddBrr = 0; // Traverse the array, arr[] for (int i = 0; i < N; i++) { // If arr[i] is // an odd number if (arr[i] & 1) { // Update cntOddArr cntOddArr += 1; } } // Traverse the array, brr[] for (int i = 0; i < M; i++) { // If brr[i] is // an odd number if (brr[i] & 1) { // Update cntOddArr cntOddBrr += 1; } } // Return pairs whose product // is an even number return (N * M) - (cntOddArr * cntOddBrr);}// Driver Codeint main(){ int arr[] = { 1, 2, 3 }; int N = sizeof(arr) / sizeof(arr[0]); int brr[] = { 1, 2 }; int M = sizeof(brr) / sizeof(brr[0]); cout << cntPairsInTwoArray(arr, brr, N, M); return 0;} |
Java
// Java program to implement// the above approachimport java.util.*;class GFG{// Function to count pairs (arr[i], brr[j])// whose product is an even numberstatic int cntPairsInTwoArray(int arr[], int brr[], int N, int M){ // Stores count of odd // numbers in arr[] int cntOddArr = 0; // Stores count of odd // numbers in brr[] int cntOddBrr = 0; // Traverse the array, arr[] for (int i = 0; i < N; i++) { // If arr[i] is // an odd number if (arr[i] % 2 == 1) { // Update cntOddArr cntOddArr += 1; } } // Traverse the array, brr[] for (int i = 0; i < M; i++) { // If brr[i] is // an odd number if (brr[i] % 2 == 1) { // Update cntOddArr cntOddBrr += 1; } } // Return pairs whose product // is an even number return (N * M) - (cntOddArr * cntOddBrr);}// Driver Codepublic static void main(String[] args){ int arr[] = { 1, 2, 3 }; int N = arr.length; int brr[] = { 1, 2 }; int M = brr.length; System.out.print(cntPairsInTwoArray(arr, brr, N, M));}}// This code is contributed by 29AjayKumar |
Python3
# Python3 program to implement# the above approach# Function to count pairs (arr[i], brr[j])# whose product is an even numberdef cntPairsInTwoArray(arr, brr, N, M): # Stores count of odd # numbers in arr[] cntOddArr = 0 # Stores count of odd # numbers in brr[] cntOddBrr = 0 # Traverse the array, arr[] for i in range(N): # If arr[i] is # an odd number if (arr[i] & 1): # Update cntOddArr cntOddArr += 1 # Traverse the array, brr[] for i in range(M): # If brr[i] is # an odd number if (brr[i] & 1): # Update cntOddArr cntOddBrr += 1 # Return pairs whose product # is an even number return (N * M) - (cntOddArr * cntOddBrr)# Driver Codeif __name__ == '__main__': arr = [ 1, 2, 3 ] N = len(arr) brr = [ 1, 2 ] M = len(brr) print(cntPairsInTwoArray(arr, brr, N, M))# This code is contributed by mohit kumar 29 |
C#
// C# program to implement// the above approach using System; class GFG{ // Function to count pairs (arr[i], brr[j])// whose product is an even numberstatic int cntPairsInTwoArray(int[] arr, int[] brr, int N, int M){ // Stores count of odd // numbers in arr[] int cntOddArr = 0; // Stores count of odd // numbers in brr[] int cntOddBrr = 0; // Traverse the array, arr[] for(int i = 0; i < N; i++) { // If arr[i] is // an odd number if (arr[i] % 2 == 1) { // Update cntOddArr cntOddArr += 1; } } // Traverse the array, brr[] for(int i = 0; i < M; i++) { // If brr[i] is // an odd number if (brr[i] % 2 == 1) { // Update cntOddArr cntOddBrr += 1; } } // Return pairs whose product // is an even number return (N * M) - (cntOddArr * cntOddBrr);} // Driver Codepublic static void Main(){ int[] arr = { 1, 2, 3 }; int N = arr.Length; int[] brr = { 1, 2 }; int M = brr.Length; Console.Write(cntPairsInTwoArray( arr, brr, N, M));}}// This code is contributed by code_hunt |
4
Time Complexity: O(N)
Space Complexity: O(1)
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