# Size of the largest divisible subset in an Array

Given an array arr[] of size N. The task is to find the size of the set of numbers from the given array such that each number divides another or is divisible by another.

Examples:

Input : arr[] = {3, 4, 6, 8, 10, 18, 21, 24}
Output : 3
One of the possible set with maximum size is {3, 6, 18}

Input : arr[] = {2, 3, 4, 8, 16}
Output : 5

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

1. Let’s take all the numbers in increasing order.
2. Note that set X = xi, …, ?xk} is acceptable iff xi divides xi+1 for (1 ≤ i ≤ k – 1).
3. Therefore, dp[x] is equal to the length of the longest suitable increasing subsequence starting in a number x.
4. DP Relation: dp[x] = max(dp[x], 1 + dp[y]) if x divides y.

Below is the implementation of the above approach:

## CPP

 `// C++ implementation of the above approach ` `#include ` `using` `namespace` `std; ` ` `  `#define N 1000005 ` ` `  `// Function to find the size of the  ` `//largest divisible subarray ` `int` `maximum_set(``int` `a[], ``int` `n) ` `{ ` `    ``int` `dp[N] = { 0 }; ` ` `  `    ``// Mark all elements of the array ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``dp[a[i]] = 1; ` ` `  `    ``int` `ans = 1; ` ` `  `    ``// Traverse reverse ` `    ``for` `(``int` `i = N - 1; i >= 1; i--) { ` ` `  `        ``if` `(dp[i] != 0) { ` `            ``// For all multiples of i ` `            ``for` `(``int` `j = 2 * i; j < N; j += i) { ` `                ``dp[i] = max(dp[i], 1 + dp[j]); ` `                ``ans = max(ans, dp[i]); ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// Return the required answer ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 3, 4, 6, 8, 10, 18, 21, 24 }; ` ` `  `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]); ` ` `  `    ``// Function call ` `    ``cout << maximum_set(arr, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the above approach  ` `class` `GFG ` `{ ` `     `  `    ``final` `static` `int` `N = ``1000005` `; ` `     `  `    ``// Function to find the size of the  ` `    ``//largest divisible subarray  ` `    ``static` `int` `maximum_set(``int` `a[], ``int` `n)  ` `    ``{  ` `        ``int` `dp[] = ``new` `int``[N] ;  ` `     `  `        ``// Mark all elements of the array  ` `        ``for` `(``int` `i = ``0``; i < n; i++)  ` `            ``dp[a[i]] = ``1``;  ` `     `  `        ``int` `ans = ``1``;  ` `     `  `        ``// Traverse reverse  ` `        ``for` `(``int` `i = N - ``1``; i >= ``1``; i--)  ` `        ``{  ` `     `  `            ``if` `(dp[i] != ``0``)  ` `            ``{  ` `                ``// For all multiples of i  ` `                ``for` `(``int` `j = ``2` `* i; j < N; j += i)  ` `                ``{  ` `                    ``dp[i] = Math.max(dp[i], ``1` `+ dp[j]);  ` `                    ``ans = Math.max(ans, dp[i]);  ` `                ``}  ` `            ``}  ` `        ``}  ` `     `  `        ``// Return the required answer  ` `        ``return` `ans;  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `main (String[] args) ` `    ``{  ` `        ``int` `arr[] = { ``3``, ``4``, ``6``, ``8``, ``10``, ``18``, ``21``, ``24` `};  ` `     `  `        ``int` `n = arr.length;  ` `     `  `        ``// Function call  ` `        ``System.out.println(maximum_set(arr, n));  ` `    ``}  ` `} ` ` `  `// This code is contributed by AnkitRai01 `

## Python

 `# Python3 implementation of the above approach ` ` `  `N ``=` `1000005` ` `  `# Function to find the size of the ` `# largest divisible subarray ` `def` `maximum_set(a, n): ` `    ``dp ``=` `[``0` `for` `i ``in` `range``(N)] ` ` `  `    ``# Mark all elements of the array ` `    ``for` `i ``in` `a: ` `        ``dp[i] ``=` `1` ` `  `    ``ans ``=` `1` ` `  `    ``# Traverse reverse ` `    ``for` `i ``in` `range``(N ``-` `1``, ``0``, ``-``1``): ` ` `  `        ``if` `(dp[i] !``=` `0``): ` `             `  `            ``# For all multiples of i ` `            ``for` `j ``in` `range``(``2` `*` `i, N, i): ` `                ``dp[i] ``=` `max``(dp[i], ``1` `+` `dp[j]) ` `                ``ans ``=` `max``(ans, dp[i]) ` ` `  `    ``# Return the required answer ` `    ``return` `ans ` ` `  `# Driver code ` ` `  `arr ``=` `[``3``, ``4``, ``6``, ``8``, ``10``, ``18``, ``21``, ``24``] ` ` `  `n ``=` `len``(arr) ` ` `  `# Function call ` `print``(maximum_set(arr, n)) ` ` `  `# This code is contributed by mohit kumar 29 `

## C#

 `// C# implementation of the above approach  ` `using` `System; ` ` `  `class` `GFG  ` `{  ` `    ``static` `int` `N = 1000005 ;  ` `     `  `    ``// Function to find the size of the  ` `    ``//largest divisible subarray  ` `    ``static` `int` `maximum_set(``int` `[]a, ``int` `n)  ` `    ``{  ` `        ``int` `[]dp = ``new` `int``[N] ;  ` `     `  `        ``// Mark all elements of the array  ` `        ``for` `(``int` `i = 0; i < n; i++)  ` `            ``dp[a[i]] = 1;  ` `     `  `        ``int` `ans = 1;  ` `     `  `        ``// Traverse reverse  ` `        ``for` `(``int` `i = N - 1; i >= 1; i--)  ` `        ``{  ` `     `  `            ``if` `(dp[i] != 0)  ` `            ``{  ` `                ``// For all multiples of i  ` `                ``for` `(``int` `j = 2 * i; j < N; j += i)  ` `                ``{  ` `                    ``dp[i] = Math.Max(dp[i], 1 + dp[j]);  ` `                    ``ans = Math.Max(ans, dp[i]);  ` `                ``}  ` `            ``}  ` `        ``}  ` `     `  `        ``// Return the required answer  ` `        ``return` `ans;  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `Main()  ` `    ``{  ` `        ``int` `[]arr = { 3, 4, 6, 8, 10, 18, 21, 24 };  ` `        ``int` `n = arr.Length;  ` `     `  `        ``// Function call  ` `        ``Console.WriteLine(maximum_set(arr, n));  ` `    ``}  ` `}  ` ` `  `// This code is contributed by AnkitRai01  `

Output:

```3
```

Time Complexity: O(n*sqrt(n))

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Improved By : mohit kumar 29, AnkitRai01