# Longest sub-sequence with minimum LCM

Given an array arr[] of length N, the task is to find the length of the longest sub-sequence with minimum possible LCM.

Examples:

Input: arr[] = {1, 3, 1}
Output: 2
{1} and {1} are the subsequences
with the minimum possible LCM.

Input: arr[] = {3, 4, 5, 3, 2, 3}
Output: 1
{2} is the required subsequence.

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

Approach: The minimum possible LCM from the array will be equal to the value of the smallest element in the array. Now, to maximize the length of the resulting subsequence, find the number of elements with a value equal to this smallest value in the array and the count of these elements is the required answer.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the length ` `// of the largest subsequence with ` `// minimum possible LCM ` `int` `maxLen(``int``* arr, ``int` `n) ` `{ ` `    ``// Minimum value from the array ` `    ``int` `min_val = *min_element(arr, arr + n); ` ` `  `    ``// To store the frequency of the ` `    ``// minimum element in the array ` `    ``int` `freq = 0; ` ` `  `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``// If current element is equal ` `        ``// to the minimum element ` `        ``if` `(arr[i] == min_val) ` `            ``freq++; ` `    ``} ` ` `  `    ``return` `freq; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 3, 1 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(``int``); ` ` `  `    ``cout << maxLen(arr, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.Arrays; ` ` `  `class` `GFG  ` `{ ` ` `  `// Function to return the length ` `// of the largest subsequence with ` `// minimum possible LCM ` `static` `int` `maxLen(``int``[] arr, ``int` `n) ` `{ ` `    ``// Minimum value from the array ` `    ``int` `min_val = Arrays.stream(arr).min().getAsInt(); ` ` `  `    ``// To store the frequency of the ` `    ``// minimum element in the array ` `    ``int` `freq = ``0``; ` ` `  `    ``for` `(``int` `i = ``0``; i < n; i++)  ` `    ``{ ` ` `  `        ``// If current element is equal ` `        ``// to the minimum element ` `        ``if` `(arr[i] == min_val) ` `            ``freq++; ` `    ``} ` ` `  `    ``return` `freq; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String []args) ` `{ ` `    ``int` `arr[] = { ``1``, ``3``, ``1` `}; ` `    ``int` `n = arr.length; ` ` `  `    ``System.out.println(maxLen(arr, n)); ` `} ` `} ` ` `  `// This code is contributed by PrinciRaj1992 `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function to return the length  ` `# of the largest subsequence with  ` `# minimum possible LCM  ` `def` `maxLen(arr, n) : ` ` `  `    ``# Minimum value from the array  ` `    ``min_val ``=` `min``(arr);  ` ` `  `    ``# To store the frequency of the  ` `    ``# minimum element in the array  ` `    ``freq ``=` `0``;  ` ` `  `    ``for` `i ``in` `range``(n) : ` ` `  `        ``# If current element is equal  ` `        ``# to the minimum element  ` `        ``if` `(arr[i] ``=``=` `min_val) : ` `            ``freq ``+``=` `1``; ` ` `  `    ``return` `freq;  ` ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `:  ` ` `  `    ``arr ``=` `[ ``1``, ``3``, ``1` `];  ` `     `  `    ``n ``=` `len``(arr);  ` ` `  `    ``print``(maxLen(arr, n));  ` ` `  `# This code is contributed by AnkitRai01 `

## C#

 `// C# implementation of the approach ` `using` `System; ` `using` `System.Linq; ` `     `  `class` `GFG  ` `{ ` ` `  `// Function to return the length ` `// of the largest subsequence with ` `// minimum possible LCM ` `static` `int` `maxLen(``int``[] arr, ``int` `n) ` `{ ` `    ``// Minimum value from the array ` `    ``int` `min_val = arr.Min(); ` ` `  `    ``// To store the frequency of the ` `    ``// minimum element in the array ` `    ``int` `freq = 0; ` ` `  `    ``for` `(``int` `i = 0; i < n; i++)  ` `    ``{ ` ` `  `        ``// If current element is equal ` `        ``// to the minimum element ` `        ``if` `(arr[i] == min_val) ` `            ``freq++; ` `    ``} ` ` `  `    ``return` `freq; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String []args) ` `{ ` `    ``int` `[]arr = { 1, 3, 1 }; ` `    ``int` `n = arr.Length; ` ` `  `    ``Console.WriteLine(maxLen(arr, n)); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

Output:

```2
```

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