Minimum operations required to make all the elements distinct in an array

• Difficulty Level : Easy
• Last Updated : 21 Jun, 2021

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:
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 using namespace std; // Function that returns minimum number of changesint minimumOperations(int a[], int n){     // Hash-table to store frequency    unordered_map 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 Codeint main(){    int a[] = { 2, 1, 2, 3, 3, 4, 3 };    int n = sizeof(a) / sizeof(a);     cout << minimumOperations(a, n);    return 0;}

Java

 // Java program to find Minimum number// of changes to make array distinctimport java.util.*; class geeks{     // Function that returns minimum number of changes    public static int minimumOperations(int[] a, int n)    {         // Hash-table to store frequency        HashMap 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 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 changesdef 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 Codea = [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 distinctusing System;using System.Collections.Generic; class geeks{     // Function that returns minimum number of changes    public static int minimumOperations(int[] a, int n)    {         // Hash-table to store frequency        Dictionary mp = new Dictionary();        // 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 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


Output:
3

Time Complexity: O(N)

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