# Count of elements on the left which are divisible by current element

Given an array A[] of N integers, the task is to generate an array B[] such that B[i] contains the count of indices j in A[] such that j < i and A[j] % A[i] = 0

Examples:

Input: arr[] = {3, 5, 1}
Output: 0 0 2
3 and 5 do not divide any element on
their left but 1 divides 3 and 5.

Input: arr[] = {8, 1, 28, 4, 2, 6, 7}
Output: 0 1 0 2 3 0 1

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

Approach: Run a loop from the first element to the last element of the array and for the current element, run another loop for the elements on its left and check how many element on its left are divisible by it.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Utility function to print the ` `// elements of the array ` `void` `printArr(``int` `arr[], ``int` `n) ` `{ ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``cout << arr[i] << ``" "``; ` `} ` ` `  `// Function to generate and print ` `// the required array ` `void` `generateArr(``int` `A[], ``int` `n) ` `{ ` `    ``int` `B[n]; ` ` `  `    ``// For every element of the array ` `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``// To store the count of elements ` `        ``// on the left that the current ` `        ``// element divides ` `        ``int` `cnt = 0; ` `        ``for` `(``int` `j = 0; j < i; j++) { ` `            ``if` `(A[j] % A[i] == 0) ` `                ``cnt++; ` `        ``} ` `        ``B[i] = cnt; ` `    ``} ` ` `  `    ``// Print the generated array ` `    ``printArr(B, n); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `A[] = { 3, 5, 1 }; ` `    ``int` `n = ``sizeof``(A) / ``sizeof``(A); ` ` `  `    ``generateArr(A, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of above approach ` `class` `GFG ` `{ ` ` `  `// Utility function to print the ` `// elements of the array ` `static` `void` `printArr(``int` `arr[], ``int` `n) ` `{ ` `    ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``System.out.print(arr[i] + ``" "``); ` `} ` ` `  `// Function to generate and print ` `// the required array ` `static` `void` `generateArr(``int` `A[], ``int` `n) ` `{ ` `    ``int` `[]B = ``new` `int``[n]; ` ` `  `    ``// For every element of the array ` `    ``for` `(``int` `i = ``0``; i < n; i++)  ` `    ``{ ` ` `  `        ``// To store the count of elements ` `        ``// on the left that the current ` `        ``// element divides ` `        ``int` `cnt = ``0``; ` `        ``for` `(``int` `j = ``0``; j < i; j++) ` `        ``{ ` `            ``if` `(A[j] % A[i] == ``0``) ` `                ``cnt++; ` `        ``} ` `        ``B[i] = cnt; ` `    ``} ` ` `  `    ``// Print the generated array ` `    ``printArr(B, n); ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String args[]) ` `{ ` `    ``int` `A[] = { ``3``, ``5``, ``1` `}; ` `    ``int` `n = A.length; ` ` `  `    ``generateArr(A, n); ` `} ` `} ` ` `  `// This code is contributed by PrinciRaj1992 `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Utility function to print the ` `# elements of the array ` `def` `printArr(arr, n): ` ` `  `    ``for` `i ``in` `arr: ` `        ``print``(i, end ``=` `" "``) ` ` `  `# Function to generate and print ` `# the required array ` `def` `generateArr(A, n): ` `    ``B ``=` `[``0` `for` `i ``in` `range``(n)] ` ` `  `    ``# For every element of the array ` `    ``for` `i ``in` `range``(n): ` ` `  `        ``# To store the count of elements ` `        ``# on the left that the current ` `        ``# element divides ` `        ``cnt ``=` `0` `        ``for` `j ``in` `range``(i): ` `            ``if` `(A[j] ``%` `A[i] ``=``=` `0``): ` `                ``cnt ``+``=` `1` ` `  `        ``B[i] ``=` `cnt ` ` `  `    ``# Print the generated array ` `    ``printArr(B, n) ` ` `  `# Driver code ` `A ``=` `[``3``, ``5``, ``1``] ` `n ``=` `len``(A) ` ` `  `generateArr(A, n) ` ` `  `# This code is contributed by Mohit Kumar  `

## C#

 `// C# implementation of above approach ` `using` `System; ` `     `  `class` `GFG ` `{ ` ` `  `// Utility function to print the ` `// elements of the array ` `static` `void` `printArr(``int` `[]arr, ``int` `n) ` `{ ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``Console.Write(arr[i] + ``" "``); ` `} ` ` `  `// Function to generate and print ` `// the required array ` `static` `void` `generateArr(``int` `[]A, ``int` `n) ` `{ ` `    ``int` `[]B = ``new` `int``[n]; ` ` `  `    ``// For every element of the array ` `    ``for` `(``int` `i = 0; i < n; i++)  ` `    ``{ ` ` `  `        ``// To store the count of elements ` `        ``// on the left that the current ` `        ``// element divides ` `        ``int` `cnt = 0; ` `        ``for` `(``int` `j = 0; j < i; j++) ` `        ``{ ` `            ``if` `(A[j] % A[i] == 0) ` `                ``cnt++; ` `        ``} ` `        ``B[i] = cnt; ` `    ``} ` ` `  `    ``// Print the generated array ` `    ``printArr(B, n); ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String []args) ` `{ ` `    ``int` `[]A = { 3, 5, 1 }; ` `    ``int` `n = A.Length; ` ` `  `    ``generateArr(A, n); ` `} ` `} ` ` `  `// This code is contributed by Rajput-Ji `

Output:

```0 0 2
```

GeeksforGeeks has prepared a complete interview preparation course with premium videos, theory, practice problems, TA support and many more features. Please refer Placement 100 for details

My Personal Notes arrow_drop_up If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.