# Count of Array elements divisible by either their product or sum of digits

Given an array arr[]. The task is to count the elements in the array that are divisible by either their product of digits or the sum of digits.

Example: Input: arr[] = {123, 25, 36, 7}
Output: 2
Explanation: Following are the elements that follows the given conditions
Sum of digits of 123 is 6 and the product of digits is also 6.
Since 123 is not divisible by 6, so it is not counted.
The sum of digits of 36 is 9 and productof digits is 18. Since 36 is divisible by both, so it is considered in answer.
Similarly, sum of digits and product of digits of 7 is 7 itself, hence it is also considered.
Hence, 2 is the final answer.

Input: arr[] = {10, 22, 15}
Output: 2
Explanation:  The sum of digits in 10 is 1, and 10 is divisible by 1, hence we consider it. The product of digits of 15 is 5, and 15 is divisible by 5, we consider it in our output.

Approach: This problem is observation-based. Follow the steps below to solve the given problem.

• Declare a variable say count = 0, to store the number of elements satisfying given conditions.
• For every number in the array arr[], find the sum of its digits and the product of its digits.
• Check whether the number is divisible by either its sum of digits or the product of digits.
• Return the count as the final answer.

Below is the implementation of the above approach.

## C++

 `#include ``using` `namespace` `std;` `int` `solve(``int` `arr[], ``int` `n)``{` `    ``    ``    ``int` `count = 0;` `    ``    ``for` `(``int` `i = 0; i < n; i++) {``        ``int` `p = arr[i];``        ``int` `sum = 0, prod = 1;``        ``while` `(p > 0) {``            ``int` `r = p % 10;``            ``sum += r;``            ``prod *= r;``            ``p /= 10;``        ``}` `        ``        ``if` `((sum > 0 && arr[i] % sum == 0)``            ``|| (prod > 0 && arr[i] % prod == 0))``            ``count++;``    ``}` `    ``    ``return` `count;``}` `int` `main()``{``    ``int` `arr[] = { 123, 25, 36, 7 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);` `    ``    ``cout << solve(arr, N) << endl;` `    ``return` `0;``}`

## Python3

 `def` `solve(arr, n):` `        ``        ``    ``count ``=` `0` `    ``    ``for` `i ``in` `range``(``0``, n):``        ``p ``=` `arr[i]``        ``sum` `=` `0``        ``prod ``=` `1``        ``while` `(p > ``0``):``            ``r ``=` `p ``%` `10``            ``sum` `+``=` `r``            ``prod ``*``=` `r``            ``p ``/``/``=` `10` `            ``        ``if` `((``sum` `> ``0` `and` `arr[i] ``%` `sum` `=``=` `0``)``                ``or` `(prod > ``0` `and` `arr[i] ``%` `prod ``=``=` `0``)):``            ``count ``+``=` `1` `        ``    ``return` `count` `if` `__name__ ``=``=` `"__main__"``:` `    ``arr ``=` `[``123``, ``25``, ``36``, ``7``]``    ``N ``=` `len``(arr)` `    ``    ``print``(solve(arr, N))` `    `

Time Complexity: O(N)
Auxiliary Space: O(1)

## What do you think?

### Minimize operations to convert (0, 0) to (N, M) by incrementing either or both by K 