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 subarrayint 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 codeint 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 approachN = 1000005# Function to find the size of the# largest divisible subarraydef 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 codearr = [3, 4, 6, 8, 10, 18, 21, 24]n = len(arr)# Function callprint(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 |
Javascript
<script>// Javascript implementation of the above approachlet N = 1000005// Function to find the size of the//largest divisible subarrayfunction 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 codelet arr = [3, 4, 6, 8, 10, 18, 21, 24];let n = arr.length;// Function calldocument.write(maximum_set(arr, n));</script> |
Output:
3
Time Complexity: O(n*sqrt(n))
Auxiliary Space: O(n), where n is the size of the given array.
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