# Largest divisible pairs subset

Given an array of n distinct elements, find length of the largest subset such that every pair in the subset is such that the larger element of the pair is divisible by smaller element.

Examples:

```Input : arr[] = {10, 5, 3, 15, 20}
Output : 3
Explanation: The largest subset is 10, 5, 20.
10 is divisible by 5, and 20 is divisible by 10.

Input : arr[] = {18, 1, 3, 6, 13, 17}
Output : 4
Explanation: The largest subset is 18, 1, 3, 6,
In the subsequence, 3 is divisible by 1,
6 by 3 and 18 by 6.
```

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

This can be solved using Dynamic Programming. We first sort the array so that the largest element is at the end. Then we traverse the sorted array from end. For every element a[i], we compute dp[i] where dp[i] indicates size of largest divisible subset where a[i] is the smallest element. We can compute dp[i] in a sorted array using values from dp[i+1] to dp[n-1]. Finally we return maximum value from dp[].

Below is the implementation of the above approach:

## C++

 `// CPP program to find the largest subset which ` `// where each pair is divisible. ` `#include ` `using` `namespace` `std; ` ` `  `// function to find the longest Subsequence ` `int` `largestSubset(``int` `a[], ``int` `n) ` `{ ` `    ``// Sort array in increasing order ` `    ``sort(a, a + n); ` ` `  `    ``// dp[i] is going to store size of largest ` `    ``// divisible subset beginning with a[i]. ` `    ``int` `dp[n]; ` ` `  `    ``// Since last element is largest, d[n-1] is 1 ` `    ``dp[n - 1] = 1; ` ` `  `    ``// Fill values for smaller elements. ` `    ``for` `(``int` `i = n - 2; i >= 0; i--) { ` ` `  `        ``// Find all multiples of a[i] and consider ` `        ``// the multiple that has largest subset ` `        ``// beginning with it. ` `        ``int` `mxm = 0; ` `        ``for` `(``int` `j = i + 1; j < n; j++) ` `            ``if` `(a[j] % a[i] == 0) ` `                ``mxm = max(mxm, dp[j]); ` ` `  `        ``dp[i] = 1 + mxm; ` `    ``} ` ` `  `    ``// Return maximum value from dp[] ` `    ``return` `*max_element(dp, dp + n); ` `} ` ` `  `// driver code to check the above function ` `int` `main() ` `{ ` `    ``int` `a[] = { 1, 3, 6, 13, 17, 18 }; ` `    ``int` `n = ``sizeof``(a) / ``sizeof``(a); ` `    ``cout << largestSubset(a, n) << endl; ` `    ``return` `0; ` `} `

## Java

 `import` `java.util.Arrays; ` ` `  `// Java program to find the largest ` `// subset which where each pair ` `// is divisible. ` `class` `GFG { ` ` `  `    ``// function to find the longest Subsequence ` `    ``static` `int` `largestSubset(``int``[] a, ``int` `n) ` `    ``{ ` `        ``// Sort array in increasing order ` `        ``Arrays.sort(a); ` ` `  `        ``// dp[i] is going to store size of largest ` `        ``// divisible subset beginning with a[i]. ` `        ``int``[] dp = ``new` `int``[n]; ` ` `  `        ``// Since last element is largest, d[n-1] is 1 ` `        ``dp[n - ``1``] = ``1``; ` ` `  `        ``// Fill values for smaller elements. ` `        ``for` `(``int` `i = n - ``2``; i >= ``0``; i--) { ` ` `  `            ``// Find all multiples of a[i] and consider ` `            ``// the multiple that has largest subset ` `            ``// beginning with it. ` `            ``int` `mxm = ``0``; ` `            ``for` `(``int` `j = i + ``1``; j < n; j++) { ` `                ``if` `(a[j] % a[i] == ``0``) { ` `                    ``mxm = Math.max(mxm, dp[j]); ` `                ``} ` `            ``} ` ` `  `            ``dp[i] = ``1` `+ mxm; ` `        ``} ` ` `  `        ``// Return maximum value from dp[] ` `        ``return` `Arrays.stream(dp).max().getAsInt(); ` `    ``} ` ` `  `    ``// driver code to check the above function ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int``[] a = { ``1``, ``3``, ``6``, ``13``, ``17``, ``18` `}; ` `        ``int` `n = a.length; ` `        ``System.out.println(largestSubset(a, n)); ` `    ``} ` `} ` ` `  `/* This JAVA code is contributed by Rajput-Ji*/`

## Python3

 `# Python program to find the largest  ` `# subset where each pair is divisible. ` ` `  `# function to find the longest Subsequence ` `def` `largestSubset(a, n): ` `     `  `    ``# Sort array in ` `    ``# increasing order ` `    ``a.sort() ` `     `  `    ``# dp[i] is going to store size  ` `    ``# of largest divisible subset  ` `    ``# beginning with a[i]. ` `    ``dp ``=` `[``0` `for` `i ``in` `range``(n)] ` `     `  `    ``# Since last element is largest, ` `    ``# d[n-1] is 1 ` `    ``dp[n ``-` `1``] ``=` `1``;  ` ` `  `    ``# Fill values for smaller elements ` `    ``for` `i ``in` `range``(n ``-` `2``, ``-``1``, ``-``1``): ` `         `  `        ``# Find all multiples of a[i]  ` `        ``# and consider the multiple  ` `        ``# that has largest subset      ` `        ``# beginning with it.  ` `        ``mxm ``=` `0``; ` `        ``for` `j ``in` `range``(i ``+` `1``, n): ` `            ``if` `a[j] ``%` `a[i] ``=``=` `0``: ` `                ``mxm ``=` `max``(mxm, dp[j]) ` `        ``dp[i] ``=` `1` `+` `mxm ` `         `  `    ``# Return maximum value from dp[]  ` `    ``return` `max``(dp) ` ` `  `# Driver Code ` `a ``=` `[ ``1``, ``3``, ``6``, ``13``, ``17``, ``18` `] ` `n ``=` `len``(a) ` `print``(largestSubset(a, n)) ` ` `  `# This code is contributed by ` `# sahil shelangia `

## C#

 `// C# program to find the largest ` `// subset which where each pair ` `// is divisible. ` `using` `System; ` `using` `System.Linq; ` ` `  `public` `class` `GFG { ` ` `  `    ``// function to find the longest Subsequence ` `    ``static` `int` `largestSubset(``int``[] a, ``int` `n) ` `    ``{ ` `        ``// Sort array in increasing order ` `        ``Array.Sort(a); ` ` `  `        ``// dp[i] is going to store size of largest ` `        ``// divisible subset beginning with a[i]. ` `        ``int``[] dp = ``new` `int``[n]; ` ` `  `        ``// Since last element is largest, d[n-1] is 1 ` `        ``dp[n - 1] = 1; ` ` `  `        ``// Fill values for smaller elements. ` `        ``for` `(``int` `i = n - 2; i >= 0; i--) { ` ` `  `            ``// Find all multiples of a[i] and consider ` `            ``// the multiple that has largest subset ` `            ``// beginning with it. ` `            ``int` `mxm = 0; ` `            ``for` `(``int` `j = i + 1; j < n; j++) ` `                ``if` `(a[j] % a[i] == 0) ` `                    ``mxm = Math.Max(mxm, dp[j]); ` ` `  `            ``dp[i] = 1 + mxm; ` `        ``} ` ` `  `        ``// Return maximum value from dp[] ` `        ``return` `dp.Max(); ` `    ``} ` ` `  `    ``// driver code to check the above function ` `    ``static` `public` `void` `Main() ` `    ``{ ` `        ``int``[] a = { 1, 3, 6, 13, 17, 18 }; ` `        ``int` `n = a.Length; ` `        ``Console.WriteLine(largestSubset(a, n)); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

## PHP

 `= 0; ``\$i``--)  ` `    ``{ ` ` `  `        ``// Find all multiples of ` `        ``// a[i] and consider ` `        ``// the multiple that  ` `        ``// has largest subset  ` `        ``// beginning with it. ` `        ``\$mxm` `= 0; ` `        ``for` `(``\$j` `= ``\$i` `+ 1; ``\$j` `< ``\$n``; ``\$j``++) ` `            ``if` `(``\$a``[``\$j``] % ``\$a``[``\$i``] == 0) ` `                ``\$mxm` `= max(``\$mxm``, ``\$dp``[``\$j``]); ` ` `  `        ``\$dp``[``\$i``] = 1 + ``\$mxm``; ` `    ``} ` ` `  `    ``// Return maximum value ` `    ``// from dp[] ` `    ``return` `max(``\$dp``); ` `} ` ` `  `    ``// Driver Code ` `    ``\$a` `= ``array``(1, 3, 6, 13, 17, 18); ` `    ``\$n` `= ``count``(``\$a``); ` `    ``echo` `largestSubset(``\$a``, ``\$n``); ` `     `  `// This code is contributed by anuj_67. ` `?> `

Output:

```4
```

Time Complexity: O(n*n)