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# Maximize MEX by adding or subtracting K from Array elements

Given an arr[] of size N and an integer, K, the task is to find the maximum possible value of MEX by adding or subtracting K any number of times from the array elements.

MEX is the minimum non-negative integer that is not present in the array

Examples:

Input: arr[]={1, 3, 4}, K = 2
Output: 2
Explanation: After subtracting K from arr twice,
the final array will be {1, 3, 0}.
So the MEX is 2 which is maximum possible answer.

Input: arr[] = {0, 1, 2, 1, 3}, K = 3
Output: 5
Explanation: After adding K to arr, the final array will be {0, 4, 2, 1, 3}.
So the MEX is 5 which is maximum possible answer.

Approach: Follow the below idea to solve the problem:

The maximum MEX which can be achieved from an array of size N is N. For this, we need to have all the elements from 0 to N – 1 by doing some operations.  If there is a number that is not achievable by doing some operation then this is the answer.

We can generate a number x from a number p by doing addition or subtraction if both remainders are same by diving it with K [i.e., the difference between them is divisible by K].

Follow the steps to solve the problem:

• Create a map that stores the frequency of the remainder of all the elements of the array.
• Traverse from 0 to N-1 and check if there is a need to generate the number x then x % K must be present in the map.
• If it is present in the map then decrease the frequency of x % K and continue.
• Else, return the number as the answer.

Below is the implementation for the above approach:

## C++

 `// C++ code for the above approach`   `#include ` `using` `namespace` `std;`   `// Function to find maximum MEX of the array` `// after doing some addition and subtraction` `int` `mex(``int` `arr[], ``int` `n, ``int` `K)` `{` `    ``// Create a map to store the frequency of` `    ``// remainder of all element by K` `    ``unordered_map<``int``, ``int``> mp;`   `    ``for` `(``int` `i = 0; i < n; i++) {` `        ``mp[arr[i] % K]++;` `    ``}`   `    ``for` `(``int` `i = 0; i < n; i++) {`   `        ``// In order to generate i find an` `        ``// element whose modulo value is equal` `        ``// to i%K, return i as answer if no` `        ``// such value is found` `        ``if` `(mp.find(i % K) == mp.end()) {` `            ``return` `i;` `        ``}`   `        ``// If we find element whose modulo` `        ``// value equal to i%K` `        ``mp[i % K]--;` `        ``if` `(mp[i % K] == 0)` `            ``mp.erase(i % K);` `    ``}`   `    ``return` `n;` `}`   `// Driver code` `int` `main()` `{` `    ``int` `arr[] = { 0, 1, 2, 1, 3 };` `    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);` `    ``int` `K = 3;`   `    ``// Function call` `    ``cout << mex(arr, N, K) << endl;` `    ``return` `0;` `}`

## Java

 `/*package whatever //do not write package name here */` `import` `java.io.*;` `import` `java.util.*;` `import` `java.util.HashMap;`   `class` `GFG {`   `    ``// Function to find maximum MEX of the array` `    ``// after doing some addition and subtraction` `    ``public` `static` `int` `mex(``int``[] arr, ``int` `n, ``int` `K)` `    ``{`   `        ``Map mp` `            ``= ``new` `HashMap();`   `        ``for` `(``int` `v : arr) {` `            ``mp.putIfAbsent(v % K, ``0``);` `            ``mp.put(v % K, mp.get(v % K) + ``1``);` `        ``}`   `        ``for` `(``int` `i = ``0``; i < n; i++) {` `            ``if` `(!mp.containsKey(i % K)) {` `                ``return` `i;` `            ``}`   `            ``mp.put(i % K, mp.get(i % K) - ``1``);`   `            ``if` `(mp.get(i % K) <= ``0``) {` `                ``mp.remove(i % K);` `            ``}` `        ``}` `        ``return` `n;` `    ``}`   `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``int` `arr[] = { ``0``, ``1``, ``2``, ``1``, ``3` `};` `        ``int` `N = arr.length;` `        ``int` `K = ``3``;`   `        ``// Function call` `        ``System.out.println(mex(arr, N, K));` `    ``}` `}`   `// This code is contributed by akashish__`

## Python3

 `# Python code to implement the approach` `from` `collections ``import` `defaultdict`   `# Function to find maximum MEX of the array` `# after doing some addition and subtraction` `def` `mex(arr, n, K) :` `    `  `    ``# Create a map to store the frequency of` `    ``# remainder of all element by K` `    ``mp ``=` `defaultdict(``lambda` `: ``0``)` ` `  `    ``# Traverse the array` `    ``for` `i ``in` `range``(n):` ` `  `        ``# Update frequency of arr[i]` `        ``mp[arr[i] ``%` `K] ``+``=` `1``;` `    `  `    ``for` `i ``in` `range``(n):`   `        ``# In order to generate i find an` `        ``# element whose modulo value is equal` `        ``# to i%K, return i as answer if no` `        ``# such value is found` `        ``if` `((i ``%` `K) ``not` `in` `mp) :` `            ``return` `i` `        `  `        ``# If we find element whose modulo` `        ``# value equal to i%K` `        ``mp[i ``%` `K] ``-``=` `1` `        ``if` `(mp[i ``%` `K] ``=``=` `0``) :` `            ``del` `mp[i ``%` `K] ` `    `  `    ``return` `n`   `# Driver code` `if` `__name__ ``=``=` `"__main__"``:` `    `  `    ``arr ``=` `[ ``0``, ``1``, ``2``, ``1``, ``3` `]` `    ``N ``=` `len``(arr)` `    ``K ``=` `3`   `    ``# Function call` `    ``print``(mex(arr, N, K)) `   `# This code is contributed by sanjoy_62.`

## C#

 `/*package whatever //do not write package name here */` `using` `System;` `using` `System.Collections.Generic;`   `public` `class` `GFG {`   `  ``// Function to find maximum MEX of the array` `  ``// after doing some addition and subtraction` `  ``public` `static` `int` `mex(``int``[] arr, ``int` `n, ``int` `K)` `  ``{` `    `  `    ``// Create a map to store the frequency of` `    ``// remainder of all element by K` `    ``Dictionary<``int``, ``int``> mp` `      ``= ``new` `Dictionary<``int``, ``int``>();`   `    ``for` `(``int` `i = 0; i < n; i++) {` `      ``mp.Add(i, 0);` `    ``}` `    ``for` `(``int` `i = 0; i < n; i++) {` `        ``if``(mp.ContainsKey(arr[i] % K))` `              ``mp[mp[arr[i] % K]]=` `             ``mp[arr[i] % K] + 1;` `        ``else` `            ``mp.Add(mp[arr[i] % K],` `               ``1);` `    ``}`   `    ``for` `(``int` `i = 0; i < n; i++) {`   `      ``// In order to generate i find an` `      ``// element whose modulo value is equal` `      ``// to i%K, return i as answer if no` `      ``// such value is found` `      ``if` `(mp[i%K] > 1 ){` `        ``return` `i;` `      ``}`   `      ``// If we find element whose modulo` `      ``// value equal to i%K` `    ``if``(mp.ContainsKey(arr[i] % K))` `      ``mp[mp[arr[i] % K]]=` `             ``mp[arr[i] % K] - 1;` `      ``if` `(mp.ContainsKey(i % K) && mp[i % K] <= 0)` `        ``mp[mp[arr[i] % K]]=` `               ``0;` `    ``}` `    ``return` `n;` `  ``}`   `  ``public` `static` `void` `Main(String[] args)` `  ``{` `    ``int` `[]arr = { 0, 1, 2, 1, 3 };` `    ``int` `N = arr.Length;` `    ``int` `K = 3;`   `    ``// Function call` `    ``Console.WriteLine(mex(arr, N, K));` `  ``}` `}`     `// This code contributed by shikhasingrajput`

## Javascript

 `  ```

Output

`5`

Time Complexity: O(N)
Auxiliary Space: O(N)

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