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

• Last Updated : 03 Aug, 2022

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

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 elements 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`

## Javascript

 ``

Output:

`0 0 2`

Time Complexity: O(n2), Since there are two nested loops inside each other in the worst case if the first loop run for all n elements, then for each element the inner loop also runs n*n times.

Auxiliary Space: O(n), since there is an extra array involved thus it takes O(n) extra space.

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