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Length of array pair formed where one contains all distinct elements and other all same elements
  • Last Updated : 28 Sep, 2020

Given an array arr[], the task is to determine the maximum possible size of two arrays of the same size that can be created using the given array elements such that in one array all the elements are distinct and in the other, all elements are the same
Note: It is not mandatory to use all the elements of the given array to build the two arrays.
Examples: 

Input: a[] = { 4, 2, 4, 1, 4, 3, 4 } 
Output:
Explanation: 
The maximum possible size would be 3 – {1 2 3} and {4 4 4}
Input: a[] = { 2, 1, 5, 4, 3 } 
Output:
Explanation: 
The maximum possible size would be 1 – {1} and {2} 
 

Approach: To solve the problem mentioned above follow the steps given below: 

  • At first count, the frequency of all the numbers in the given array and also count the number of distinct elements in that array. Then count the maximum frequency among all of them maxfrq
  • It is clear, that the maximum possible size of one array(that contains the same elements) would be maxfrq because it is the highest possible value of frequency of a number. And, the maximum possible size of another array(that contains all distinct elements) would be dist which is the total number of distinct elements. Now we have to consider 2 cases: 
    1. If we take all the elements having frequency maxfrq in one array(that contains the same elements), then the total number of distinct elements left would be dist – 1, so the other array can only contain dist – 1 number of elements. In this case, the answer would be, ans1 = min(dist – 1, maxfrq)
    2. If we take all the elements having frequency maxfrq except one element in one array(that contains the same elements), then the total number of distinct elements left would be dist, so the other array can only contain dist number of elements. In this case, the answer would be, ans2 = min(dist, maxfrq – 1).
  • Hence, the actual answer is 

max( min(dist – 1, maxfrq), min(dist, maxfrq – 1) )

  • Below is the implementation of the above approach: 
     

C++




// C++ implementation to Divide the array
// into two arrays having same size,
// one with distinct elements
// and other with same elements
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the max size possible
int findMaxSize(int a[], int n)
{
    vector<int> frq(n + 1);
 
    for (int i = 0; i < n; ++i)
        frq[a[i]]++;
 
    // Counting the maximum frequency
    int maxfrq
        = *max_element(
            frq.begin(), frq.end());
 
    // Counting total distinct elements
    int dist
        = n + 1 - count(
                      frq.begin(),
                      frq.end(), 0);
 
    int ans1 = min(maxfrq - 1, dist);
 
    int ans2 = min(maxfrq, dist - 1);
 
    // Find max of both the answer
    int ans = max(ans1, ans2);
 
    return ans;
}
 
// Driver code
int main()
{
 
    int arr[] = { 4, 2, 4, 1, 4, 3, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << findMaxSize(arr, n);
 
    return 0;
}

Java




// Java implementation to Divide the array
// into two arrays having same size,
// one with distinct elements
// and other with same elements
import java.io.*;
import java.util.*;
 
class GFG {
     
// Function to find the max size possible
static int findMaxSize(int a[], int n)
{
    ArrayList<Integer> frq = new ArrayList<Integer>(n+1);
     
    for(int i = 0; i <= n; i++)
       frq.add(0);
        
    for(int i = 0; i < n; ++i)
       frq.set(a[i], frq.get(a[i]) + 1);
        
    // Counting the maximum frequency
    int maxfrq = Collections.max(frq);
 
    // Counting total distinct elements
    int dist = n + 1 - Collections.frequency(frq, 0);
     
    int ans1 = Math.min(maxfrq - 1, dist);
    int ans2 = Math.min(maxfrq, dist - 1);
 
    // Find max of both the answer
    int ans = Math.max(ans1, ans2);
 
    return ans;
}
     
// Driver code
public static void main(String[] args)
{
    int arr[] = { 4, 2, 4, 1, 4, 3, 4 };
    int n = arr.length;
     
    System.out.println(findMaxSize(arr, n));
}
}
 
// This code is contributed by coder001

Python3




# Python3 implementation to divide the 
# array into two arrays having same 
# size, one with distinct elements
# and other with same elements
 
# Function to find the max size possible
def findMaxSize(a, n):
 
    frq = [0] * (n + 1)
 
    for i in range(n):
        frq[a[i]] += 1
 
    # Counting the maximum frequency
    maxfrq = max(frq)
         
    # Counting total distinct elements
    dist = n + 1 - frq.count(0)
    ans1 = min(maxfrq - 1, dist)
    ans2 = min(maxfrq, dist - 1)
 
    # Find max of both the answer
    ans = max(ans1, ans2)
 
    return ans
 
# Driver code
arr = [ 4, 2, 4, 1, 4, 3, 4 ]
n = len(arr)
 
print(findMaxSize(arr, n))
 
# This code is contributed by divyamohan123

C#




// C# implementation to Divide the array
// into two arrays having same size,
// one with distinct elements
// and other with same elements
using System;
using System.Collections;
class GFG{
      
// Function to find the max size possible
static int findMaxSize(int []a, int n)
{
  ArrayList frq = new ArrayList(n + 1);
 
  for(int i = 0; i <= n; i++)
    frq.Add(0);
 
  for(int i = 0; i < n; ++i)
    frq[a[i]] = (int)frq[a[i]] + 1;
 
  // Counting the maximum frequency
  int maxfrq = int.MinValue;
  for(int i = 0; i < frq.Count; i++)
  {
    if(maxfrq < (int)frq[i])
    {
      maxfrq = (int)frq[i];
    }
  }
 
  // Counting total distinct elements
  int dist = n + 1;
  for(int i = 0; i < frq.Count; i++)
  {
    if((int)frq[i] == 0)
    {
      dist--;
    }
  }
 
  int ans1 = Math.Min(maxfrq - 1, dist);
  int ans2 = Math.Min(maxfrq, dist - 1);
 
  // Find max of both the answer
  int ans = Math.Max(ans1, ans2);
 
  return ans;
}
      
// Driver code
public static void Main(string[] args)
{
  int []arr = {4, 2, 4, 1, 4, 3, 4};
  int n = arr.Length;
  Console.Write(findMaxSize(arr, n));
}
}
 
// This code is contributed by Rutvik_56
Output: 
3

 

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