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

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:

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
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[0]);
  
    generateArr(A, n);
  
    return 0;
}
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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 
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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
chevron_right

Output:
0 0 2



Check out this Author's contributed articles.

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.





Article Tags :
Practice Tags :