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[2] twice, Â
the final array will be {1, 3, 0}.Â
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[1], the final array will be {0, 4, 2, 1, 3}.Â
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.Â
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 approachÂ
#include <bits/stdc++.h>using namespace std;Â
// Function to find maximum MEX of the array// after doing some addition and subtractionint 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;Â
    for (int i = 0; i < n; i++) {        mp[arr[i] % K]++;    }Â
    for (int i = 0; i < n; i++) {Â
        // 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;        }Â
        // If we find element whose modulo        // value equal to i%K        mp[i % K]--;        if (mp[i % K] == 0)            mp.erase(i % K);    }Â
    return n;}Â
// Driver codeint main(){Â Â Â Â int arr[] = { 0, 1, 2, 1, 3 };Â Â Â Â int N = sizeof(arr) / sizeof(arr[0]);Â Â Â Â int K = 3;Â
    // 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;Â
class GFG {Â
    // Function to find maximum MEX of the array    // after doing some addition and subtraction    public static int mex(int[] arr, int n, int K)    {Â
        Map<Integer, Integer> mp            = new HashMap<Integer, Integer>();Â
        for (int v : arr) {            mp.putIfAbsent(v % K, 0);            mp.put(v % K, mp.get(v % K) + 1);        }Â
        for (int i = 0; i < n; i++) {            if (!mp.containsKey(i % K)) {                return i;            }Â
            mp.put(i % K, mp.get(i % K) - 1);Â
            if (mp.get(i % K) <= 0) {                mp.remove(i % K);            }        }        return n;    }Â
    public static void main(String[] args)    {        int arr[] = { 0, 1, 2, 1, 3 };        int N = arr.length;        int K = 3;Â
        // Function call        System.out.println(mex(arr, N, K));    }}Â
// This code is contributed by akashish__ |
Python3
# Python code to implement the approachfrom collections import defaultdictÂ
# Function to find maximum MEX of the array# after doing some addition and subtractiondef 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):Â
        # 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 nÂ
# Driver codeif __name__ == "__main__":Â Â Â Â Â Â Â Â Â arr = [ 0, 1, 2, 1, 3 ]Â Â Â Â N = len(arr)Â Â Â Â K = 3Â
    # Function call    print(mex(arr, N, K)) Â
# This code is contributed by sanjoy_62. |
C#
/*package whatever //do not write package name here */using System;using System.Collections.Generic;Â
public class GFG {Â
  // 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>();Â
    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);    }Â
    for (int i = 0; i < n; i++) {Â
      // 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;      }Â
      // 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;  }Â
  public static void Main(String[] args)  {    int []arr = { 0, 1, 2, 1, 3 };    int N = arr.Length;    int K = 3;Â
    // Function call    Console.WriteLine(mex(arr, N, K));  }}Â
Â
// This code contributed by shikhasingrajput |
Javascript
  <script>// Javascript code for the above approachÂ
// Function to find maximum MEX of the array// after doing some addition and subtractionfunction mex(arr,n,K){    // Create a map to store the frequency of    // remainder of all element by K        let mp=new Map();    for (let i = 0; i < n; i++) {        mp[arr[i] % K]++;    }Â
    for (let i = 0; i < n; i++) {Â
        // 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.has(i % K)) {            return i;        }Â
        // If we find element whose modulo        // value equal to i%K        mp[i % K]--;        if (mp[i % K] == 0)            mp.delete(i % K);    }Â
    return n;}Â
// Driver codeÂ
    let arr = [ 0, 1, 2, 1, 3 ];    let N = arr.length;    let K = 3;Â
    // Function call    document.write(mex(arr, N, K));         // This code is contributed by satwik4409.    </script> |
Output
5
Time Complexity: O(N)
Auxiliary Space: O(N)
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