Given a positive integer N > 1, the task is to find the maximum LCM among all the pairs (i, j) such that i < j ? N.
Examples:
Input: N = 3
Output: 6
LCM(1, 2) = 2
LCM(1, 3) = 3
LCM(2, 3) = 6
Input: N = 4
Output: 12
Approach: Since the LCM of two consecutive elements is equal to their multiples then it is obvious that the maximum LCM will be of the pair (N, N – 1) i.e. (N * (N – 1)).
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Function to return the maximum LCM // among all the pairs(i, j) of // first n natural numbers int maxLCM( int n)
{ return (n * (n - 1));
} // Driver code int main()
{ int n = 3;
cout << maxLCM(n);
return 0;
} |
Java
// Java implementation of the approach class GFG
{ // Function to return the maximum LCM // among all the pairs(i, j) of // first n natural numbers static int maxLCM( int n)
{ return (n * (n - 1 ));
} // Driver code public static void main(String[] args)
{ int n = 3 ;
System.out.println(maxLCM(n));
} } // This code is contributed by Code_Mech |
Python3
# Python3 implementation of the approach # Function to return the maximum LCM # among all the pairs(i, j) of # first n natural numbers def maxLCM(n) :
return (n * (n - 1 ));
# Driver code if __name__ = = "__main__" :
n = 3 ;
print (maxLCM(n));
# This code is contributed by AnkitRai01 |
C#
// C# implementation of the approach using System;
class GFG
{ // Function to return the maximum LCM // among all the pairs(i, j) of // first n natural numbers static int maxLCM( int n)
{ return (n * (n - 1));
} // Driver code public static void Main(String[] args)
{ int n = 3;
Console.WriteLine(maxLCM(n));
} } // This code is contributed by Rajput-Ji |
Javascript
<script> // Javascript implementation of the approach // Function to return the maximum LCM // among all the pairs(i, j) of // first n natural numbers function maxLCM(n)
{ return (n * (n - 1));
} // Driver code var n = 3;
document.write(maxLCM(n)); </script> |
Output:
6
Time Complexity: O(1)
Auxiliary Space: O(1)