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Subsequence pair from given Array having all unique and all same elements respectively
  • Last Updated : 28 Aug, 2020

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

Input: arr[] = {1, 2, 3, 1, 2, 3, 3, 3} 
Output:
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:
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 size of first subsequence cannot exceed d and size of second subsequence cannot exceeds f.
  4. The maximum value of subsequence is given on the basis of below two cases: 
    • The subsequence with distinct elements must have an element with maximum frequency. Therefore minimum of (d, f – 1) is the possible length of subsequence with given property.
    • If the length of a subsequence with distinct element is greater than maximum frequency then, the minimum of (d – 1, f) is the possible length of subsequence with given property.
  5. The maximum value in the above two cases is the maximum length of subsequence with the given property.

Below is the implementation of the above approach:

C++

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// 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
    // elment 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;
}

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Java

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// 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
    // elment 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

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Python3

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# 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
    # elment 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

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C#

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// 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
    // elment 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

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Output: 

4



 

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

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