# Maximum number of multiples in an array before any element

• Difficulty Level : Hard
• Last Updated : 25 May, 2022

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:
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:

Approach:

1. Use a map to store all the divisors of each array element.
2. Generate all the divisors of an element in sqrt(n) time using the approach discussed in this article.
3. 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 ``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

 ``

Output:

`3`

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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