# Largest subset with maximum difference as 1

Given an array arr[] of n positive integers. The task is to find the size of the subset formed from the elements of the given array and the absolute difference between any two elements of the set is less than equal to 1.

Examples :

Input : arr[] = {8, 9, 8, 7, 8, 9, 10, 11}
Output : 5
If we make subset with elements {8, 9, 8, 8, 9}.
Each pair in the subset has an absolute
difference <= 1

Input : arr[] = {4, 5, 2, 4, 4, 4}
Output : 5
Subset is {4, 5, 4, 4, 4}

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

Observe, since we want the absolute difference between any two elements to be less than equal to 1, then there can be a maximum of two distinct numbers. So, the subset we choose will be in the form {a, a, a, ….., b, b, b} or {a, a, a, a, …..}.
Now, to find the size of such subset we will find the frequency of each element say c1, c2, c3, …., cj, …., cmaximum element in arr. Then our answer will be maximal value of ci + ci+1.

Below is the implementation of this approach:

## C++

 // CPP Program to find the size of // the subset formed from the elements  // of the given array such that the  // maximum difference is 1 #include using namespace std;    // Return the maximum size of subset with  // absolute difference between any element  // is less than 1. int maxsizeSubset(int arr[], int n) {     // Inserting elements and their     // frequencies in a hash table.     unordered_map mp;     for (int i = 0; i < n; i++)          mp[arr[i]]++;         // Traverse through map, for every element     // x in map, find if x+1 also exists in map.      // If exists, see if sum of their frequencies     // is more than current result.     int res = 0;     for (auto x : mp)      if (mp.find(x.first + 1) != mp.end())     {         res = max(res, mp[x.first] + mp[x.first+1]);      }     else     {         res=max(res,mp[x.first]);     }      return res; }    // Driven Program int main() {     int arr[] = {1, 2, 2, 3, 1, 2};     int n = sizeof(arr) / sizeof(arr[0]);     cout << maxsizeSubset(arr, n) << endl;     return 0; }

## Java

 // Java Program to find the size of  // the subset formed from the elements  // of the given array such that the  // maximum difference is 1  import java.util.*;    class GFG  {        // Return the maximum size of subset with      // absolute difference between any element      // is less than 1.      static int maxsizeSubset(int arr[], int n)      {         // Inserting elements and their          // frequencies in a hash table.          Map mp = new HashMap<>();         for (int i = 0; i < n; i++)          {             if (mp.containsKey(arr[i]))             {                 mp.put(arr[i], mp.get(arr[i]) + 1);             }              else              {                 mp.put(arr[i], 1);             }         }            // Traverse through map, for every element          // x in map, find if x+1 also exists in map.          // If exists, see if sum of their frequencies          // is more than current result.          int res = 0;         for (Map.Entry x : mp.entrySet())          {             if (mp.containsKey(x.getKey() + 1))              {                 res = Math.max(res, mp.get(x.getKey()) + mp.get(x.getKey() + 1));             }              else             {                 res = Math.max(res, mp.get(x.getKey()));             }         }         return res;     }        // Driver code      public static void main(String[] args)     {         int arr[] = {1, 2, 2, 3, 1, 2};         int n = arr.length;         System.out.println(maxsizeSubset(arr, n));     } }    /* This code is contributed by PrinciRaj1992 */

## C#

 // C# Program to find the size of  // the subset formed from the elements  // of the given array such that the  // maximum difference is 1  using System; using System.Collections.Generic;    class GFG  {         // Return the maximum size of subset with      // absolute difference between any element      // is less than 1.      static int maxsizeSubset(int []arr, int n)      {          // Inserting elements and their          // frequencies in a hash table.          Dictionary mp = new Dictionary();         for (int i = 0 ; i < n; i++)         {             if(mp.ContainsKey(arr[i]))             {                 var val = mp[arr[i]];                 mp.Remove(arr[i]);                 mp.Add(arr[i], val + 1);              }             else             {                 mp.Add(arr[i], 1);             }         }             // Traverse through map, for every element          // x in map, find if x+1 also exists in map.          // If exists, see if sum of their frequencies          // is more than current result.          int res = 0;          foreach(KeyValuePair x in mp)         {              if (mp.ContainsKey(x.Key + 1))              {                  res = Math.Max(res, mp[x.Key] + mp[x.Key + 1]);              }              else             {                  res = Math.Max(res, mp[x.Key]);              }          }          return res;      }         // Driver code      public static void Main(String[] args)      {          int []arr = {1, 2, 2, 3, 1, 2};          int n = arr.Length;          Console.WriteLine(maxsizeSubset(arr, n));      }  }     // This code is contributed by 29AjayKumar

Output:

5

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

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