Maximum number of multiples in an array before any element

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Given an array arr[], the task is to find the maximum number of indices j < i such that (arr[j] % arr[i]) = 0 among all the array elements.

Example:

Input: arr[] = {8, 1, 28, 4, 2, 6, 7}
Output: 3
No of multiples for each element before itself –
N(8) = 0 ()
N(1) = 1 (8)
N(28) = 0 ()
N(4) = 2 (28, 8)
N(2) = 3 (4, 28, 8)
N(6) = 0 ()
N(7) = 1 (28)
Maximum out of these multiples is – 3

Input: arr[] = {8, 12, 56, 32, 10, 3, 2, 4}
Output: 5

Approach:

Use a map to store all the divisors of each array element.
Generate all the divisors of an element in sqrt(n) time using the approach discussed in this article.
Now, take the maximum of all the stored divisors for each element and update it.

Below is the implementation of the above approach:

C++14

// C++ implementation of the approach
 
#include <bits/stdc++.h>
using namespace std;
 
const int MAX = 100000;
 
// Map to store the divisor count
int divisors[MAX];
 
// Function to generate the divisors
// of all the array elements
int generateDivisors(int n)
{
    for (int i = 1; i <= sqrt(n); i++) {
        if (n % i == 0) {
            if (n / i == i) {
                divisors[i]++;
            }
            else {
                divisors[i]++;
                divisors[n / i]++;
            }
        }
    }
}
 
// Function to find the maximum number
// of multiples in an array before it
int findMaxMultiples(int* arr, int n)
{
    // To store the maximum divisor count
    int ans = 0;
 
    for (int i = 0; i < n; i++) {
 
        // Update ans if more number
        // of divisors are found
        ans = max(divisors[arr[i]], ans);
 
        // Generating all the divisors of
        // the next element of the array
        generateDivisors(arr[i]);
    }
    return ans;
}
 
// Driver code
int main()
{
    int arr[] = { 8, 1, 28, 4, 2, 6, 7 };
    int n = sizeof(arr) / sizeof(int);
 
    cout << findMaxMultiples(arr, n);
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
 
class GFG
{
 
static int MAX = 100000;
 
// Map to store the divisor count
static int []divisors = new int[MAX];
 
// Function to generate the divisors
// of all the array elements
static void generateDivisors(int n)
{
    for (int i = 1; i <= Math.sqrt(n); i++) 
    {
        if (n % i == 0) 
        {
            if (n / i == i)
            {
                divisors[i]++;
            }
            else
            {
                divisors[i]++;
                divisors[n / i]++;
            }
        }
    }
}
 
// Function to find the maximum number
// of multiples in an array before it
static int findMaxMultiples(int []arr, int n)
{
    // To store the maximum divisor count
    int ans = 0;
 
    for (int i = 0; i < n; i++)
    {
 
        // Update ans if more number
        // of divisors are found
        ans = Math.max(divisors[arr[i]], ans);
 
        // Generating all the divisors of
        // the next element of the array
        generateDivisors(arr[i]);
    }
    return ans;
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 8, 1, 28, 4, 2, 6, 7 };
    int n = arr.length;
 
    System.out.print(findMaxMultiples(arr, n));
}
}
 
// This code is contributed by PrinciRaj1992

Python3

# Python3 implementation of the approach
from math import ceil,sqrt
MAX = 100000
 
# Map to store the divisor count
divisors = [0] * MAX
 
# Function to generate the divisors
# of all the array elements
def generateDivisors(n):
    for i in range(1,ceil(sqrt(n)) + 1):
        if (n % i == 0):
            if (n // i == i):
                divisors[i]+=1
            else:
                divisors[i] += 1
                divisors[n // i] += 1
 
# Function to find the maximum number
# of multiples in an array before it
def findMaxMultiples(arr, n):
     
    # To store the maximum divisor count
    ans = 0
    for i in range(n):
 
        # Update ans if more number
        # of divisors are found
        ans = max(divisors[arr[i]], ans)
 
        # Generating all the divisors of
        # the next element of the array
        generateDivisors(arr[i])
    return ans
 
# Driver code
arr = [8, 1, 28, 4, 2, 6, 7]
n = len(arr)
 
print(findMaxMultiples(arr, n))
 
# This code is contributed by mohit kumar 29

C#

// C# implementation of the approach
using System;
 
class GFG
{
 
static int MAX = 100000;
 
// Map to store the divisor count
static int []divisors = new int[MAX];
 
// Function to generate the divisors
// of all the array elements
static void generateDivisors(int n)
{
    for (int i = 1; i <= Math.Sqrt(n); i++) 
    {
        if (n % i == 0) 
        {
            if (n / i == i)
            {
                divisors[i]++;
            }
            else
            {
                divisors[i]++;
                divisors[n / i]++;
            }
        }
    }
}
 
// Function to find the maximum number
// of multiples in an array before it
static int findMaxMultiples(int []arr, int n)
{
    // To store the maximum divisor count
    int ans = 0;
 
    for (int i = 0; i < n; i++)
    {
 
        // Update ans if more number
        // of divisors are found
        ans = Math.Max(divisors[arr[i]], ans);
 
        // Generating all the divisors of
        // the next element of the array
        generateDivisors(arr[i]);
    }
    return ans;
}
 
// Driver code
public static void Main(String[] args)
{
    int []arr = { 8, 1, 28, 4, 2, 6, 7 };
    int n = arr.Length;
 
    Console.Write(findMaxMultiples(arr, n));
}
}
 
// This code is contributed by 29AjayKumar

Javascript

<script>
 
// JavaScript implementation of the approach
const MAX = 100000;
 
// Map to store the divisor count
var divisors = new Array(MAX).fill(0);
 
// Function to generate the divisors
// of all the array elements
function generateDivisors(n) 
{
    for(var i = 1; i <= Math.sqrt(n); i++) 
    {
        if (n % i == 0) 
        {
            if (n / i == i)
            {
                divisors[i]++;
            } 
            else
            {
                divisors[i]++;
                divisors[n / i]++;
            }
        }
    }
}
 
// Function to find the maximum number
// of multiples in an array before it
function findMaxMultiples(arr, n) 
{
     
    // To store the maximum divisor count
    var ans = 0;
     
    for(var i = 0; i < n; i++)
    {
         
        // Update ans if more number
        // of divisors are found
        ans = Math.max(divisors[arr[i]], ans);
         
        // Generating all the divisors of
        // the next element of the array
        generateDivisors(arr[i]);
    }
    return ans;
}
 
// Driver code
var arr = [ 8, 1, 28, 4, 2, 6, 7 ];
var n = arr.length;
 
document.write(findMaxMultiples(arr, n));
 
// This code is contributed by rdtank
 
</script>

Output:

Time Complexity: O(N*sqrt(val)), where N is the length of the array and val is the maximum value of the array elements.

Auxiliary Space: O(100000), as we are using extra space.

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