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

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  • Last Updated : 03 Aug, 2022
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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 <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;
}


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




<script>
// Javascript implementation of the approach
 
// Utility function to print the
// elements of the array
function printArr(arr, n)
{
    for (let i = 0; i < n; i++)
        document.write(arr[i] + " ");
}
 
// Function to generate and print
// the required array
function generateArr(A, n)
{
    let B = new Array(n);
 
    // For every element of the array
    for (let i = 0; i < n; i++) {
 
        // To store the count of elements
        // on the left that the current
        // element divides
        let cnt = 0;
        for (let j = 0; j < i; j++) {
            if (A[j] % A[i] == 0)
                cnt++;
        }
        B[i] = cnt;
    }
 
    // Print the generated array
    printArr(B, n);
}
 
// Driver code
    let A = [ 3, 5, 1 ];
    let n = A.length;
 
    generateArr(A, n);
 
</script>


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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