Given an array arr[] consisting of N distinct integers and an integer K, the task is to find the maximum size of a subset possible such that no element in the subset is K times any other element of the subset(i.e. no such pair {n, m} should be present in the subset such that either m = n * K or n = m * K).
Examples:
Input: arr[] = {2, 8, 6, 5, 3}, K = 2
Output: 4
Explanation:
Only possible pair existing in the array with an element being K( = 2) times the other is {6, 3}.
Hence, all possible subsets which does not contain both the elements of the pair {6, 3} together can be considered.
Therefore, the longest possible subset can be of length 4.Input: arr[] = {1, 4, 3, 2}, K = 3
output: 3
Approach:
Follow the steps below to solve the problem:
- Find the number of pairs possible such that one element is K times the other from the given array
- Sort the array in increasing order of elements.
- Traverse the array and store the frequenciindices of array elements in Map.
- Initialize an array visited to mark for every index, whether that element is included(0) or not(1) in the subset.
- Traverse the array again and for every index having vis[i] = 0, check if arr[i] * K is present in the Map or not. If found, then increase the count of pairs and set vis[mp[arr[i] * K]] = 1.
- Finally, print N – count of pairs as the answer.
Below is implementation of above approach:
C++
// C++ implementation of // the above aproach #include <bits/stdc++.h> #define ll long long using namespace std; // Function to find the maximum // size of the required subset int findMaxLen(vector<ll>& a, ll k) { // Size of the array int n = a.size(); // Sort the array sort(a.begin(), a.end()); // Stores which index is // included or excluded vector<bool> vis(n, 0); // Stores the indices of // array elements map<int, int> mp; for (int i = 0; i < n; i++) { mp[a[i]] = i; } // Count of pairs int c = 0; // Iterate through all // the element for (int i = 0; i < n; ++i) { // If element is included if (vis[i] == false) { int check = a[i] * k; // Check if a[i] * k is present // in the array or not if (mp.find(check) != mp.end()) { // Increase count of pair c++; // Exclude the pair vis[mp[check]] = true; } } } return n - c; } // Driver code int main() { int K = 3; vector<ll> arr = { 1, 4, 3, 2 }; cout << findMaxLen(arr, K); } |
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Java
// Java implementation of // the above aproach import java.util.*; class GFG{ // Function to find the maximum // size of the required subset static int findMaxLen(int[] a, int k) { // Size of the array int n = a.length; // Sort the array Arrays.sort(a); // Stores which index is // included or excluded boolean []vis = new boolean[n]; // Stores the indices of // array elements HashMap<Integer, Integer> mp = new HashMap<Integer, Integer>(); for(int i = 0; i < n; i++) { mp.put(a[i], i); } // Count of pairs int c = 0; // Iterate through all // the element for(int i = 0; i < n; ++i) { // If element is included if (vis[i] == false) { int check = a[i] * k; // Check if a[i] * k is present // in the array or not if (mp.containsKey(check)) { // Increase count of pair c++; // Exclude the pair vis[mp.get(check)] = true; } } } return n - c; } // Driver code public static void main(String[] args) { int K = 3; int []arr = { 1, 4, 3, 2 }; System.out.print(findMaxLen(arr, K)); } } // This code is contributed by amal kumar choubey |
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Python3
# Python3 implementation of # the above approach # Function to find the maximum # size of the required subset def findMaxLen(a, k): # Size of the array n = len(a) # Sort the array a.sort() # Stores which index is # included or excluded vis = [0] * n # Stores the indices of # array elements mp = {} for i in range(n): mp[a[i]] = i # Count of pairs c = 0 # Iterate through all # the element for i in range(n): # If element is included if(vis[i] == False): check = a[i] * k # Check if a[i] * k is present # in the array or not if(check in mp.keys()): # Increase count of pair c += 1 # Exclude the pair vis[mp[check]] = True return n - c # Driver Code if __name__ == '__main__': K = 3 arr = [ 1, 4, 3, 2 ] print(findMaxLen(arr, K)) # This code is contributed by Shivam Singh |
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C#
// C# implementation of // the above aproach using System; using System.Collections.Generic; class GFG{ // Function to find the maximum // size of the required subset static int findMaxLen(int[] a, int k) { // Size of the array int n = a.Length; // Sort the array Array.Sort(a); // Stores which index is // included or excluded bool []vis = new bool[n]; // Stores the indices of // array elements Dictionary<int, int> mp = new Dictionary<int, int>(); for(int i = 0; i < n; i++) { mp.Add(a[i], i); } // Count of pairs int c = 0; // Iterate through all // the element for(int i = 0; i < n; ++i) { // If element is included if (vis[i] == false) { int check = a[i] * k; // Check if a[i] * k is present // in the array or not if (mp.ContainsKey(check)) { // Increase count of pair c++; // Exclude the pair vis[mp[check]] = true; } } } return n - c; } // Driver code public static void Main(String[] args) { int K = 3; int []arr = { 1, 4, 3, 2 }; Console.Write(findMaxLen(arr, K)); } } // This code is contributed by gauravrajput1 |
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Output:
3
Time Complexity: O(N)
Auxiliary Space: O(N)
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