给定数组中所有对(i，j)中的最大LCM

给定一个数组arr [] ，任务是在将数组中的元素成对使用时找到最大LCM。

例子：

Input: arr[] = {17, 3, 8, 6} 
Output: 136 
Explanation: 
Respective Pairs with their LCM are: 
{8, 17} has LCM 136, 
{3, 17} has LCM 51, 
{6, 17} has LCM 102, 
{3, 8} has LCM 24, 
{3, 6} has LCM 6, and 
{6, 8} has LCM 24. 
Maximum LCM among these =136.
Input: array[] = {1, 8, 12, 9} 
Output: 72 
Explanation: 
72 is the highest LCM among all the pairs of the given array. 

天真的方法：使用两个循环来生成数组的所有可能的元素对，并计算它们的LCM。每当我们获得更高的价值时，就更新LCM。
时间复杂度： O(N ² )
下面是上述方法的实现：

C++

// C++ implementation to find the maximum
// LCM of pairs in an array
 
#include 
using namespace std;
 
// Function comparing all LCM pairs
int maxLcmOfPairs(int arr[], int n)
{
    // To store the highest LCM
    int maxLCM = -1;
 
    // To generate all pairs from array
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
 
            // Find LCM of the pair
            // Update the maxLCM if this is
            // greater than its existing value
            maxLCM
                = max(maxLCM, (arr[i] * arr[j])
                                  / __gcd(arr[i], arr[j]));
        }
    }
 
    // Return the highest value of LCM
    return maxLCM;
}
 
// Driver code
int main()
{
    int arr[] = { 17, 3, 8, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << maxLcmOfPairs(arr, n);
 
    return 0;
}

Java

// Java implementation to find the maximum
// LCM of pairs in an array
import java.util.*;
class GFG {
 
    // Function comparing all LCM pairs
    static int maxLcmOfPairs(int arr[], int n)
    {
        // To store the highest LCM
        int maxLCM = -1;
 
        // To generate all pairs from array
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
 
                // Find LCM of the pair
                // Update the maxLCM if this is
                // greater than its existing value
                maxLCM = Math.max(
                    maxLCM, (arr[i] * arr[j])
                                / __gcd(arr[i], arr[j]));
            }
        }
 
        // Return the highest value of LCM
        return maxLCM;
    }
 
    static int __gcd(int a, int b)
    {
        return b == 0 ? a : __gcd(b, a % b);
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 17, 3, 8, 6 };
        int n = arr.length;
 
        System.out.print(maxLcmOfPairs(arr, n));
    }
}
 
// This code is contributed by sapnasingh4991

Python3

# Python3 implementation to find the
# maximum LCM of pairs in an array
from math import gcd
 
# Function comparing all LCM pairs
 
 
def maxLcmOfPairs(arr, n):
 
    # To store the highest LCM
    maxLCM = -1
 
    # To generate all pairs from array
    for i in range(n):
        for j in range(i + 1, n, 1):
 
            # Find LCM of the pair
            # Update the maxLCM if this is
            # greater than its existing value
            maxLCM = max(maxLCM, (arr[i] * arr[j]) //
                         gcd(arr[i], arr[j]))
 
    # Return the highest value of LCM
    return maxLCM
 
 
# Driver code
if __name__ == '__main__':
 
    arr = [17, 3, 8, 6]
    n = len(arr)
 
    print(maxLcmOfPairs(arr, n))
 
# This code is contributed by hupendraSingh

C#

// C# implementation to find the maximum
// LCM of pairs in an array
using System;
class GFG {
 
    // Function comparing all LCM pairs
    static int maxLcmOfPairs(int[] arr, int n)
    {
        // To store the highest LCM
        int maxLCM = -1;
 
        // To generate all pairs from array
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
 
                // Find LCM of the pair
                // Update the maxLCM if this is
                // greater than its existing value
                maxLCM = Math.Max(
                    maxLCM, (arr[i] * arr[j])
                                / __gcd(arr[i], arr[j]));
            }
        }
 
        // Return the highest value of LCM
        return maxLCM;
    }
 
    static int __gcd(int a, int b)
    {
        return b == 0 ? a : __gcd(b, a % b);
    }
 
    // Driver code
    public static void Main()
    {
        int[] arr = { 17, 3, 8, 6 };
        int n = arr.Length;
 
        Console.Write(maxLcmOfPairs(arr, n));
    }
}
 
// This code is contributed by Code_Mech

Javascript

// javascript implementation to find the maximum
// LCM of pairs in an array
 
    // Function comparing all LCM pairs
    function maxLcmOfPairs(arr , n)
    {
     
        // To store the highest LCM
        var maxLCM = -1;
 
        // To generate all pairs from array
        for (i = 0; i < n; i++)
        {
            for (j = i + 1; j < n; j++)
            {
 
                // Find LCM of the pair
                // Update the maxLCM if this is
                // greater than its existing value
                maxLCM = Math.max(maxLCM, (arr[i] * arr[j]) / __gcd(arr[i], arr[j]));
            }
        }
 
        // Return the highest value of LCM
        return maxLCM;
    }
 
    function __gcd(a , b) {
        return b == 0 ? a : __gcd(b, a % b);
    }
 
    // Driver code
        var arr = [ 17, 3, 8, 6 ];
        var n = arr.length;
        document.write(maxLcmOfPairs(arr, n));
 
// This code is contributed by umadevi9616

输出

136

另一种方法：
我们可以使用贪婪方法。为了应用贪婪方法，我们必须对给定的数组进行排序，然后比较该数组元素对的LCM，最后计算LCM的最大值。
下面是上述方法的实现：

C++

// C++ implementation to find the maximum
// LCM of pairs in an array
 
#include 
using namespace std;
 
// Function for the highest value of LCM pairs
int greedyLCM(int arr[], int n)
{
    // Sort the given array
    sort(arr, arr + n);
 
    // Compute the highest LCM
    int maxLCM = arr[n - 1];
 
    for (int i = n - 1; i >= 0; i--) {
        if (arr[i] * arr[i] < maxLCM)
            break;
 
        for (int j = i - 1; j >= 0; j--) {
 
            if (arr[i] * arr[j] < maxLCM)
                break;
 
            else
 
                // Find LCM of the pair
                // Update the maxLCM if this is
                // greater than its existing value
                maxLCM = max(maxLCM,
                             (arr[i] * arr[j])
                                 / __gcd(arr[i], arr[j]));
        }
    }
 
    // return the maximum lcm
    return maxLCM;
}
 
// Driver code
int main()
{
    int arr[] = { 17, 3, 8, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << greedyLCM(arr, n);
 
    return 0;
}

Java

// Java implementation to find the
// maximum LCM of pairs in an array
import java.util.*;
 
class GFG {
 
    // Function for the highest value
    // of LCM pairs
    static int greedyLCM(int arr[], int n)
    {
 
        // Sort the given array
        Arrays.sort(arr);
 
        // Compute the highest LCM
        int maxLCM = arr[n - 1];
 
        for (int i = n - 1; i >= 0; i--) {
            if (arr[i] * arr[i] < maxLCM)
                break;
 
            for (int j = i - 1; j >= 0; j--) {
                if (arr[i] * arr[j] < maxLCM)
                    break;
                else
 
                    // Find LCM of the pair
                    // Update the maxLCM if this is
                    // greater than its existing value
                    maxLCM = Math.max(
                        maxLCM,
                        (arr[i] * arr[j])
                            / __gcd(arr[i], arr[j]));
            }
        }
 
        // Return the maximum lcm
        return maxLCM;
    }
 
    static int __gcd(int a, int b)
    {
        return b == 0 ? a : __gcd(b, a % b);
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 17, 3, 8, 6 };
        int n = arr.length;
 
        System.out.print(greedyLCM(arr, n));
    }
}
 
// This code is contributed by Amit Katiyar

Python3

# Python3 implementation to
# find the maximum LCM of
# pairs in an array
from math import gcd
 
# Function for the highest
# value of LCM pairs
 
 
def greedyLCM(arr, n):
 
    # Sort the given array
    arr.sort()
 
    # Compute the highest LCM
    maxLCM = arr[n - 1]
 
    for i in range(n - 1, -1, -1):
        if (arr[i] * arr[i] < maxLCM):
            break
 
        for j in range(i - 1, -1, -1):
            if (arr[i] * arr[j] < maxLCM):
                break
 
            else:
                # Find LCM of the pair
                # Update the maxLCM if this is
                # greater than its existing value
                maxLCM = max(maxLCM,
                             (arr[i] * arr[j]) //
                             gcd(arr[i], arr[j]))
 
    # Return the maximum lcm
    return maxLCM
 
 
# Driver code
arr = [17, 3, 8, 6]
n = len(arr)
 
print(greedyLCM(arr, n))
 
# This code is contributed by divyeshrabadiya07

C＃

// C# implementation to find the
// maximum LCM of pairs in an array
using System;
 
class GFG {
 
    // Function for the highest value
    // of LCM pairs
    static int greedyLCM(int[] arr, int n)
    {
 
        // Sort the given array
        Array.Sort(arr);
 
        // Compute the highest LCM
        int maxLCM = arr[n - 1];
 
        for (int i = n - 1; i >= 0; i--) {
            if (arr[i] * arr[i] < maxLCM)
                break;
 
            for (int j = i - 1; j >= 0; j--) {
                if (arr[i] * arr[j] < maxLCM)
                    break;
                else
 
                    // Find LCM of the pair
                    // Update the maxLCM if this is
                    // greater than its existing value
                    maxLCM = Math.Max(
                        maxLCM,
                        (arr[i] * arr[j])
                            / __gcd(arr[i], arr[j]));
            }
        }
 
        // Return the maximum lcm
        return maxLCM;
    }
 
    static int __gcd(int a, int b)
    {
        return b == 0 ? a : __gcd(b, a % b);
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int[] arr = { 17, 3, 8, 6 };
        int n = arr.Length;
 
        Console.Write(greedyLCM(arr, n));
    }
}
 
// This code is contributed by Amit Katiyar

输出

136

时间复杂度： O(N ² )