Size of the largest divisible subset in an Array
Last Updated :
06 Feb, 2023
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 sets with a maximum size is {3, 6, 18}
Input : arr[] = {2, 3, 4, 8, 16}
Output : 4
Approach:
- Let's take all the numbers in increasing order.
- Note that set X = xi, ..., ?xk} is acceptable if xi divides xi+1 for (1 ? i ? k - 1).
- Therefore, dp[x] is equal to the length of the longest suitable increasing subsequence starting at the number x.
- 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 <bits/stdc++.h>
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 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
# 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# 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
<script>
// Javascript implementation of the above approach
let N = 1000005
// Function to find the size of the
//largest divisible subarray
function maximum_set(a, n) {
let dp = new Array(N).fill(0);
// Mark all elements of the array
for (let i = 0; i < n; i++)
dp[a[i]] = 1;
let ans = 1;
// Traverse reverse
for (let i = N - 1; i >= 1; i--) {
if (dp[i] != 0) {
// For all multiples of i
for (let 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
let arr = [3, 4, 6, 8, 10, 18, 21, 24];
let n = arr.length;
// Function call
document.write(maximum_set(arr, n));
</script>
Time Complexity: O(n*sqrt(n))
Auxiliary Space: O(n), where n is the size of the given array.
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