# Count of elements on the left which are divisible by current element | Set 2

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
Explanation:
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.

Naive Approach: This approach is already discuss here. But the complexity of this approach is O(N2).

Efficient Approach:

• We can say that if number A divides a number B then A is a factor of B. So, we need to find the number of previous elements whose factor is the current element.
• We will maintain a count array which contains the count of the factor of each element.
• Now, Iterate over the array and for each element

1. Make the answer of current element equal to count [ arr[i] ] and
2. Increment the frequency of each factor of arr[i] in count array.

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 increment the count ` `// for each factor of given val ` `void` `IncrementFactors(``int` `count[], ` `                      ``int` `val) ` `{ ` ` `  `    ``for` `(``int` `i = 1; i * i <= val; i++) { ` `        ``if` `(val % i == 0) { ` `            ``if` `(i == val / i) { ` `                ``count[i]++; ` `            ``} ` `            ``else` `{ ` `                ``count[i]++; ` `                ``count[val / i]++; ` `            ``} ` `        ``} ` `    ``} ` `} ` ` `  `// Function to generate and print ` `// the required array ` `void` `generateArr(``int` `A[], ``int` `n) ` `{ ` `    ``int` `B[n]; ` ` `  `    ``// Find max element of array ` `    ``int` `maxi = *max_element(A, A + n); ` ` `  `    ``// Create count array of maxi size ` `    ``int` `count[maxi + 1] = { 0 }; ` ` `  `    ``// For every element of the array ` `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``// Count[ A[i] ] denotes how many ` `        ``// previous elements are there whose ` `        ``// factor is the current element. ` `        ``B[i] = count[A[i]]; ` ` `  `        ``// Increment in count array for ` `        ``// factors of A[i] ` `        ``IncrementFactors(count, A[i]); ` `    ``} ` ` `  `    ``// Print the generated array ` `    ``printArr(B, n); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 8, 1, 28, 4, 2, 6, 7 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]); ` ` `  `    ``generateArr(arr, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.*; ` ` `  `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 increment the count ` `// for each factor of given val ` `static` `void` `IncrementFactors(``int` `count[], ` `                             ``int` `val) ` `{ ` `    ``for``(``int` `i = ``1``; i * i <= val; i++) ` `    ``{ ` `       ``if` `(val % i == ``0``) ` `       ``{ ` `           ``if` `(i == val / i) ` `           ``{ ` `               ``count[i]++; ` `           ``} ` `           ``else` `           ``{ ` `               ``count[i]++; ` `               ``count[val / i]++; ` `           ``} ` `       ``} ` `    ``} ` `} ` ` `  `// Function to generate and print ` `// the required array ` `static` `void` `generateArr(``int` `A[], ``int` `n) ` `{ ` `    ``int` `[]B = ``new` `int``[n]; ` ` `  `    ``// Find max element of array ` `    ``int` `maxi = Arrays.stream(A).max().getAsInt(); ` ` `  `    ``// Create count array of maxi size ` `    ``int` `count[] = ``new` `int``[maxi + ``1``]; ` ` `  `    ``// For every element of the array ` `    ``for``(``int` `i = ``0``; i < n; i++) ` `    ``{ ` `        `  `       ``// Count[ A[i] ] denotes how many ` `       ``// previous elements are there whose ` `       ``// factor is the current element. ` `       ``B[i] = count[A[i]]; ` `        `  `       ``// Increment in count array for ` `       ``// factors of A[i] ` `       ``IncrementFactors(count, A[i]); ` `    ``} ` ` `  `    ``// Print the generated array ` `    ``printArr(B, n); ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `arr[] = { ``8``, ``1``, ``28``, ``4``, ``2``, ``6``, ``7` `}; ` `    ``int` `n = arr.length; ` ` `  `    ``generateArr(arr, n); ` `} ` `} ` ` `  `// This code is contributed by Amit Katiyar `

## Python3

 `# Python3 implementation of the approach  ` ` `  `# Utility function to prthe ` `# elements of the array ` `def` `printArr(arr, n): ` `     `  `    ``for` `i ``in` `range``(n): ` `        ``print``(arr[i], end ``=` `" "``) ` ` `  `# Function to increment the count ` `# for each factor of given val ` `def` `IncrementFactors(count, val): ` ` `  `    ``i ``=` `1` `    ``while``(i ``*` `i <``=` `val): ` `        ``if` `(val ``%` `i ``=``=` `0``): ` `            ``if` `(i ``=``=` `val ``/``/` `i): ` `                ``count[i] ``+``=` `1` `             `  `            ``else``: ` `                ``count[i] ``+``=` `1` `                ``count[val ``/``/` `i] ``+``=` `1` `                 `  `        ``i ``+``=` `1` ` `  `# Function to generate and print ` `# the required array ` `def` `generateArr(A, n): ` `     `  `    ``B ``=` `[``0``] ``*` `n ` ` `  `    ``# Find max element of arr ` `    ``maxi ``=` `max``(A) ` ` `  `    ``# Create count array of maxi size ` `    ``count ``=` `[``0``] ``*` `(maxi ``+` `1``) ` ` `  `    ``# For every element of the array ` `    ``for` `i ``in` `range``(n): ` ` `  `        ``# Count[ A[i] ] denotes how many ` `        ``# previous elements are there whose ` `        ``# factor is the current element. ` `        ``B[i] ``=` `count[A[i]] ` ` `  `        ``# Increment in count array for ` `        ``# factors of A[i] ` `        ``IncrementFactors(count, A[i]) ` `     `  `    ``# Print the generated array ` `    ``printArr(B, n) ` ` `  `# Driver code ` `arr ``=` `[ ``8``, ``1``, ``28``, ``4``, ``2``, ``6``, ``7` `] ` `n ``=` `len``(arr) ` ` `  `generateArr(arr, n) ` ` `  `# This code is contributed by code_hunt `

## C#

 `// C# implementation of the approach ` `using` `System; ` `using` `System.Linq; ` ` `  `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 increment the count ` `// for each factor of given val ` `static` `void` `IncrementFactors(``int` `[]count, ` `                             ``int` `val) ` `{ ` `    ``for``(``int` `i = 1; i * i <= val; i++) ` `    ``{ ` `       ``if` `(val % i == 0) ` `       ``{ ` `           ``if` `(i == val / i) ` `           ``{ ` `               ``count[i]++; ` `           ``} ` `           ``else` `           ``{ ` `               ``count[i]++; ` `               ``count[val / i]++; ` `           ``} ` `       ``} ` `    ``} ` `} ` ` `  `// Function to generate and print ` `// the required array ` `static` `void` `generateArr(``int` `[]A, ``int` `n) ` `{ ` `    ``int` `[]B = ``new` `int``[n]; ` ` `  `    ``// Find max element of array ` `    ``int` `maxi = A.Max(); ` ` `  `    ``// Create count array of maxi size ` `    ``int` `[]count = ``new` `int``[maxi + 1]; ` ` `  `    ``// For every element of the array ` `    ``for``(``int` `i = 0; i < n; i++) ` `    ``{ ` `        `  `       ``// Count[ A[i] ] denotes how many ` `       ``// previous elements are there whose ` `       ``// factor is the current element. ` `       ``B[i] = count[A[i]]; ` `        `  `       ``// Increment in count array for ` `       ``// factors of A[i] ` `       ``IncrementFactors(count, A[i]); ` `    ``} ` ` `  `    ``// Print the generated array ` `    ``printArr(B, n); ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `[]arr = { 8, 1, 28, 4, 2, 6, 7 }; ` `    ``int` `n = arr.Length; ` ` `  `    ``generateArr(arr, n); ` `} ` `} ` ` `  `// This code is contributed by Amit Katiyar `

Output:

```0 1 0 2 3 0 1
```

Time Complexity: O(N * root(MaxElement))

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