Maximum count of elements divisible on the left for any element

Given an array arr[] of N elements. The good value of an element arr[i] is the number of valid indices j<i such that arr[j] is divisible by arr[i].

Example:

Input: arr[] = {9, 6, 2, 3}
Output: 2
9 doesn’t has any element on its left.
6 doesn’t divide any element on its left.
2 divides 6.
3 divides 6 and 9.



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

Naive approach: For every element, find the count of numbers divisible by it on its left and print the maximum of these values.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the maximum count 
// of required elements 
int findMax(int arr[], int n)
    int res = 0;
    int i, j;
      
    // For every element in the array starting 
    // from the second element 
    for(i = 0; i < n ; i++)
    {
          
        // Check all the elements on the left 
        // of current element which are divisible 
        // by the current element 
        int count = 0;
        for(j = 0; j < i; j++)
        
            if (arr[j] % arr[i] == 0) 
                count += 1;
        }
        res = max(count, res);
    }
    return res;
}
  
// Driver code
int main()
{
    int arr[] = { 8, 1, 28, 4, 2, 6, 7 };
    int n = sizeof(arr) / sizeof(int);
  
    cout << findMax(arr, n);
  
    return 0;
}
  
// This code is contributed by Rajput-Ji

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Java

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// Java implementation of the approach 
class GFG 
{
  
// Function to return the maximum count 
// of required elements 
static int findMax(int arr[], int n)
    int res = 0;
    int i, j;
      
    // For every element in the array starting 
    // from the second element 
    for(i = 0; i < n ; i++)
    {
          
        // Check all the elements on the left 
        // of current element which are divisible 
        // by the current element 
        int count = 0;
        for(j = 0; j < i; j++)
        
            if (arr[j] % arr[i] == 0
                count += 1;
        }
        res = Math.max(count, res);
    }
    return res;
}
  
// Driver Code
public static void main (String[] args)
{
    int arr[] = {8, 1, 28, 4, 2, 6, 7};
    int n = arr.length; 
    System.out.println(findMax(arr, n));
}
}
  
// This code is contributed by AnkitRai01

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Python3

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# Python3 implementation of the approach
  
# Function to return the maximum count 
# of required elements
def findMax(arr, n):
    res = 0
      
    # For every element in the array starting 
    # from the second element
    for i in range(1, n):
          
        # Check all the elements on the left 
        # of current element which are divisible 
        # by the current element
        count = 0
        for j in range(0, i):
            if arr[j] % arr[i] == 0:
                count += 1
        res = max(count, res)
    return res
  
# Driver code
arr = [8, 1, 28, 4, 2, 6, 7]
n = len(arr)
print(findMax(arr, n))

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C#

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// C# implementation of the above approach 
using System;
      
class GFG 
{
  
// Function to return the maximum count 
// of required elements 
static int findMax(int []arr, int n)
    int res = 0;
    int i, j;
      
    // For every element in the array  
    // starting from the second element 
    for(i = 0; i < n ; i++)
    {
          
        // Check all the elements on the left 
        // of current element which are divisible 
        // by the current element 
        int count = 0;
        for(j = 0; j < i; j++)
        
            if (arr[j] % arr[i] == 0) 
                count += 1;
        }
        res = Math.Max(count, res);
    }
    return res;
}
  
// Driver Code
public static void Main (String[] args)
{
    int []arr = {8, 1, 28, 4, 2, 6, 7};
    int n = arr.Length; 
    Console.WriteLine(findMax(arr, n));
}
}
  
// This code is contributed by PrinciRaj1992

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Output:

3

Time Complexity: O(N2)

Efficient approach: It can be observed that for any element pair (arr[i], arr[j]) where i < j and (arr[i] % arr[j]) = 0, if the count of elements divisible by arr[i] on its left is X then the count of elements divisible by arr[j] on its left will definitely be greater than X as all the elements which are divisible by arr[i] will also be divisible by arr[j]. So, for every element which is divisible by any other element on its right, the count of elements on its left which are divisible by it doesn’t need to be calculated which will improve the time complexity of the overall program but it should be noted that for the input where no element is divisible by any other element (for example, when all the lements are prime), the worst-case time complexity would still be O(N2).

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the maximum count
// of required elements
int findMax(int arr[], int n)
{
  
    // divisible[i] will store true
    // if arr[i] is divisible by
    // any element on its right
    bool divisible[n] = { false };
  
    // To store the maximum required count
    int res = 0;
  
    // For every element of the array
    for (int i = n - 1; i > 0; i--) {
  
        // If the current element is
        // divisible by any element
        // on its right
        if (divisible[i])
            continue;
  
        // Find the count of element
        // on the left which are divisible
        // by the current element
        int cnt = 0;
        for (int j = 0; j < i; j++) {
  
            // If arr[j] is divisible then
            // set divisible[j] to true
            if ((arr[j] % arr[i]) == 0) {
                divisible[j] = true;
                cnt++;
            }
        }
  
        // Update the maximum required count
        res = max(res, cnt);
    }
  
    return res;
}
  
// Driver code
int main()
{
    int arr[] = { 8, 1, 28, 4, 2, 6, 7 };
    int n = sizeof(arr) / sizeof(int);
  
    cout << findMax(arr, n);
  
    return 0;
}

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Java

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// Java implementation of the approach
class GFG
{
  
// Function to return the maximum count
// of required elements
static int findMax(int arr[], int n)
{
  
    // divisible[i] will store true
    // if arr[i] is divisible by
    // any element on its right
    boolean []divisible = new boolean[n];
  
    // To store the maximum required count
    int res = 0;
  
    // For every element of the array
    for (int i = n - 1; i > 0; i--)
    {
  
        // If the current element is
        // divisible by any element
        // on its right
        if (divisible[i])
            continue;
  
        // Find the count of element
        // on the left which are divisible
        // by the current element
        int cnt = 0;
        for (int j = 0; j < i; j++) 
        {
  
            // If arr[j] is divisible then
            // set divisible[j] to true
            if ((arr[j] % arr[i]) == 0)
            {
                divisible[j] = true;
                cnt++;
            }
        }
  
        // Update the maximum required count
        res = Math.max(res, cnt);
    }
    return res;
}
  
// Driver code
public static void main(String[] args) 
{
    int arr[] = { 8, 1, 28, 4, 2, 6, 7 };
    int n = arr.length;
  
    System.out.println(findMax(arr, n));
}
}
  
// This code is contributed by Rajput-Ji

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Python3

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# Python3 implementation of the approach 
  
# Function to return the maximum count 
# of required elements 
def findMax(arr, n) : 
  
    # divisible[i] will store true 
    # if arr[i] is divisible by 
    # any element on its right 
    divisible = [ False ] * n; 
  
    # To store the maximum required count 
    res = 0
  
    # For every element of the array 
    for i in range(n - 1, -1, -1) : 
  
        # If the current element is 
        # divisible by any element 
        # on its right 
        if (divisible[i]) :
            continue
  
        # Find the count of element 
        # on the left which are divisible 
        # by the current element 
        cnt = 0
        for j in range(i) :
  
            # If arr[j] is divisible then 
            # set divisible[j] to true 
            if ((arr[j] % arr[i]) == 0) :
                divisible[j] = True
                cnt += 1
  
        # Update the maximum required count 
        res = max(res, cnt); 
  
    return res; 
  
# Driver code 
if __name__ == "__main__"
  
    arr = [ 8, 1, 28, 4, 2, 6, 7 ]; 
    n = len(arr); 
  
    print(findMax(arr, n)); 
  
# This code is contributed by kanugargng

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C#

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// C# implementation of the approach
using System;
  
class GFG
{
  
// Function to return the maximum count
// of required elements
static int findMax(int []arr, int n)
{
  
    // divisible[i] will store true
    // if arr[i] is divisible by
    // any element on its right
    bool []divisible = new bool[n];
  
    // To store the maximum required count
    int res = 0;
  
    // For every element of the array
    for (int i = n - 1; i > 0; i--)
    {
  
        // If the current element is
        // divisible by any element
        // on its right
        if (divisible[i])
            continue;
  
        // Find the count of element
        // on the left which are divisible
        // by the current element
        int cnt = 0;
        for (int j = 0; j < i; j++) 
        {
  
            // If arr[j] is divisible then
            // set divisible[j] to true
            if ((arr[j] % arr[i]) == 0)
            {
                divisible[j] = true;
                cnt++;
            }
        }
  
        // Update the maximum required count
        res = Math.Max(res, cnt);
    }
    return res;
}
  
// Driver code
public static void Main(String[] args) 
{
    int []arr = { 8, 1, 28, 4, 2, 6, 7 };
    int n = arr.Length;
  
    Console.WriteLine(findMax(arr, n));
}
}
  
// This code is contributed by Rajput-Ji

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Output:

3


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