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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

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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 geeks
{
 
    // 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 geeks
{
 
    // 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>
Output: 
3

 

Time Complexity: O(N) 
 




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