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# Count Pairs from two arrays with even sum in C++

We are given two arrays of integer type elements let’s say, arr_1[] and arr_2[] and the task is to pick one element from arr_1[] and another element from arr_[] to form a pair then calculate the sum of elements in the pair and check whether the resultant sum is even or not.

**Input**

int arr_1[] = {2, 3, 7, 1, 4} int arr_2[] = { 2, 4, 1, 3}

**Output**

Count Pairs from two arrays with even sum are: 10

**Explanation**

We will form the pairs using both the arrays and the pairs so formed are-: (2, 2) = 4(valid), (2, 4) = 6(valid), (2, 1) = 3(invalid), (2, 3) = 5(invalid), (3, 2) = 5(invalid), (3, 4) = 7(invalid), (3, 1) = 4(valid), (3, 3) = 5(valid), (7, 2) = 9(invalid), (7, 4) = 11(invalid), (7, 1) = 8(valid), (7, 3) = 10(valid), (1, 2) = 3(invalid), (1, 4) = 5(invalid), (1, 1) = 2(valid), (1, 3) = 4(valid), (4, 2) = 6(valid), (4, 4) = 8(valid), (4, 1) = 5(invalid), (4, 3) = 7(invalid). There are 10 valid pairs formed using given two arrays that are even sums.

**Input**

int arr_1[] = {3, 1, 2} int arr_2[] = { 2, 4}

**Output**

Count Pairs from two arrays with even sum are: 2

**Explanation**

We will form the pairs using both the arrays and the pairs so formed are-: (3, 2) = 5(invalid), (3, 4) = 7(invalid), (1, 2) = 3(invalid), (1, 4) = 5(invalid), (2, 2) = 4(valid), (2, 4) = 6(valid), . There are 2 valid pairs formed using given two arrays that are even sums.

## Approach used in the below program is as follows

Input the two arrays of integer type elements and calculate the size of both the arrays and pass the data to the function for further processing.

Take a temporary variable as count to store the count of pairs with even sum

Start loop FOR from i to 0 till the size of array 1

Inside the loop, start another loop FOR from j to 0 till the size of array 2

Now store the sum of arr_1[i] and arr_2[j] in an integer variable let’s say sum

Check IF sum % 2 ==0 i.e. sum is even or not. IF yes then increment the count by 1.

Return the count

Print the result.

Input the two arrays of integer type elements and calculate the size of both the arrays and pass the data to the function for further processing.

Take a temporary variable as count to store the count of pairs with even sum

Start loop FOR from i to 0 till the size of array 1

Inside the loop, start another loop FOR from j to 0 till the size of array 2

Now store the sum of arr_1[i] and arr_2[j] in an integer variable let’s say sum

Check IF sum % 2 ==0 i.e. sum is even or not. IF yes then increment the count by 1.

Return the count

Print the result.

## Example

#include <iostream> using namespace std; int even_pair(int arr_1[], int size_arr1, int arr_2[], int size_arr2){ int count = 0; int odd = 0; for(int i = 0 ;i <size_arr1 ; i++){ for(int j = 0; j<size_arr2 ; j++){ int even = arr_1[i] + arr_2[j]; if(even % 2 == 0){ count++; } } } return count; } int main(){ int arr_1[] = {2, 3, 7, 1, 4}; int arr_2[] = { 2, 4, 1, 3}; int size_arr1 = sizeof(arr_1) / sizeof(arr_1[0]); int size_arr2 = sizeof(arr_2) / sizeof(arr_2[0]); cout<<"Count Pairs from two arrays with even sum are: "<<even_pair(arr_1, size_arr1, arr_2, size_arr2); return 0; }

## Output

If we run the above code it will generate the following output −

Count Pairs from two arrays with even sum are: 10

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