# 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 : 4Explanation: The largest subset is 18, 1, 3, 6,In the subsequence, 3 is divisible by 1, 6 by 3 and 18 by 6.`

Brute Force Approach: The brute force approach generates all possible subsets of the input array using bit manipulation and checks if each subset contains only pairs of elements that are divisible. It keeps track of the largest subset that satisfies this condition and returns it as the result.

• Initialize maxSubset = 1
• For each subset S of the input array a:
•  Initialize subsetSize = 0 and validSubset = true
• For each pair of elements (a[i], a[j]) in S:
• If a[i] % a[j] != 0 and a[j] % a[i] != 0, set validSubset = false and break out of the loop.
•  Otherwise, continue to the next pair.
• If validSubset is still true, increment subsetSize by 1
• If subsetSize is greater than maxSubset, set maxSubset = subsetSize
• Return maxSubset as the result

## C++

 `#include ``using` `namespace` `std;` `int` `largestSubset(``int` `a[], ``int` `n)``{``    ``int` `maxSubset``        ``= 1; ``// variable to store the maximum subset size``             ``// found so far, initialized to 1``    ``for` `(``int` `i = 0; i < (1 << n);``         ``i++) { ``// loop through all possible subsets of the``                ``// input array``        ``int` `subsetSize``            ``= 0; ``// variable to store the size of the``                 ``// current subset being considered``        ``bool` `validSubset``            ``= ``true``; ``// variable to keep track of whether the``                    ``// current subset is valid or not``        ``for` `(``int` `j = 0; j < n;``             ``j++) { ``// loop through all elements in the``                    ``// input array``            ``if` `(i & (1 << j)) { ``// check if the j-th element``                                ``// is present in the current``                                ``// subset being considered``                ``for` `(``int` `k = j + 1; k < n;``                     ``k++) { ``// loop through all elements``                            ``// after the j-th element``                    ``if` `(i``                        ``& (1``                           ``<< k)) { ``// check if the k-th``                                    ``// element is present in``                                    ``// the current subset``                                    ``// being considered``                        ``if` `(a[j] % a[k] != 0``                            ``&& a[k] % a[j]``                                   ``!= 0) { ``// check if the``                                           ``// pair (a[j],``                                           ``// a[k]) is not``                                           ``// divisible``                            ``validSubset``                                ``= ``false``; ``// if the pair is``                                         ``// not divisible,``                                         ``// mark the current``                                         ``// subset as``                                         ``// invalid``                            ``break``; ``// break out of the inner``                                   ``// loop since we don't``                                   ``// need to check any more``                                   ``// pairs``                        ``}``                    ``}``                ``}``                ``if` `(validSubset) { ``// if the current subset``                                   ``// is still valid after``                                   ``// checking all pairs,``                                   ``// increment the subset``                                   ``// size``                    ``subsetSize++;``                ``}``                ``else` `{ ``// if the current subset is not``                       ``// valid, break out of the outer loop``                       ``// since we don't need to consider``                       ``// this subset anymore``                    ``break``;``                ``}``            ``}``        ``}``        ``maxSubset``            ``= max(maxSubset,``                  ``subsetSize); ``// update the maximum subset``                               ``// size found so far``    ``}``    ``return` `maxSubset; ``// return the maximum subset size``                      ``// found``}` `int` `main()``{``    ``int` `a[] = {10, 5, 3, 15, 20 }; ``// sample input array``    ``int` `n = ``sizeof``(a)``            ``/ ``sizeof``(a[0]); ``// size of the input array``    ``cout << largestSubset(a, n)``         ``<< endl; ``// call the function and print the result``    ``return` `0;``}`

## Java

 `import` `java.util.*;` `class` `Main {``    ``public` `static` `int` `largestSubset(``int``[] a, ``int` `n) {``        ``int` `maxSubset = ``1``; ``// variable to store the maximum subset size found so far, initialized to 1``        ``for` `(``int` `i = ``0``; i < (``1` `<< n); i++) { ``// loop through all possible subsets of the input array``            ``int` `subsetSize = ``0``; ``// variable to store the size of the current subset being considered``            ``boolean` `validSubset = ``true``; ``// variable to keep track of whether the current subset is valid or not``            ``for` `(``int` `j = ``0``; j < n; j++) { ``// loop through all elements in the input array``                ``if` `((i & (``1` `<< j)) != ``0``) { ``// check if the j-th element is present in the current subset being considered``                    ``for` `(``int` `k = j + ``1``; k < n; k++) { ``// loop through all elements after the j-th element``                        ``if` `((i & (``1` `<< k)) != ``0``) { ``// check if the k-th element is present in the current subset being considered``                            ``if` `(a[j] % a[k] != ``0` `&& a[k] % a[j] != ``0``) { ``// check if the pair (a[j], a[k]) is not divisible``                                ``validSubset = ``false``; ``// if the pair is not divisible, mark the current subset as invalid``                                ``break``; ``// break out of the inner loop since we don't need to check any more pairs``                            ``}``                        ``}``                    ``}``                    ``if` `(validSubset) { ``// if the current subset is still valid after checking all pairs, increment the subset size``                        ``subsetSize++;``                    ``} ``else` `{ ``// if the current subset is not valid, break out of the outer loop since we don't need to consider this subset anymore``                        ``break``;``                    ``}``                ``}``            ``}``            ``maxSubset = Math.max(maxSubset, subsetSize); ``// update the maximum subset size found so far``        ``}``        ``return` `maxSubset; ``// return the maximum subset size found``    ``}` `    ``public` `static` `void` `main(String[] args) {``        ``int``[] a = {``10``, ``5``, ``3``, ``15``, ``20``}; ``// sample input array``        ``int` `n = a.length; ``// size of the input array``        ``System.out.println(largestSubset(a, n)); ``// call the function and print the result``    ``}``}`

## Python3

 `# Implementation of largestSubset function in Python`  `def` `largestSubset(a, n):``    ``maxSubset ``=` `1`  `# variable to store the maximum subset size``    ``# found so far, initialized to 1``    ``for` `i ``in` `range``(``1` `<< n):  ``# loop through all possible subsets of the``                            ``# input array``        ``subsetSize ``=` `0`  `# variable to store the size of the``        ``# current subset being considered``        ``validSubset ``=` `True`  `# variable to keep track of whether the``        ``# current subset is valid or not``        ``for` `j ``in` `range``(n):  ``# loop through all elements in the``                           ``# input array``            ``if` `i & (``1` `<< j):  ``# check if the j-th element``                             ``# is present in the current``                             ``# subset being considered``                ``for` `k ``in` `range``(j ``+` `1``, n):  ``# loop through all elements``                                         ``# after the j-th element``                    ``if` `i & (``1` `<< k):  ``# check if the k-th``                                     ``# element is present in``                                     ``# the current subset``                                     ``# being considered``                        ``if` `a[j] ``%` `a[k] !``=` `0` `and` `a[k] ``%` `a[j] !``=` `0``:``                            ``# check if the pair (a[j], a[k]) is not``                            ``# divisible``                            ``validSubset ``=` `False`  `# if the pair is``                            ``# not divisible,``                            ``# mark the current``                            ``# subset as``                            ``# invalid``                            ``break`  `# break out of the inner``                            ``# loop since we don't``                            ``# need to check any more``                            ``# pairs``                ``if` `validSubset:  ``# if the current subset``                                 ``# is still valid after``                                 ``# checking all pairs,``                                 ``# increment the subset``                                 ``# size``                    ``subsetSize ``+``=` `1``                ``else``:  ``# if the current subset is not``                      ``# valid, break out of the outer loop``                      ``# since we don't need to consider``                      ``# this subset anymore``                    ``break``        ``maxSubset ``=` `max``(maxSubset, subsetSize)  ``# update the maximum subset``        ``# size found so far``    ``return` `maxSubset  ``# return the maximum subset size``    ``# found`  `# Sample input array``a ``=` `[``10``, ``5``, ``3``, ``15``, ``20``]``n ``=` `len``(a)  ``# size of the input array``print``(largestSubset(a, n))  ``# call the function and print the result`

## C#

 `using` `System;` `class` `GFG``{``    ``static` `int` `LargestSubset(``int``[] a, ``int` `n)``    ``{``        ``int` `maxSubset = 1; ``// variable to store the maximum subset size``                           ``// found so far, initialized to 1``        ``for` `(``int` `i = 0; i < (1 << n); i++)``        ``{``            ``// loop through all possible subsets of the input array``            ``int` `subsetSize = 0; ``// variable to store the size of the``                                ``// current subset being considered``            ``bool` `validSubset = ``true``; ``// variable to keep track of whether the``                                     ``// current subset is valid or not``            ``for` `(``int` `j = 0; j < n; j++)``            ``{``                ``// loop through all elements in the input array``                ``if` `((i & (1 << j)) != 0)``                ``{``                    ``// check if the j-th element is present in the current``                    ``// subset being considered``                    ``for` `(``int` `k = j + 1; k < n; k++)``                    ``{``                        ``// loop through all elements after the j-th element``                        ``if` `((i & (1 << k)) != 0)``                        ``{``                            ``// check if the k-th element is present in``                            ``// the current subset being considered``                            ``if` `(a[j] % a[k] != 0 && a[k] % a[j] != 0)``                            ``{``                                ``// check if the pair (a[j], a[k]) is not divisible``                                ``validSubset = ``false``; ``// if the pair is not divisible,``                                                     ``// mark the current subset as invalid``                                ``break``; ``// break out of the inner loop since we don't``                                       ``// need to check any more pairs``                            ``}``                        ``}``                    ``}``                    ``if` `(validSubset)``                    ``{``                        ``// if the current subset is still valid after``                        ``// checking all pairs, increment the subset size``                        ``subsetSize++;``                    ``}``                    ``else``                    ``{``                        ``// if the current subset is not valid, break out of the outer loop``                        ``// since we don't need to consider this subset anymore``                        ``break``;``                    ``}``                ``}``            ``}``            ``maxSubset = Math.Max(maxSubset, subsetSize); ``// update the maximum subset``                                                        ``// size found so far``        ``}``        ``return` `maxSubset; ``// return the maximum subset size found``    ``}` `    ``static` `void` `Main()``    ``{``        ``int``[] a = { 10, 5, 3, 15, 20 }; ``// sample input array``        ``int` `n = a.Length; ``// size of the input array``        ``Console.WriteLine(LargestSubset(a, n)); ``// call the function and print the result``    ``}``}`

## Javascript

 `function` `largestSubset(a) {``    ``let maxSubset = 1; ``// variable to store the maximum subset size found so far, initialized to 1` `    ``for` `(let i = 0; i < (1 << a.length); i++) { ``// loop through all possible subsets of the input array``        ``let subsetSize = 0; ``// variable to store the size of the current subset being considered``        ``let validSubset = ``true``; ``// variable to keep track of whether the current subset is valid or not` `        ``for` `(let j = 0; j < a.length; j++) { ``// loop through all elements in the input array``            ``if` `(i & (1 << j)) { ``// check if the j-th element is present in the current subset being considered``                ``for` `(let k = j + 1; k < a.length; k++) { ``// loop through all elements after the j-th element``                    ``if` `(i & (1 << k)) { ``// check if the k-th element is present in the current subset being considered``                        ``if` `(a[j] % a[k] !== 0 && a[k] % a[j] !== 0) { ``// check if the pair (a[j], a[k]) is not divisible``                            ``validSubset = ``false``; ``// if the pair is not divisible, mark the current subset as invalid``                            ``break``; ``// break out of the inner loop since we don't need to check any more pairs``                        ``}``                    ``}``                ``}``                ``if` `(validSubset) { ``// if the current subset is still valid after checking all pairs, increment the subset size``                    ``subsetSize++;``                ``} ``else` `{ ``// if the current subset is not valid, break out of the outer loop since we don't need to consider this subset anymore``                    ``break``;``                ``}``            ``}``        ``}``        ``maxSubset = Math.max(maxSubset, subsetSize); ``// update the maximum subset size found so far``    ``}` `    ``return` `maxSubset; ``// return the maximum subset size found``}` `const a = [10, 5, 3, 15, 20]; ``// sample input array``console.log(largestSubset(a)); ``// call the function and print the result`

Output
```3

```

Time Complexity: O(2n*n2)
Space Complexity:  O(1)

Dynamic programming Approach: This can be solved using Dynamic Programming. We traverse the sorted array from the 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 array using values from dp[i+1] to dp[n-1]. Finally, we return the 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)``{``    ``// 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 || a[i] % a[j] == 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[0]);``    ``cout << largestSubset(a, n) << endl;``    ``return` `0;``}`

## Java

 `import` `java.util.Arrays;` `// Java program to find the largest``// subset which was each pair``// is divisible.``class` `GFG {` `    ``// function to find the longest Subsequence``    ``static` `int` `largestSubset(``int``[] a, ``int` `n)``    ``{``        ``// 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` `|| a[i] % a[j] == ``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):``    ` `    ``# 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` `or` `a[i] ``%` `a[j] ``=``=` `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)``    ``{``        ``// 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 | a[i] % a[j] == 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.`

## Javascript

 ``

## 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 ``or` `\$a``[``\$i``] % ``\$a``[``\$j``] == 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(n2)
Space Complexity: O(n)

Largest divisible pairs subset in c:

Approach

1. Create an array dp of size n to store the size of the largest divisible subset ending with each element of the array arr.

2. Initialize all elements of the dp array to 1, as every element is a divisible subset of itself.

3. For every element arr[i] in the array, check all previous elements arr[j] (where j < i) to see if arr[i] is divisible by arr[j] or arr[j] is divisible by arr[i].

4. If arr[i] is divisible by arr[j] or arr[j] is divisible by arr[i], then we can extend the divisible subset ending with arr[j] by including the arr[i] element as well. We will choose the arr[j] element which gives the largest divisible subset ending with arr[j].

5. Update the dp array with the size of the largest divisible subset ending with arr[i].

6. Keep track of the maximum element in the dp array, which gives the largest divisible subset among all elements.

7. Return the maximum element in the dp array.

8. In the main() function, create an array arr of integers and call the largest_divisible_pairs_subset() function with the array and its size.

9. Print the result returned by the largest_divisible_pairs_subset() function, which gives the size of the largest divisible subset.

## C++

 `#include ``using` `namespace` `std;` `int` `largest_divisible_pairs_subset(``int` `arr[], ``int` `n) {``    ``vector<``int``> dp(n, 1);``    ``int` `max_dp = 1;` `    ``// Compute dp array``    ``for` `(``int` `i = 1; i < n; i++) {``        ``for` `(``int` `j = 0; j < i; j++) {``            ``if` `(arr[i] % arr[j] == 0 || arr[j] % arr[i] == 0) {``                ``dp[i] = max(dp[j] + 1, dp[i]);``            ``}``        ``}``        ``max_dp = max(dp[i], max_dp);``    ``}` `    ``return` `max_dp;``}` `int` `main() {``    ``int` `arr[] = {3, 5, 10, 20, 21, 33};``    ``int` `n = ``sizeof``(arr)/``sizeof``(arr[0]);``    ``cout << largest_divisible_pairs_subset(arr, n);``    ``return` `0;``}`

## C

 `#include ``#include ` `int` `largest_divisible_pairs_subset(``int` `arr[], ``int` `n) {``    ``int` `dp[n];``    ``int` `i, j, max_dp = 1;``    ` `    ``// Initialize dp array``    ``for` `(i = 0; i < n; i++) {``        ``dp[i] = 1;``    ``}``    ` `    ``// Compute dp array``    ``for` `(i = 1; i < n; i++) {``        ``for` `(j = 0; j < i; j++) {``            ``if` `(arr[i] % arr[j] == 0 || arr[j] % arr[i] == 0) {``                ``dp[i] = dp[j] + 1 > dp[i] ? dp[j] + 1 : dp[i];``            ``}``        ``}``        ``max_dp = dp[i] > max_dp ? dp[i] : max_dp;``    ``}``    ` `    ``return` `max_dp;``}` `int` `main() {``    ``int` `arr[] = {3, 5, 10, 20, 21, 33};``    ``int` `n = ``sizeof``(arr)/``sizeof``(arr[0]);``    ``printf``(``"%d"``, largest_divisible_pairs_subset(arr, n));``    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.util.*;` `public` `class` `Main {` `    ``// Function to find the largest divisible``    ``// subset in a given array``    ``public` `static` `int` `largestDivisiblePairsSubset(``int` `arr[],``                                                  ``int` `n)``    ``{``        ``// Stores the recurring state``        ``int``[] dp = ``new` `int``[n];``        ``Arrays.fill(dp, ``1``);``        ``int` `maxDp = ``1``;` `        ``// Compute dp array``        ``for` `(``int` `i = ``1``; i < n; i++) {``            ``for` `(``int` `j = ``0``; j < i; j++) {``                ``if` `(arr[i] % arr[j] == ``0``                    ``|| arr[j] % arr[i] == ``0``) {``                    ``dp[i] = Math.max(dp[j] + ``1``, dp[i]);``                ``}``            ``}``            ``maxDp = Math.max(dp[i], maxDp);``        ``}` `        ``// Return the maximum value``        ``return` `maxDp;``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr[] = { ``3``, ``5``, ``10``, ``20``, ``21``, ``33` `};``        ``int` `n = arr.length;` `        ``System.out.println(``            ``largestDivisiblePairsSubset(arr, n));``    ``}``}`

## Python3

 `import` `sys` `def` `largest_divisible_pairs_subset(arr, n):``    ``dp ``=` `[``1``] ``*` `n``    ``max_dp ``=` `1` `    ``# Compute dp array``    ``for` `i ``in` `range``(``1``, n):``        ``for` `j ``in` `range``(i):``            ``if` `arr[i] ``%` `arr[j] ``=``=` `0` `or` `arr[j] ``%` `arr[i] ``=``=` `0``:``                ``dp[i] ``=` `max``(dp[j] ``+` `1``, dp[i])``        ``max_dp ``=` `max``(dp[i], max_dp)` `    ``return` `max_dp` `if` `__name__ ``=``=` `'__main__'``:``    ``arr ``=` `[``3``, ``5``, ``10``, ``20``, ``21``, ``33``]``    ``n ``=` `len``(arr)``    ``print``(largest_divisible_pairs_subset(arr, n))` `# This code is contributed by shivhack9999`

## C#

 `using` `System;` `class` `GFG {``    ``static` `int` `largest_divisible_pairs_subset(``int``[] arr,``                                              ``int` `n)``    ``{``        ``// Initialize dp array``        ``int``[] dp = ``new` `int``[n];``        ``for` `(``int` `i = 0; i < n; i++)``            ``dp[i] = 1;``        ``int` `max_dp = 1;` `        ``// Compute dp array``        ``for` `(``int` `i = 1; i < n; i++) {``            ``for` `(``int` `j = 0; j < i; j++) {``                ``if` `(arr[i] % arr[j] == 0``                    ``|| arr[j] % arr[i] == 0) {``                    ``dp[i] = Math.Max(dp[j] + 1, dp[i]);``                ``}``            ``}``            ``max_dp = Math.Max(dp[i], max_dp);``        ``}` `        ``return` `max_dp;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``int``[] arr = { 3, 5, 10, 20, 21, 33 };``        ``int` `n = arr.Length;``        ``Console.WriteLine(``            ``largest_divisible_pairs_subset(arr, n));``    ``}``}`

## Javascript

 `function` `largest_divisible_pairs_subset(arr, n) {``    ``const dp = ``new` `Array(n).fill(1);``    ``let max_dp = 1;` `    ``// Compute dp array``    ``for` `(let i = 1; i < n; i++) {``        ``for` `(let j = 0; j < i; j++) {``            ``if` `(arr[i] % arr[j] === 0 || arr[j] % arr[i] === 0) {``                ``dp[i] = Math.max(dp[j] + 1, dp[i]);``            ``}``        ``}``        ``max_dp = Math.max(dp[i], max_dp);``    ``}` `    ``return` `max_dp;``}` `const arr = [3, 5, 10, 20, 21, 33];``const n = arr.length;``console.log(largest_divisible_pairs_subset(arr, n));`

Output
```3

```

Time Complexity:
The time complexity of the above algorithm is O(n^2), as we need to check all previous elements for each element.

Auxiliary Space:
The space complexity of the above algorithm is O(n), as we are only using an array of size n to store the largest divisible subset ending with each element.

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