# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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; ` ` `  `// Fucntion 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); ` ` `  `    ``cout << minimumOperations(a, n); ` `    ``return` `0; ` `} `

## Java

 `// Java program to find Minimum number  ` `// of changes to make array distinct  ` `import` `java.util.*; ` ` `  `class` `geeks ` `{ ` ` `  `    ``// Fucntion 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 ` ` `  `# Fucntion 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 ` `{ ` ` `  `    ``// Fucntion 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 */`

Output:

```3
```

Time Complexity: O(N) My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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