Subsecuencia más larga con LCM mínimo

Dada una array arr[] de longitud N , la tarea es encontrar la longitud de la subsecuencia más larga con el mínimo LCM posible .
Ejemplos: 
 

Entrada: arr[] = {1, 3, 1} 
Salida: 2 
{1} y {1} son las subsecuencias 
con el mínimo LCM posible.
Entrada: arr[] = {3, 4, 5, 3, 2, 3} 
Salida: 1 
{2} es la subsecuencia requerida. 
 

Enfoque: El LCM mínimo posible de la array será igual al valor del elemento más pequeño de la array. Ahora, para maximizar la longitud de la subsecuencia resultante, busque el número de elementos con un valor igual a este valor más pequeño en la array y el recuento de estos elementos es la respuesta requerida.
A continuación se muestra la implementación del enfoque anterior: 
 

// C++ implementation of the approach
#include <bits/stdc++.h>
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 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 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# 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

<script>
 
// Javascript implementation of the approach
 
// Function to return the length
// of the largest subsequence with
// minimum possible LCM
function maxLen(arr, n)
{
    // Minimum value from the array
    var min_val = arr.reduce((a, b) => Math.min(a,b))
 
    // To store the frequency of the
    // minimum element in the array
    var freq = 0;
 
    for (var i = 0; i < n; i++) {
 
        // If current element is equal
        // to the minimum element
        if (arr[i] == min_val)
            freq++;
    }
 
    return freq;
}
 
// Driver code
var arr = [ 1, 3, 1 ];
var n = arr.length;
document.write( maxLen(arr, n));
 
// This code is contributed by itsok.
</script>

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