最小LCM的最长子序列

给定长度为N的数组arr [] ，任务是找到具有最小可能LCM的最长子序列的长度。

例子：

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.

方法：来自数组的最小LCM等于数组中最小元素的值。现在，要最大化结果子序列的长度，请找到数组中具有等于此最小值的值的元素数量，并且这些元素的计数是必需的答案。

下面是上述方法的实现：

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

输出：

2