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.

**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:- 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. - We
**don’t change**the value**X**if the value**X**is present only once on the array. - 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.

- We
- 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 <bits/stdc++.h> ` `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; ` `} ` |

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## 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 ` |

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## 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)) ` |

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## 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 ` |

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**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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