Maximum LCM among all pairs (i, j) from the given Array

Given an array arr[], the task is to find the maximum LCM when the elements of the array are taken in pairs.

Examples:

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.

Naive Approach: Use two loops to generate all possible pairs of elements of the array and calculate LCM of them. Update the LCM whenever we get a higher value.



Time Complexity: O(N2)

Below is the implementation of the above approach:

C++

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// C++ implementation to find the maximum
// LCM of pairs in an array
  
#include <bits/stdc++.h>
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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

136

Efficient Approach:

To optimize the above method we can use Greedy Method. For applying the greedy approach we have to sort the given array and then comparing LCM of pairs of elements of the array and finally compute the maximum value of LCM.

Below is the implementation of the above approach:

C++

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// C++ implementation to find the maximum
// LCM of pairs in an array
  
#include <bits/stdc++.h>
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;
}

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Java

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// 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

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C#

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// 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

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Output:

136

Time Complexity: O(N * log(N))

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