Divide array in two maximum equal length arrays of similar and dissimilar elements

Given an array arr of natural numbers up to n, find the maximum size for which array arr can be divided into two equal-sized arrays, such that the first array contains all same elements while the second array contains all distinct elements.
Examples:

Input :n = 8, arr[] ={7, 3, 7, 1, 7, 7} 
Output :
Maximum size is : 3 
arr1[] ={7, 7, 7} 
arr2[] ={1, 3, 7} 
Explanation :
It is possible to construct two arrays of size 3. 
The first array is [7, 7, 7] and the second array is [1, 3, 7].
Input : n = 7, arr[] ={1, 2, 1, 5, 1, 6, 7, 2} 
Output :
Maximum size is : 3 
arr1[] ={1, 1, 1} 
arr2[] ={2, 5, 6}

Approach: 
To solve the problem mentioned above the main idea is to use hashing to find the frequency of every element present in the array.

• Find the element with maximum frequency present in the array arr[] using hash vector v.
• Find the total unique elements present in array arr[].
• There are two cases for the element with maximum frequency: the maximum frequency element will go to the first array then the sizes of the array are at most diff1 – 1 and max1 correspondingly. Otherwise, at least one element of maximum frequency goes to the second array and the sizes are at most diff1 and max1 ? 1 correspondingly. Then find the max-size to which array can be splitted as max(min(diff1 ? 1, max1), min(diff1, max1 ? 1)).
• Find the first array of similar elements using max_size and element with maximum frequency max1.
• Find the second array of unique elements using max_size and hash vector v.

Below is the implementation of the above approach:

C++

// C++ program to find the max-size to which
// an array can be divided into 2 equal parts
// such that one part contains unique elements
// while another contains similar elements
#include <bits/stdc++.h>
using namespace std;

// Function to find the max-size to which an array
// can be divided into 2 equal parts
void Solve(int arr[], int size, int n)
{

    vector<int> v(n + 1);

    // Vector to find the frequency of
    // each element of array
    for (int i = 0; i < size; i++)
        v[arr[i]]++;

    // Find the maximum frequency
    // element present in array arr[]
    int max1 = (max_element(v.begin(), v.end())
                                       - v.begin());
    // Find total unique elements
    // present in array arr[]
    int diff1 = n + 1 - count(v.begin(), v.end(), 0);

    // Find the Max-Size to which
    // an array arr[] can be splitted
    int max_size = max(min(v[max1] - 1, diff1),
                            min(v[max1], diff1 - 1));

    cout << "Maximum size is :" << max_size << "\n";

    // Find the first array
    // containing same elements
    cout << "The First Array Is : \n";
    for (int i = 0; i < max_size; i++) {
        cout << max1 << " ";
        v[max1] -= 1;
    }

    cout << "\n";

    // Find the second array
    // containing unique elements
    cout << "The Second Array Is : \n";
    for (int i = 0; i < (n + 1); i++) {
        if (v[i] > 0) {
            cout << i << " ";
            max_size--;
        }
        if (max_size < 1)
            break;
    }

    cout << "\n";
}

// Driver code
int main()
{
    // initialise n
    int n = 7;

    // array declaration
    int arr[] = { 1, 2, 1, 5, 1, 6, 7, 2 };

    // size of array
    int size = sizeof(arr) / sizeof(arr[0]);

    Solve(arr, size, n);

    return 0;
}

Java

// Java program to find the
// max-size to which an array
// can be divided into 2 equal parts
// such that one part contains unique
// elements while another contains
// similar elements
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG{

  // Function to find the max-size to
  // which an array can be divided into
  // 2 equal parts
  static void Solve(int arr[],
                    int size, int n)
  {
    int[] v = new int[n + 1];

    // Array to find the frequency of
    // each element of array
    for (int i = 0; i < size; i++)
      v[arr[i]]++;

    // Find the index maximum frequency
    // element present in array arr[]
    int max1 = -1, mx = -1;
    for (int i = 0; i < v.length; i++)
    {
      if (v[i] > mx)
      {
        mx = v[i];
        max1 = i;
      }
    }
    // Find total unique elements
    // present in array arr[]
    int cnt = 0;
    for (int i : v)
    {
      if (i == 0)
        ++cnt;
    }
    int diff1 = n + 1 - cnt;

    // Find the Max-Size to which
    // an array arr[] can be splitted
    int max_size = Math.max(Math.min(v[max1] - 1,
                                     diff1),
                            Math.min(v[max1],
                                     diff1 - 1));
    System.out.println("Maximum size is: " +
                        max_size);

    // Find the first array
    // containing same elements
    System.out.println("First Array is");
    for (int i = 0; i < max_size; i++)
    {
      System.out.print(max1 + " ");
      v[max1] -= 1;
    }
    System.out.println();

    // Find the second array
    // containing unique elements
    System.out.println("The Second Array Is :");
    for (int i = 0; i < (n + 1); i++)
    {
      if (v[i] > 0)
      {
        System.out.print(i + " ");
        max_size--;
      }
      if (max_size < 1)
        break;
    }
    System.out.println();
  }

  // Driver Code
  public static void main(String[] args)
  {
    // initialise n
    int n = 7;

    // array declaration
    int arr[] = new int[] {1, 2, 1, 5,
                           1, 6, 7, 2};

    // size of array
    int size = arr.length;

    Solve(arr, size, n);
  }
}

// This code is contributed by Sri_srajit

Python3

# Python3 program to find the max-size to which
# an array can be divided into 2 equal parts
# such that one part contains unique elements
# while another contains similar elements

# Function to find the max-size to which an
# array can be divided into 2 equal parts
def Solve(arr, size, n):

    v = [0] * (n + 1);

    # Vector to find the frequency of
    # each element of list
    for i in range(size):
        v[arr[i]] += 1

    # Find the maximum frequency
    # element present in list arr
    max1 = max(set(arr), key = v.count)

    # Find total unique elements
    # present in list arr
    diff1 = n + 1 - v.count(0)

    # Find the Max-Size to which
    # an array arr[] can be splitted
    max_size = max(min(v[max1] - 1, diff1),
                   min(v[max1], diff1 - 1))

    print("Maximum size is :", max_size)

    # Find the first array
    # containing same elements
    print("The First Array Is : ")
    for i in range(max_size):
        print(max1, end = " ")
        v[max1] -= 1

    print()

    # Find the second array
    # containing unique elements
    print("The Second Array Is : ")
    for i in range(n + 1):
        if (v[i] > 0):
            print(i, end = " ")
            max_size -= 1

        if (max_size < 1):
            break

    print()

# Driver code
if __name__ == "__main__":

    # Initialise n
    n = 7

    # Array declaration
    arr = [ 1, 2, 1, 5, 1, 6, 7, 2 ]

    # Size of array
    size = len(arr)

    Solve(arr, size, n)

# This code is contributed by chitranayal

C#

// C# program to find the max-size
// to which an array can be divided
// into 2 equal parts such that one
// part contains unique elements
// while another contains similar
// elements
using System;

class GFG{

// Function to find the max-size to
// which an array can be divided into
// 2 equal parts
static void Solve(int []arr,
                  int size, int n)
{
    int[] v = new int[n + 1];

    // Array to find the frequency of
    // each element of array
    for(int i = 0; i < size; i++)
        v[arr[i]]++;

    // Find the index maximum frequency
    // element present in array arr[]
    int max1 = -1, mx = -1;
    for(int i = 0; i < v.Length; i++)
    {
        if (v[i] > mx)
        {
            mx = v[i];
            max1 = i;
        }
    }

    // Find total unique elements
    // present in array arr[]
    int cnt = 0;
    foreach(int i in v)
    {
        if (i == 0)
            ++cnt;
    }

    int diff1 = n + 1 - cnt;

    // Find the Max-Size to which
    // an array arr[] can be splitted
    int max_size = Math.Max(Math.Min(v[max1] - 1,
                                     diff1),
                            Math.Min(v[max1],
                                     diff1 - 1));

    Console.Write("Maximum size is :" +
                   max_size + "\n");

    // Find the first array
    // containing same elements
    Console.Write("The First Array Is :\n");

    for(int i = 0; i < max_size; i++)
    {
        Console.Write(max1 + " ");
        v[max1] -= 1;
    }
    Console.Write("\n");

    // Find the second array
    // containing unique elements
    Console.Write("The Second Array Is :\n");
    for(int i = 0; i < (n + 1); i++)
    {
        if (v[i] > 0)
        {
            Console.Write(i + " ");
            max_size--;
        }
        if (max_size < 1)
            break;
    }
    Console.Write("\n");
}

// Driver Code
public static void Main(string[] args)
{

    // Initialise n
    int n = 7;

    // Array declaration
    int []arr = new int[] { 1, 2, 1, 5,
                            1, 6, 7, 2 };

    // Size of array
    int size = arr.Length;

    Solve(arr, size, n);
}
}

// This code is contributed by rutvik_56

Javascript

<script>

// Javascript program to find the
// max-size to which an array
// can be divided into 2 equal parts
// such that one part contains unique
// elements while another contains
// similar elements

  // Function to find the max-size to
  // which an array can be divided into
  // 2 equal parts
  function Solve(arr, size, n)
  {
    let v = Array.from({length: n+1}, (_, i) => 0);

    // Array to find the frequency of
    // each element of array
    for (let i = 0; i < size; i++)
      v[arr[i]]++;

    // Find the index maximum frequency
    // element present in array arr[]
    let max1 = -1, mx = -1;
    for (let i = 0; i < v.length; i++)
    {
      if (v[i] > mx)
      {
        mx = v[i];
        max1 = i;
      }
    }
    // Find total unique elements
    // present in array arr[]
    let cnt = 0;
    for (let i in v)
    {
      if (i == 0)
        ++cnt;
    }
    let diff1 = n + 1 - cnt;

    // Find the Max-Size to which
    // an array arr[] can be splitted
    let max_size = Math.max(Math.min(v[max1] - 1,
                                     diff1),
                            Math.min(v[max1],
                                     diff1 - 1));
    document.write("Maximum size is: " +
                        max_size + "<br/>");

    // Find the first array
    // containing same elements
    document.write("First Array is" + "<br/>");
    for (let i = 0; i < max_size; i++)
    {
      document.write(max1 + " ");
      v[max1] -= 1;
    }
    document.write("<br/>");

    // Find the second array
    // containing unique elements
    document.write("The Second Array Is :" + "<br/>");
    for (let i = 0; i < (n + 1); i++)
    {
      if (v[i] > 0)
      {
        document.write(i + " ");
        max_size--;
      }
      if (max_size < 1)
        break;
    }
    document.write("<br/>");
  }

// Driver code

      // initialise n
    let n = 7;

    // array declaration
    let arr = [1, 2, 1, 5,
                     1, 6, 7, 2];

    // size of array
    let size = arr.length;

    Solve(arr, size, n);

   // This code is contributed by code_hunt.
</script>

Output:

Maximum size is :3
The First Array Is : 
1 1 1 
The Second Array Is : 
2 5 6

Time Complexity :O(N) 

Auxiliary Space: O(N)