# Count distinct pairs from two arrays having same sum of digits

Given two arrays arr1[] and arr2[]. The task is to find the total number of distinct pairs(formed by picking 1 element from arr1 and one element from arr2), such that both the elements of the pair have the sum of digits.

Note: Pairs occurring more than once must be counted only once.

Examples:

```Input : arr1[] = {33, 41, 59, 1, 3}
arr2[] = {3, 32, 51, 3}
Output : 3
Possible pairs are:
(33, 51), (41, 32), (3, 3)

Input : arr1[] = {1, 6, 4, 22}
arr2[] = {1, 3, 24}
Output : 2
Possible pairs are:
(1, 1), (6, 24)
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

• Run two nested loops to generate all possible pairs from the two arrays taking one element from arr1[] and one from arr2[].
• If sum of digits is equal, then insert the pair(a, b) into a set, in order to avoid duplicates where a is the smaller element and b is the larger one.
• Total pairs will be the size of the final set.

Below is the implementation of the above approach:

## C++

 `// C++ program to count total number of ` `// pairs having elements with same ` `// sum of digits ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function for returning ` `// sum of digits of a number ` `int` `digitSum(``int` `n) ` `{ ` `    ``int` `sum = 0; ` `    ``while` `(n > 0) { ` `        ``sum += n % 10; ` `        ``n = n / 10; ` `    ``} ` `    ``return` `sum; ` `} ` ` `  `// Function to return the total pairs ` `// of elements with equal sum of digits ` `int` `totalPairs(``int` `arr1[], ``int` `arr2[], ``int` `n, ``int` `m) ` `{ ` ` `  `    ``// set is used to avoid duplicate pairs ` `    ``set > s; ` ` `  `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``for` `(``int` `j = 0; j < m; j++) { ` ` `  `            ``// check sum of digits ` `            ``// of both the elements ` `            ``if` `(digitSum(arr1[i]) == digitSum(arr2[j])) { ` ` `  `                ``if` `(arr1[i] < arr2[j]) ` `                    ``s.insert(make_pair(arr1[i], arr2[j])); ` `                ``else` `                    ``s.insert(make_pair(arr2[j], arr1[i])); ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// return size of the set ` `    ``return` `s.size(); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr1[] = { 100, 3, 7, 50 }; ` `    ``int` `arr2[] = { 5, 1, 10, 4 }; ` `    ``int` `n = ``sizeof``(arr1) / ``sizeof``(arr1); ` `    ``int` `m = ``sizeof``(arr2) / ``sizeof``(arr2); ` ` `  `    ``cout << totalPairs(arr1, arr2, n, m); ` `    ``return` `0; ` `} `

## Java

 `// Java program to count total number of ` `// pairs having elements with same ` `// sum of digits ` `import` `java.util.*; ` ` `  `class` `GFG  ` `{ ` ` `  `static` `class` `pair ` `{  ` `    ``int` `first, second;  ` `    ``public` `pair(``int` `first, ``int` `second)  ` `    ``{  ` `        ``this``.first = first;  ` `        ``this``.second = second;  ` `    ``}  ` `} ` ` `  `// Function for returning ` `// sum of digits of a number ` `static` `int` `digitSum(``int` `n) ` `{ ` `    ``int` `sum = ``0``; ` `    ``while` `(n > ``0``) ` `    ``{ ` `        ``sum += n % ``10``; ` `        ``n = n / ``10``; ` `    ``} ` `    ``return` `sum; ` `} ` ` `  `// Function to return the total pairs ` `// of elements with equal sum of digits ` `static` `int` `totalPairs(``int` `arr1[], ``int` `arr2[], ` `                      ``int` `n, ``int` `m) ` `{ ` ` `  `    ``// set is used to avoid duplicate pairs ` `    ``Set s = ``new` `HashSet<>(); ` ` `  `    ``for` `(``int` `i = ``0``; i < n; i++)  ` `    ``{ ` `        ``for` `(``int` `j = ``0``; j < m; j++)  ` `        ``{ ` ` `  `            ``// check sum of digits ` `            ``// of both the elements ` `            ``if` `(digitSum(arr1[i]) == digitSum(arr2[j])) ` `            ``{ ` `                ``if` `(arr1[i] < arr2[j]) ` `                    ``s.add(``new` `pair(arr1[i], arr2[j])); ` `                ``else` `                    ``s.add(``new` `pair(arr2[j], arr1[i])); ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// return size of the set ` `    ``return` `s.size(); ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `arr1[] = { ``100``, ``3``, ``7``, ``50` `}; ` `    ``int` `arr2[] = { ``5``, ``1``, ``10``, ``4` `}; ` `    ``int` `n = arr1.length; ` `    ``int` `m = arr2.length; ` ` `  `    ``System.out.println(totalPairs(arr1, arr2, n, m)); ` `} ` `}  ` ` `  `// This code is contributed by Rajput-Ji `

## Python3

 `# Python3 program to count total number of  ` `# pairs having elements with same sum of digits  ` ` `  `# Function for returning  ` `# sum of digits of a number  ` `def` `digitSum(n):  ` `  `  `    ``Sum` `=` `0`  `    ``while` `n > ``0``:   ` `        ``Sum` `+``=` `n ``%` `10`  `        ``n ``=` `n ``/``/` `10`  `      `  `    ``return` `Sum`  ` `  `# Function to return the total pairs  ` `# of elements with equal sum of digits  ` `def` `totalPairs(arr1, arr2, n, m):  ` ` `  `    ``# set is used to avoid duplicate pairs  ` `    ``s ``=` `set``()  ` ` `  `    ``for` `i ``in` `range``(``0``, n):   ` `        ``for` `j ``in` `range``(``0``, m):   ` ` `  `            ``# check sum of digits  ` `            ``# of both the elements  ` `            ``if` `digitSum(arr1[i]) ``=``=` `digitSum(arr2[j]):   ` ` `  `                ``if` `arr1[i] < arr2[j]:  ` `                    ``s.add((arr1[i], arr2[j]))  ` `                ``else``: ` `                    ``s.add((arr2[j], arr1[i]))  ` `              `  `    ``# return size of the set  ` `    ``return` `len``(s)  ` ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"``: ` `  `  `    ``arr1 ``=` `[``100``, ``3``, ``7``, ``50``]   ` `    ``arr2 ``=` `[``5``, ``1``, ``10``, ``4``]  ` `    ``n ``=` `len``(arr1)  ` `    ``m ``=` `len``(arr2)  ` ` `  `    ``print``(totalPairs(arr1, arr2, n, m))  ` `     `  `# This code is contributed by Rituraj Jain `

## C#

 `// C# program to count total number of ` `// pairs having elements with same ` `// sum of digits ` `using` `System; ` `using` `System.Collections.Generic; ` ` `  `class` `GFG  ` `{ ` ` `  `public` `class` `pair ` `{  ` `    ``public` `int` `first, second;  ` `    ``public` `pair(``int` `first, ``int` `second)  ` `    ``{  ` `        ``this``.first = first;  ` `        ``this``.second = second;  ` `    ``}  ` `} ` ` `  `// Function for returning ` `// sum of digits of a number ` `static` `int` `digitSum(``int` `n) ` `{ ` `    ``int` `sum = 0; ` `    ``while` `(n > 0) ` `    ``{ ` `        ``sum += n % 10; ` `        ``n = n / 10; ` `    ``} ` `    ``return` `sum; ` `} ` ` `  `// Function to return the total pairs ` `// of elements with equal sum of digits ` `static` `int` `totalPairs(``int` `[]arr1, ``int` `[]arr2, ` `                      ``int` `n, ``int` `m) ` `{ ` ` `  `    ``// set is used to avoid duplicate pairs ` `    ``HashSet s = ``new` `HashSet(); ` ` `  `    ``for` `(``int` `i = 0; i < n; i++)  ` `    ``{ ` `        ``for` `(``int` `j = 0; j < m; j++)  ` `        ``{ ` ` `  `            ``// check sum of digits ` `            ``// of both the elements ` `            ``if` `(digitSum(arr1[i]) == digitSum(arr2[j])) ` `            ``{ ` `                ``if` `(arr1[i] < arr2[j]) ` `                    ``s.Add(``new` `pair(arr1[i], arr2[j])); ` `                ``else` `                    ``s.Add(``new` `pair(arr2[j], arr1[i])); ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// return size of the set ` `    ``return` `s.Count; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `[]arr1 = { 100, 3, 7, 50 }; ` `    ``int` `[]arr2 = { 5, 1, 10, 4 }; ` `    ``int` `n = arr1.Length; ` `    ``int` `m = arr2.Length; ` ` `  `    ``Console.WriteLine(totalPairs(arr1, arr2, n, m)); ` `} ` `} ` ` `  `// This code is contributed by Princi Singh `

Output:

```3
```

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