# 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 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: 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}.

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

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++

 `// C++ program for the above approach ` ` `  `#include ` `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); ` ` `  `    ``// Function call ` `    ``maximumSubsequence(arr, N); ` `    ``return` `0; ` `} `

## Python 3

 `# 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 `

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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