Minimum replacements required to have at most K distinct elements in the array
Given an array arr[] consisting of N positive integers and an integer K, the task is to find the minimum number of array elements required to be replaced by the other array elements such that the array contains at most K distinct elements.
Input: arr[] = { 1, 1, 2, 2, 5 }, K = 2
Output: 1
Explanation:
Replacing arr[4] with arr[0] modifies arr[] to { 1, 1, 2, 2, 1 }
Distinct array elements of the array arr[] are { 1, 2 }
Therefore, the required output is 1.Input: arr[] = { 5, 1, 3, 2, 4, 1, 1, 2, 3, 4 }, K = 3
Output: 3
Explanation:
Replacing arr[0] with arr[1] modifies arr[] to { 1, 1, 3, 2, 4, 1, 1, 2, 3, 4 }
Replacing arr[2] with arr[0] modifies arr[] to { 1, 1, 1, 2, 4, 1, 1, 2, 3, 4 }
Replacing arr[8] with arr[0] modifies arr[] to { 1, 1, 1, 2, 4, 1, 1, 2, 1, 4 }
Distinct array elements of the array arr[] are { 1, 2, 4 }
Therefore, the required output is 3.
Approach:The problem can be solved using Greedy technique. The idea is to replace the smaller frequency array elements with the higher frequency array elements. Follow the steps below to solve the problem:
- Initialize a map, say mp, to store the frequency of each distinct array element.
- Traverse the array and store the frequency of each distinct element of the array into the Map.
- Traverse the map using the key value of the map as i and insert the value of mp[i] into an array, say Freq[].
- Sort the array Freq[] in descending order.
- Initialize a variable, say cntMin, to store the minimum count of array elements required to be replaced such that the count of distinct elements in the array is at most K.
- Initialize a variable, say len, to store the size of the array Freq[].
- Iterate over the range [K, len] using variable i. For every ith value, update cntMin += Freq[i].
- Finally, print the value of cntMin.
Below is the implementation of the above approach:
C++
// C++ program to implement // the above approach #include <bits/stdc++.h> using namespace std; // Function to find minimum count of array // elements required to be replaced such that // count of distinct elements is at most K int min_elements( int arr[], int N, int K) { // Store the frequency of each // distinct element of the array map< int , int > mp; // Traverse the array for ( int i = 0; i < N; i++) { // Update frequency // of arr[i] mp[arr[i]]++; } // Store frequency of each distinct // element of the array vector< int > Freq; // Traverse the map for ( auto it : mp) { // Stores key of the map int i = it.first; // Insert mp[i] into Freq[] Freq.push_back(mp[i]); } // Sort Freq[] in descending order sort(Freq.rbegin(), Freq.rend()); // Stores size of Freq[] int len = Freq.size(); // If len is less than // or equal to K if (len <= K) { return 0; } // Stores minimum count of array elements // required to be replaced such that // count of distinct elements is at most K int cntMin = 0; // Iterate over the range [K, len] for ( int i = K; i < len; i++) { // Update cntMin cntMin += Freq[i]; } return cntMin; } // Driver Code int main() { int arr[] = { 5, 1, 3, 2, 4, 1, 1, 2, 3, 4 }; int N = sizeof (arr) / sizeof (arr[0]); int K = 3; cout << min_elements(arr, N, K); return 0; } |
Java
// Java program to implement // the above approach import java.util.*; class GFG { // Function to find minimum count of array // elements required to be replaced such that // count of distinct elements is at most K static int min_elements( int arr[], int N, int K) { // Store the frequency of each // distinct element of the array HashMap<Integer,Integer> mp = new HashMap<Integer,Integer>(); // Traverse the array for ( int i = 0 ; i < N; i++) { // Update frequency // of arr[i] if (mp.containsKey(arr[i])) { mp.put(arr[i], mp.get(arr[i])+ 1 ); } else { mp.put(arr[i], 1 ); } } // Store frequency of each distinct // element of the array Vector<Integer> Freq = new Vector<Integer>(); // Traverse the map for (Map.Entry<Integer,Integer> it : mp.entrySet()) { // Stores key of the map int i = it.getKey(); // Insert mp[i] into Freq[] Freq.add(mp.get(i)); } // Sort Freq[] in descending order Collections.sort(Freq,Collections.reverseOrder()); // Stores size of Freq[] int len = Freq.size(); // If len is less than // or equal to K if (len <= K) { return 0 ; } // Stores minimum count of array elements // required to be replaced such that // count of distinct elements is at most K int cntMin = 0 ; // Iterate over the range [K, len] for ( int i = K; i < len; i++) { // Update cntMin cntMin += Freq.get(i); } return cntMin; } // Driver Code public static void main(String[] args) { int arr[] = { 5 , 1 , 3 , 2 , 4 , 1 , 1 , 2 , 3 , 4 }; int N = arr.length; int K = 3 ; System.out.print(min_elements(arr, N, K)); } } // This code is contributed by shikhasingrajput |
Python3
# Python3 program to implement # the above approach # Function to find minimum count of array # elements required to be replaced such that # count of distinct elements is at most K def min_elements(arr, N, K) : # Store the frequency of each # distinct element of the array mp = {} # Traverse the array for i in range (N) : # Update frequency # of arr[i] if arr[i] in mp : mp[arr[i]] + = 1 else : mp[arr[i]] = 1 # Store frequency of each distinct # element of the array Freq = [] # Traverse the map for it in mp : # Stores key of the map i = it # Insert mp[i] into Freq[] Freq.append(mp[i]) # Sort Freq[] in descending order Freq.sort() Freq.reverse() # Stores size of Freq[] Len = len (Freq) # If len is less than # or equal to K if ( Len < = K) : return 0 # Stores minimum count of array elements # required to be replaced such that # count of distinct elements is at most K cntMin = 0 # Iterate over the range [K, len] for i in range (K, Len ) : # Update cntMin cntMin + = Freq[i] return cntMin # Driver code arr = [ 5 , 1 , 3 , 2 , 4 , 1 , 1 , 2 , 3 , 4 ] N = len (arr) K = 3 ; print (min_elements(arr, N, K)) # This code is contributed by divyesh072019. |
C#
// C# program to implement // the above approach using System; using System.Collections.Generic; class GFG{ // Function to find minimum count of array // elements required to be replaced such that // count of distinct elements is at most K static int min_elements( int []arr, int N, int K) { // Store the frequency of each // distinct element of the array Dictionary< int , int > mp = new Dictionary< int , int >(); // Traverse the array for ( int i = 0; i < N; i++) { // Update frequency // of arr[i] if (mp.ContainsKey(arr[i])) { mp[arr[i]] = mp[arr[i]] + 1; } else { mp.Add(arr[i], 1); } } // Store frequency of each distinct // element of the array List< int > Freq = new List< int >(); // Traverse the map foreach (KeyValuePair< int , int > it in mp) { // Stores key of the map int i = it.Key; // Insert mp[i] into Freq[] Freq.Add(mp[i]); } // Sort Freq[] in descending order Freq.Sort(); Freq.Reverse(); // Stores size of Freq[] int len = Freq.Count; // If len is less than // or equal to K if (len <= K) { return 0; } // Stores minimum count of array elements // required to be replaced such that // count of distinct elements is at most K int cntMin = 0; // Iterate over the range [K, len] for ( int i = K; i < len; i++) { // Update cntMin cntMin += Freq[i]; } return cntMin; } // Driver Code public static void Main(String[] args) { int []arr = { 5, 1, 3, 2, 4, 1, 1, 2, 3, 4 }; int N = arr.Length; int K = 3; Console.Write(min_elements(arr, N, K)); } } // This code is contributed by gauravrajput1 |
Javascript
<script> // Javascript program to implement // the above approach // Function to find minimum count of array // elements required to be replaced such that // count of distinct elements is at most K function min_elements(arr, N, K) { // Store the frequency of each // distinct element of the array let mp = new Map(); // Traverse the array for (let i = 0; i < N; i++) { // Update frequency // of arr[i] if (mp.has(arr[i])) { mp.set(arr[i], mp.get(arr[i]) + 1) } else { mp.set(arr[i], 1) } } // Store frequency of each distinct // element of the array let Freq = []; // Traverse the map for (let it of mp) { // Stores key of the map let i = it[0]; // Insert mp[i] into Freq[] Freq.push(mp.get(i)); } // Sort Freq[] in descending order Freq.sort((a, b) => b - a); // Stores size of Freq[] let len = Freq.length; // If len is less than // or equal to K if (len <= K) { return 0; } // Stores minimum count of array elements // required to be replaced such that // count of distinct elements is at most K let cntMin = 0; // Iterate over the range [K, len] for (let i = K; i < len; i++) { // Update cntMin cntMin += Freq[i]; } return cntMin; } // Driver Code let arr = [ 5, 1, 3, 2, 4, 1, 1, 2, 3, 4 ]; let N = arr.length; let K = 3; document.write(min_elements(arr, N, K)); // This code contributed by _saurabh_jaiswal </script> |
3
Time Complexity: O(N * log(N))
Auxiliary Space: O(N)
Please Login to comment...