# Count pairs from two arrays whose modulo operation yields K

Given an integer and two arrays and , the task is to count the total pairs (formed after choosing an element from and another from ) from these arrays whose modulo operation yields .

**Note:** If in a pair **(a, b)**, **a > b** then the modulo must be performed as **a % b**. Also, pairs occurring more than once will be counted only once.

**Examples:**

Input:arr1[] = {1, 3, 7}, arr2[] = {5, 3, 1}, K = 2

Output:2

(3, 5) and (7, 5) are the only possible pairs.

Since, 5 % 3 = 2 and 7 % 5 = 2

Input:arr1[] = {2, 5, 99}, arr2[] = {2, 8, 1, 4}, K = 0

Output:6

All possible pairs are (2, 2), (2, 8), (2, 4), (2, 1), (5, 1) and (99, 1).

**Approach:**

- Take one element from at a time and perform it’s modulo operation with all the other elements of one by one.
- If the result from the previous step is equal to then store the pair (a, b) in a set in order to avoid duplicates where a is the smaller element and b is the larger one.
- Total required pairs will be the size of the set in the end.

Below is the implementation of the above approach:

## C++

`// C++ implementation of above approach ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to return the total pairs ` `// of elements whose modulo yield K ` `int` `totalPairs(` `int` `arr1[], ` `int` `arr2[], ` `int` `K, ` `int` `n, ` `int` `m) ` `{ ` ` ` ` ` `// set is used to avoid duplicate pairs ` ` ` `set<pair<` `int` `, ` `int` `> > s; ` ` ` ` ` `for` `(` `int` `i = 0; i < n; i++) { ` ` ` `for` `(` `int` `j = 0; j < m; j++) { ` ` ` ` ` `// check which element is greater and ` ` ` `// proceed according to it ` ` ` `if` `(arr1[i] > arr2[j]) { ` ` ` ` ` `// check if modulo is equal to K ` ` ` `if` `(arr1[i] % arr2[j] == K) ` ` ` `s.insert(make_pair(arr1[i], arr2[j])); ` ` ` `} ` ` ` `else` `{ ` ` ` `if` `(arr2[j] % arr1[i] == K) ` ` ` `s.insert(make_pair(arr2[j], arr1[i])); ` ` ` `} ` ` ` `} ` ` ` `} ` ` ` ` ` `// return size of the set ` ` ` `return` `s.size(); ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `arr1[] = { 8, 3, 7, 50 }; ` ` ` `int` `arr2[] = { 5, 1, 10, 4 }; ` ` ` `int` `K = 3; ` ` ` `int` `n = ` `sizeof` `(arr1) / ` `sizeof` `(arr1[0]); ` ` ` `int` `m = ` `sizeof` `(arr2) / ` `sizeof` `(arr2[0]); ` ` ` ` ` `cout << totalPairs(arr1, arr2, K, n, m); ` ` ` `return` `0; ` `} ` |

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## Python3

`# Python3 implementation of above approach ` ` ` `# Function to return the total pairs ` `# of elements whose modulo yield K ` `def` `totalPairs(arr1, arr2, K, n, m): ` ` ` ` ` `# set is used to avoid duplicate pairs ` ` ` `s` `=` `{} ` ` ` ` ` `for` `i ` `in` `range` `(n): ` ` ` `for` `j ` `in` `range` `(m): ` ` ` ` ` `# check which element is greater and ` ` ` `# proceed according to it ` ` ` `if` `(arr1[i] > arr2[j]): ` ` ` ` ` `# check if modulo is equal to K ` ` ` `if` `(arr1[i] ` `%` `arr2[j] ` `=` `=` `K): ` ` ` `s[(arr1[i], arr2[j])]` `=` `1` ` ` `else` `: ` ` ` `if` `(arr2[j] ` `%` `arr1[i] ` `=` `=` `K): ` ` ` `s[(arr2[j], arr1[i])]` `=` `1` ` ` ` ` ` ` ` ` `# return size of the set ` ` ` `return` `len` `(s) ` ` ` `# Driver code ` ` ` `arr1 ` `=` `[ ` `8` `, ` `3` `, ` `7` `, ` `50` `] ` `arr2 ` `=` `[` `5` `, ` `1` `, ` `10` `, ` `4` `] ` `K ` `=` `3` `n ` `=` `len` `(arr1) ` `m ` `=` `len` `(arr2) ` ` ` `print` `(totalPairs(arr1, arr2, K, n, m)) ` ` ` `# This code is contributed by mohit kumar 29 ` |

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**Output:**

3

**Note:** To print all the pairs just print the elements of set.

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