Minimum operations required to make all the elements distinct in an array
Given an array of N integers. If a number occurs more than once, choose any number y from the array and replace the x in the array to x+y such that x+y is not in the array. The task is to find the minimum number of operations to make the array a distinct one.
Examples:
Input: a[] = {2, 1, 2}
Output: 1
Let x = 2, y = 1 then replace 2 by 3.
Performing the above step makes all the elements in the array distinct.
Input: a[] = {1, 2, 3}
Output: 0
Approach: If a number appears more than once, then the summation of (occurrences-1) for all duplicate elements will be the answer. The main logic behind this is if x is replaced by x+y where y is the largest element in the array, then x is replaced by x+y which is the largest element in the array. Use a map to store the frequency of the numbers of array. Traverse in the map, and if the frequency of an element is more than 1, add it to the count by subtracting one.
Below is the implementation of the above approach:
C++
// C++ program to find Minimum number // of changes to make array distinct #include <bits/stdc++.h> using namespace std; // Function that returns minimum number of changes int minimumOperations( int a[], int n) { // Hash-table to store frequency unordered_map< int , int > mp; // Increase the frequency of elements for ( int i = 0; i < n; i++) mp[a[i]] += 1; int count = 0; // Traverse in the map to sum up the (occurrences-1) // of duplicate elements for ( auto it = mp.begin(); it != mp.end(); it++) { if ((*it).second > 1) count += (*it).second-1; } return count; } // Driver Code int main() { int a[] = { 2, 1, 2, 3, 3, 4, 3 }; int n = sizeof (a) / sizeof (a[0]); cout << minimumOperations(a, n); return 0; } |
Java
// Java program to find Minimum number // of changes to make array distinct import java.util.*; class Beginner { // Function that returns minimum number of changes public static int minimumOperations( int [] a, int n) { // Hash-table to store frequency HashMap<Integer, Integer> mp = new HashMap<>(); // Increase the frequency of elements for ( int i = 0 ; i < n; i++) { if (mp.get(a[i]) != null ) { int x = mp.get(a[i]); mp.put(a[i], ++x); } else mp.put(a[i], 1 ); } int count = 0 ; // Traverse in the map to sum up the (occurrences-1) // of duplicate elements for (HashMap.Entry<Integer, Integer> entry : mp.entrySet()) { if (entry.getValue() > 1 ) { count += (entry.getValue() - 1 ); } } return count; } // Driver Code public static void main(String[] args) { int [] a = { 2 , 1 , 2 , 3 , 3 , 4 , 3 }; int n = a.length; System.out.println(minimumOperations(a, n)); } } // This code is contributed by // sanjeev2552 |
Python3
# Python3 program to find Minimum number # of changes to make array distinct # Function that returns minimum # number of changes def minimumOperations(a, n): # Hash-table to store frequency mp = dict () # Increase the frequency of elements for i in range (n): if a[i] in mp.keys(): mp[a[i]] + = 1 else : mp[a[i]] = 1 count = 0 # Traverse in the map to sum up the # (occurrences-1)of duplicate elements for it in mp: if (mp[it] > 1 ): count + = mp[it] - 1 return count # Driver Code a = [ 2 , 1 , 2 , 3 , 3 , 4 , 3 ] n = len (a) print (minimumOperations(a, n)) # This code is contributed # by Mohit Kumar |
C#
// C# program to find Minimum number // of changes to make array distinct using System; using System.Collections.Generic; class Beginner { // Function that returns minimum number of changes public static int minimumOperations( int [] a, int n) { // Hash-table to store frequency Dictionary< int , int > mp = new Dictionary< int , int >(); // Increase the frequency of elements for ( int i = 0 ; i < n; i++) { if (mp.ContainsKey(a[i])) { var val = mp[a[i]]; mp.Remove(a[i]); mp.Add(a[i], val + 1); } else { mp.Add(a[i], 1); } } int count = 0; // Traverse in the map to sum up the (occurrences-1) // of duplicate elements foreach (KeyValuePair< int , int > entry in mp) { if (entry.Value > 1) { count += (entry.Value - 1); } } return count; } // Driver Code public static void Main(String[] args) { int [] a = { 2, 1, 2, 3, 3, 4, 3 }; int n = a.Length; Console.WriteLine(minimumOperations(a, n)); } } /* This code is contributed by PrinciRaj1992 */ |
Javascript
<script> // JavaScript program to find Minimum number // of changes to make array distinct // Function that returns minimum // number of changes function minimumOperations(a,n) { // Hash-table to store frequency let mp = new Map(); // Increase the frequency of elements for (let i = 0; i < n; i++) { if (mp.get(a[i]) != null ) { let x = mp.get(a[i]); mp.set(a[i], ++x); } else mp.set(a[i], 1); } let count = 0; // Traverse in the map to // sum up the (occurrences-1) // of duplicate elements for (let [key, value] of mp.entries()) { if (value > 1) { count += (value - 1); } } return count; } // Driver Code let a=[2, 1, 2, 3, 3, 4, 3]; let n = a.length; document.write(minimumOperations(a, n)); // This code is contributed by patel2127 </script> |
3
Time Complexity: O(N), where N represents the size of the given array.
Auxiliary Space: O(N), where N represents the size of the given array.