Subsequence pair from given Array having all unique and all same elements respectively

Given an array arr[] of N integers, the task is to choose the two subsequences of equal lengths such that the first subsequence must have all the unique elements and the second subsequence must have all the same elements. Print the maximum length of the subsequence pair.

Examples:

Input: arr[] = {1, 2, 3, 1, 2, 3, 3, 3} 
Output: 3 
Explanation: 
The first subsequence consists of elements {1, 2, 3}. 
The second subsequence consists of elements {3, 3, 3}.

Input: arr[] = {2, 2, 2, 3} 
Output: 2 
Explanation: 
The first subsequence consists of elements {2, 3}. 
The second subsequence consists of elements {2, 2}.

Approach:

1. Count the maximum frequency(say f) of the element in the given array.
2. Count the distinct elements(say d) present in the array.
3. To make both the subsequence of equal length with the given property, then the size of the first subsequence cannot exceed d, and the size of the second subsequence cannot exceed f.
4. The maximum value of the subsequence is given on the basis of two cases below: 
   
   • The subsequence with distinct elements must have an element with maximum frequency. Therefore, the minimum of (d, f - 1) is the possible length of a subsequence with the given property.
   • If the length of a subsequence with a distinct element is greater than maximum frequency, then, the minimum of (d - 1, f) is the possible length of a subsequence with the given property.
5. The maximum value in the above two cases is the maximum length of a subsequence with the given property.

Below is the implementation of the above approach:

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to find the maximum length
// of subsequences with given property
int maximumSubsequence(int arr[], int N)
{
    // To store the frequency
    unordered_map<int, int> M;

    // Traverse the array to store the
    // frequency
    for (int i = 0; i < N; i++) {
        M[arr[i]]++;
    }

    // M.size() given count of distinct
    // element in arr[]
    int distinct_size = M.size();
    int maxFreq = 1;

    // Traverse map to find max frequency
    for (auto& it : M) {

        maxFreq = max(maxFreq, it.second);
    }

    // Find the maximum length on the basis
    // of two cases in the approach

    cout << max(min(distinct_size, maxFreq - 1),
                min(distinct_size - 1, maxFreq));
}

// Driver Code
int main()
{
    int arr[] = { 1, 2, 3, 4, 4, 4, 4, 4, 5 };
    int N = sizeof(arr) / sizeof(arr[0]);

    // Function call
    maximumSubsequence(arr, N);
    return 0;
}

// Java program for the above approach
import java.util.*;

class GFG{

// Function to find the maximum length
// of subsequences with given property
static void maximumSubsequence(int arr[], int N)
{
    
    // To store the frequency
    HashMap<Integer,
            Integer> M = new HashMap<Integer,
                                     Integer>();

    // Traverse the array to store the
    // frequency
    for(int i = 0; i < N; i++)
    {
        if(M.containsKey(arr[i]))
        {
            M.put(arr[i], M.get(arr[i]) + 1);
        }
        else
        {
            M.put(arr[i], 1);
        }
    }

    // M.size() given count of distinct
    // element in arr[]
    int distinct_size = M.size();
    int maxFreq = 1;

    // Traverse map to find max frequency
    for(Map.Entry<Integer, Integer> it : M.entrySet())
    {
        maxFreq = Math.max(maxFreq, it.getValue());
    }

    // Find the maximum length on the basis
    // of two cases in the approach

    System.out.print(Math.max(
        Math.min(distinct_size, maxFreq - 1),
        Math.min(distinct_size - 1, maxFreq)));
}

// Driver Code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 3, 4, 4, 4, 4, 4, 5 };
    int N = arr.length;

    // Function call
    maximumSubsequence(arr, N);
}
}

// This code is contributed by amal kumar choubey 

# Python 3 program for the above approach

# Function to find the maximum length
# of subsequences with given property
def maximumSubsequence(arr, N):
    
    # To store the frequency
    M = {i : 0 for i in range(100)}

    # Traverse the array to store the
    # frequency
    for i in range(N):
        M[arr[i]] += 1

    # M.size() given count of distinct
    # element in arr[]
    distinct_size = len(M)
    maxFreq = 1

    # Traverse map to find max frequency
    for value in M.values():
        maxFreq = max(maxFreq, value)

    # Find the maximum length on the basis
    # of two cases in the approach

    print(max(min(distinct_size,  maxFreq - 1), 
          min(distinct_size - 1, maxFreq)))

# Driver Code
if __name__ == '__main__':
    
    arr = [ 1, 2, 3, 4, 4, 4, 4, 5 ]
    N = len(arr)

    # Function call
    maximumSubsequence(arr, N)

# This code is contributed by Samarth

// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{

// Function to find the maximum length
// of subsequences with given property
static void maximumSubsequence(int []arr, int N)
{
    
    // To store the frequency
    Dictionary<int,
               int> M = new Dictionary<int,
                                       int>();

    // Traverse the array to store the
    // frequency
    for(int i = 0; i < N; i++)
    {
        if(M.ContainsKey(arr[i]))
        {
            M[arr[i]] =  M[arr[i]] + 1;
        }
        else
        {
            M.Add(arr[i], 1);
        }
    }

    // M.Count given count of distinct
    // element in []arr
    int distinct_size = M.Count;
    int maxFreq = 1;

    // Traverse map to find max frequency
    foreach(KeyValuePair<int, int> m in M)
    {
        maxFreq = Math.Max(maxFreq, m.Value);
    }

    // Find the maximum length on the basis
    // of two cases in the approach

    Console.Write(Math.Max(
        Math.Min(distinct_size, maxFreq - 1),
        Math.Min(distinct_size - 1, maxFreq)));
}

// Driver Code
public static void Main(String[] args)
{
    int []arr = { 1, 2, 3, 4, 4, 4, 4, 4, 5 };
    int N = arr.Length;

    // Function call
    maximumSubsequence(arr, N);
}
}

// This code is contributed by Rohit_ranjan 

<script>
// Javascript program for the above approach

// Function to find the maximum length
// of subsequences with given property
function maximumSubsequence(arr, N) {
    // To store the frequency
    let M = new Map();

    // Traverse the array to store the
    // frequency
    for (let i = 0; i < N; i++) {
        if (M.has(arr[i])) {
            M.set(arr[i], M.get(arr[i]) + 1);
        }
        else {
            M.set(arr[i], 1);
        }
    }

    // M.size() given count of distinct
    // element in arr[]
    let distinct_size = M.size;
    let maxFreq = 1;

    // Traverse map to find max frequency
    for (let it of M) {

        maxFreq = Math.max(maxFreq, it[1]);
    }

    // Find the maximum length on the basis
    // of two cases in the approach

    document.write(Math.max(Math.min(distinct_size, maxFreq - 1), Math.min(distinct_size - 1, maxFreq)));
}

// Driver Code

let arr = [1, 2, 3, 4, 4, 4, 4, 4, 5];
let N = arr.length;

// Function call
maximumSubsequence(arr, N);

// This code is contributed by _saurabh_jaiswal
</script>

4

Time Complexity: O(N), where N is the number of elements in the array. 
Auxiliary Space Complexity: O(N), where N is the number of elements in the array.