# Maximum number of unique values in the array after performing given operations

Given an array arr[] of size N and every element in the array arr[] is in the range [1, N] and the array may contain duplicates. The task is to find the maximum number of unique values that can be obtained such that the value at any index i can either be:

• Increased by 1.
• Decreased by 1.
• Left as it is.

Note: The operation can be performed only once with each index and have to be performed for all the indices in the array arr[].

Examples:

Input: arr[] = {1, 2, 4, 4}
Output: 4
Explanation:
One way is to perform the following operations for the index:
1: Leave the value at the first index(1) as it is.
2: Leave the value at the second index(2) as it is.
3: Leave the value at the third index(4) as it is.
4: Increment the value at the fourth index(4) by 1.
Then the array becomes {1, 2, 4, 5} and there are 4 unique values.

Input: arr[]={3, 3, 3, 3}
Output: 3
Explanation:
One way is to perform the following operations for the index:
1: Leave the value at the first index(3) as it is.
2: Decrement the value at the second index(3) by 1.
3: Leave the value at the third index(3) as it is.
4: Increment the value at the fourth index(3) by 1.
Then the array becomes {3, 2, 3, 4} and there are 3 unique values.

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

Approach:

• For some arbitrary element X present in the array at index i, we decide what operation to perform on it by taking the following things into consideration:
1. We decrement the value X by 1 if the value (X – 1) is not present in the array and there are one or more other X’s present in the array at different indices.
2. We don’t change the value X if the value X is present only once on the array.
3. We increment the value X by 1 if the value (X + 1) is not present in the array and there are one or more other X’s present in the array at different indices.
• By taking the above decisions for every element, we can be sure that the final count of unique elements which we get is the maximum.
• However, to perform the above steps for every index and count the occurrences of the element X and continuously update the array arr[], the time taken would be quadratic which is not feasible for large-sized arrays.
• One alternative to reduce the time complexity is to initially sort the array. By sorting, all the elements in the array are grouped and all the repeated values come together.
• After sorting the array, since the range of the numbers is already given and it is fixed, a hash map can be used where the keys of the hash are the numbers in the range [1, N] and the value for each key is boolean which is used to determine if the key is present in the array or not.
• In this problem, since the indices themselves are the keys of the hash, an array freq[] of size (N + 2) is used to implement the hash.

Below is the implementation of the above approach:

## C++

 `// C++ program to find the maximum number of ` `// unique values in the array ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to find the maximum number of ` `// unique values in the array ` `int` `uniqueNumbers(``int` `arr[], ``int` `n) ` `{ ` `    ``// Sorting the given array ` `    ``sort(arr, arr + n); ` ` `  `    ``// This array will store the frequency ` `    ``// of each number in the array ` `    ``// after performing the given operation ` `    ``int` `freq[n + 2]; ` ` `  `    ``// Initialising the array with all zeroes ` `    ``memset``(freq, 0, ``sizeof``(freq)); ` ` `  `    ``// Loop to apply operation on ` `    ``// each element of the array ` `    ``for` `(``int` `x = 0; x < n; x++) { ` ` `  `        ``// Incrementing the value at index x ` `        ``// if the value  arr[x] - 1 is ` `        ``// not present in the array ` `        ``if` `(freq[arr[x] - 1] == 0) { ` `            ``freq[arr[x] - 1]++; ` `        ``} ` ` `  `        ``// If arr[x] itself is not present, then it ` `        ``// is left as it is ` `        ``else` `if` `(freq[arr[x]] == 0) { ` `            ``freq[arr[x]]++; ` `        ``} ` ` `  `        ``// If both arr[x] - 1 and arr[x] are present ` `        ``// then the value is incremented by 1 ` `        ``else` `{ ` `            ``freq[arr[x] + 1]++; ` `        ``} ` `    ``} ` ` `  `    ``// Variable to store the number of unique values ` `    ``int` `unique = 0; ` ` `  `    ``// Finding the unique values ` `    ``for` `(``int` `x = 0; x <= n + 1; x++) { ` `        ``if` `(freq[x]) { ` `            ``unique++; ` `        ``} ` `    ``} ` ` `  `    ``// Returning the number of unique values ` `    ``return` `unique; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 3, 3, 3, 3 }; ` ` `  `    ``// Size of the array ` `    ``int` `n = 4; ` ` `  `    ``int` `ans = uniqueNumbers(arr, n); ` `    ``cout << ans; ` `    ``return` `0; ` `} `

## Java

 `// Java program to find the maximum number of  ` `// unique values in the array  ` `import` `java.util.*; ` ` `  `class` `GFG { ` `     `  `    ``// Function to find the maximum number of  ` `    ``// unique values in the array  ` `    ``static` `int` `uniqueNumbers(``int` `arr[], ``int` `n)  ` `    ``{  ` `        ``// Sorting the given array  ` `        ``Arrays.sort(arr);  ` `     `  `        ``// This array will store the frequency  ` `        ``// of each number in the array  ` `        ``// after performing the given operation  ` `        ``int` `freq[] = ``new` `int``[n + ``2``];  ` `     `  `        ``// Initialising the array with all zeroes  ` `        ``for``(``int` `i = ``0``; i < n + ``2``; i++) ` `            ``freq[i] = ``0``; ` ` `  `        ``// Loop to apply operation on  ` `        ``// each element of the array  ` `        ``for` `(``int` `x = ``0``; x < n; x++) {  ` `     `  `            ``// Incrementing the value at index x  ` `            ``// if the value arr[x] - 1 is  ` `            ``// not present in the array  ` `            ``if` `(freq[arr[x] - ``1``] == ``0``) {  ` `                ``freq[arr[x] - ``1``]++;  ` `            ``}  ` `     `  `            ``// If arr[x] itself is not present, then it  ` `            ``// is left as it is  ` `            ``else` `if` `(freq[arr[x]] == ``0``) {  ` `                ``freq[arr[x]]++;  ` `            ``}  ` `     `  `            ``// If both arr[x] - 1 and arr[x] are present  ` `            ``// then the value is incremented by 1  ` `            ``else` `{  ` `                ``freq[arr[x] + ``1``]++;  ` `            ``}  ` `        ``}  ` `     `  `        ``// Variable to store the number of unique values  ` `        ``int` `unique = ``0``;  ` `     `  `        ``// Finding the unique values  ` `        ``for` `(``int` `x = ``0``; x <= n + ``1``; x++) {  ` `            ``if` `(freq[x] != ``0``) {  ` `                ``unique++;  ` `            ``}  ` `        ``}  ` `     `  `        ``// Returning the number of unique values  ` `        ``return` `unique;  ` `    ``}  ` `     `  `    ``// Driver Code  ` `    ``public` `static` `void` `main (String[] args) ` `    ``{  ` `        ``int` `[]arr = { ``3``, ``3``, ``3``, ``3` `};  ` `     `  `        ``// Size of the array  ` `        ``int` `n = ``4``;  ` `     `  `        ``int` `ans = uniqueNumbers(arr, n);  ` `        ``System.out.println(ans);  ` `    ``}  ` `} ` ` `  `// This code is contributed by Yash_R `

## Python3

 `# Python program to find the maximum number of ` `# unique values in the array ` ` `  `# Function to find the maximum number of ` `# unique values in the array ` `def` `uniqueNumbers(arr, n): ` ` `  `    ``# Sorting the given array ` `    ``arr.sort() ` ` `  `    ``# This array will store the frequency ` `    ``# of each number in the array ` `    ``# after performing the given operation ` `    ``freq ``=``[``0``]``*``(n ``+` `2``) ` ` `  `    ``# Loop to apply the operation on  ` `    ``# each element of the array ` `    ``for` `val ``in` `arr: ` ` `  `        ``# Incrementing the value at index x ` `        ``# if the value  arr[x] - 1 is ` `        ``# not present in the array ` `        ``if``(freq[val``-``1``]``=``=` `0``): ` `            ``freq[val``-``1``]``+``=` `1` ` `  `        ``# If arr[x] itself is not present, then it ` `        ``# is left as it is ` `        ``elif``(freq[val]``=``=` `0``): ` `            ``freq[val]``+``=` `1` ` `  `        ``# If both arr[x] - 1 and arr[x] are present ` `        ``# then the value is incremented by 1 ` `        ``else``: ` `            ``freq[val ``+` `1``]``+``=` `1` `     `  `    ``# Variable to store the  ` `    ``# number of unique values ` `    ``unique ``=` `0` ` `  `    ``# Finding the number of unique values ` `    ``for` `val ``in` `freq: ` `        ``if``(val>``0``): ` `            ``unique``+``=` `1` `     `  `    ``return` `unique ` ` `  `# Driver code ` `if` `__name__ ``=``=` `"__main__"``: ` `    ``arr ``=``[``3``, ``3``, ``3``, ``3``] ` `    ``n ``=` `4` `    ``print``(uniqueNumbers(arr, n)) `

## C#

 `// C# program to find the maximum number of  ` `// unique values in the array  ` `using` `System; ` ` `  `class` `GFG { ` `     `  `    ``// Function to find the maximum number of  ` `    ``// unique values in the array  ` `    ``static` `int` `uniqueNumbers(``int` `[]arr, ``int` `n)  ` `    ``{  ` `        ``// Sorting the given array  ` `        ``Array.Sort(arr);  ` `     `  `        ``// This array will store the frequency  ` `        ``// of each number in the array  ` `        ``// after performing the given operation  ` `        ``int` `[]freq = ``new` `int``[n + 2];  ` `     `  `        ``// Initialising the array with all zeroes  ` `        ``for``(``int` `i = 0; i < n + 2; i++) ` `            ``freq[i] = 0; ` ` `  `        ``// Loop to apply operation on  ` `        ``// each element of the array  ` `        ``for` `(``int` `x = 0; x < n; x++) {  ` `     `  `            ``// Incrementing the value at index x  ` `            ``// if the value arr[x] - 1 is  ` `            ``// not present in the array  ` `            ``if` `(freq[arr[x] - 1] == 0) {  ` `                ``freq[arr[x] - 1]++;  ` `            ``}  ` `     `  `            ``// If arr[x] itself is not present, then it  ` `            ``// is left as it is  ` `            ``else` `if` `(freq[arr[x]] == 0) {  ` `                ``freq[arr[x]]++;  ` `            ``}  ` `     `  `            ``// If both arr[x] - 1 and arr[x] are present  ` `            ``// then the value is incremented by 1  ` `            ``else` `{  ` `                ``freq[arr[x] + 1]++;  ` `            ``}  ` `        ``}  ` `     `  `        ``// Variable to store the number of unique values  ` `        ``int` `unique = 0;  ` `     `  `        ``// Finding the unique values  ` `        ``for` `(``int` `x = 0; x <= n + 1; x++) {  ` `            ``if` `(freq[x] != 0) {  ` `                ``unique++;  ` `            ``}  ` `        ``}  ` `     `  `        ``// Returning the number of unique values  ` `        ``return` `unique;  ` `    ``}  ` `     `  `    ``// Driver Code  ` `    ``public` `static` `void` `Main (``string``[] args) ` `    ``{  ` `        ``int` `[]arr = { 3, 3, 3, 3 };  ` `     `  `        ``// Size of the array  ` `        ``int` `n = 4;  ` `     `  `        ``int` `ans = uniqueNumbers(arr, n);  ` `        ``Console.WriteLine(ans);  ` `    ``}  ` `} ` ` `  `// This code is contributed by Yash_R `

Output:

```3
```

Time Complexity Analysis:

• The time taken to sort the given array is O(N * log(N)) where N is the size of the array.
• The time taken to run a loop over the sorted array to perform the operations is O(N).
• The time taken to run a loop over the hash to count the unique values is O(N).
• Therefore, the overall time complexity is O(N * log(N)) + O(N) + O(N). Since N * log(N) is greater, the final time complexity of the above approach is O(N * log(N)).

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