# Maximize distinct elements by incrementing/decrementing an element or keeping it same

• Last Updated : 29 Sep, 2022

Given an array arr[] of N elements, the task is to maximize the count of distinct elements in the array, by either of the given operation on each element of the array:

• either increasing the element by 1
• or decreasing the element by 1
• or keeping the element as it is.

Note: No element can be less than or equal to 0.

Examples:

Input: arr = [4, 4, 5, 5, 5, 5, 6, 6]
Output:
Explanation: After modification of each element of the array in any of the three possible ways, arr[] = [3, 4, 5, 5, 5, 5, 6, 7]. Here distinct elements are 5.

Input: arr = [1, 1, 1, 8, 8, 8, 9, 9]
Output:
Explanation: After modification of each element of the array in any of the three possible ways, arr[] = [1, 1, 2, 7, 8, 8, 9, 10]. Here distinct elements are 6.

Approach: The idea is to sort the given array first, so that the elements can be checked easily, if it is distinct, by comparing with adjacent elements.

1. First, sort all elements of the array.
2. Initialize variables count and prev to 0. (To store the count of distinct elements and previous element respectively.)
3. After that keep a track of the previous element using prev variable.
4. Iterate the sorted array.
5. Decrease the current element’s value by 1 and check if the previous element is lesser than the decreased value. If it is lesser then increment the count and assign current value to prev.
6. If the decreased value of the current element is not greater than the previous element then keep the current element as it is and check if the previous element is lesser than the current element. If it is lesser then increment the count and assign current value to prev.
7. If the current value is not greater than the previous element then increment the current value by 1 and check if the previous element is lesser than the incremented current element. If it is lesser then increment the count and assign current value to prev.
8. If incremented value of current element is not lesser than previous value then skip that element.

Below is the implementation of the above approach:

## C++

 `// C++ program to Maximize distinct``// elements by incrementing/decrementing``// an element or keeping it same` `#include ``using` `namespace` `std;` `// Function that Maximize``// the count of distinct``// element``int` `max_dist_ele(``int` `arr[],``                 ``int` `n)``{` `    ``// sort the array``    ``sort(arr, arr + n);` `    ``int` `ans = 0;` `    ``// keeping track of``    ``// previous change``    ``int` `prev = 0;` `    ``for` `(``int` `i = 0;``         ``i < n; i++) {` `        ``// check the``        ``// decremented value``        ``if` `(prev < (arr[i] - 1)) {` `            ``ans++;``            ``prev = arr[i] - 1;``        ``}` `        ``// check the current``        ``// value``        ``else` `if` `(prev < (arr[i])) {` `            ``ans++;``            ``prev = arr[i];``        ``}` `        ``// check the``        ``// incremented value``        ``else` `if` `(prev < (arr[i] + 1)) {` `            ``ans++;``            ``prev = arr[i] + 1;``        ``}``    ``}``    ``return` `ans;``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 1, 1, 1, 8,``                  ``8, 8, 9, 9 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);``    ``cout << max_dist_ele(arr, n)``         ``<< endl;``    ``return` `0;``}`

## Java

 `// Java program to Maximize``// the count of distinct element` `import` `java.util.*;` `public` `class` `GFG {` `    ``// Function that Maximize``    ``// the count of distinct element``    ``static` `int` `max_dist_ele(``        ``int` `arr[], ``int` `n)``    ``{``        ``// sort the array``        ``Arrays.sort(arr);` `        ``int` `ans = ``0``;` `        ``// keeping track of``        ``// previous change``        ``int` `prev = ``0``;` `        ``for` `(``int` `i = ``0``;``             ``i < n; i++) {` `            ``// decrement is possible``            ``if` `(prev < (arr[i] - ``1``)) {` `                ``ans++;``                ``prev = arr[i] - ``1``;``            ``}` `            ``// remain as it is``            ``else` `if` `(prev < (arr[i])) {` `                ``ans++;``                ``prev = arr[i];``            ``}``            ``// increment is possible``            ``else` `if` `(prev < (arr[i] + ``1``)) {``                ``ans++;``                ``prev = arr[i] + ``1``;``            ``}``        ``}` `        ``return` `ans;``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String args[])``    ``{``        ``int` `arr[] = { ``1``, ``1``, ``1``, ``8``,``                      ``8``, ``8``, ``9``, ``9` `};``        ``int` `n = arr.length;` `        ``System.out.println(max_dist_ele(arr, n));``    ``}``}`

## Python3

 `# Python3 program to Maximize``# the count of distinct element``def` `max_dist_ele(arr, n):``    ` `    ``# Sort the list``    ``arr.sort()``    ` `    ``ans ``=` `0``    ` `    ``# Keeping track of``    ``# previous change``    ``prev ``=` `0``    ` `    ``for` `i ``in` `range``(n):``        ` `        ``# Decrement is possible``        ``if` `prev < (arr[i] ``-` `1``):``            ``ans ``+``=` `1``;``            ``prev ``=` `arr[i] ``-` `1``            ` `        ``# Remain as it is``        ``elif` `prev < (arr[i]):``            ``ans ``+``=` `1``            ``prev ``=` `arr[i]``            ` `        ``# Increment is possible``        ``elif` `prev < (arr[i] ``+` `1``):``            ``ans ``+``=` `1``            ``prev ``=` `arr[i] ``+` `1``    ` `    ``return` `ans` `# Driver Code``arr ``=` `[ ``1``, ``1``, ``1``, ``8``, ``8``, ``8``, ``9``, ``9` `]``n ``=` `len``(arr)` `print``(max_dist_ele(arr, n))` `# This code is contributed by rutvik_56`

## C#

 `// C# program to maximize the``// count of distinct element``using` `System;` `class` `GFG{` `// Function that maximize the``// count of distinct element``static` `int` `max_dist_ele(``int` `[]arr, ``int` `n)``{``    ` `    ``// Sort the array``    ``Array.Sort(arr);` `    ``int` `ans = 0;` `    ``// Keeping track of``    ``// previous change``    ``int` `prev = 0;` `    ``for``(``int` `i = 0; i < n; i++)``    ``{``       ` `       ``// Decrement is possible``       ``if` `(prev < (arr[i] - 1))``       ``{``           ``ans++;``           ``prev = arr[i] - 1;``       ``}``       ` `       ``// Remain as it is``       ``else` `if` `(prev < (arr[i]))``       ``{``           ``ans++;``           ``prev = arr[i];``       ``}``       ` `       ``// Increment is possible``       ``else` `if` `(prev < (arr[i] + 1))``       ``{``           ``ans++;``           ``prev = arr[i] + 1;``       ``}``    ``}``    ``return` `ans;``}` `// Driver Code``public` `static` `void` `Main(String []args)``{``    ``int` `[]arr = { 1, 1, 1, 8,``                  ``8, 8, 9, 9 };``    ``int` `n = arr.Length;` `    ``Console.WriteLine(max_dist_ele(arr, n));``}``}` `// This code is contributed by Amit Katiyar`

## Javascript

 ``

Output:

`6`

Time Complexity: O(N*logN)

Auxiliary Space: O(1) as it is using constant space for variables

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