C Program to find LCM of two numbers using Recursion

Given two integers N and M, the task is to find their LCM using recursion.

Examples:

Input: N = 2, M = 4
Output: 4
Explanation: LCM of 2, 4 is 4.

Input: N = 3, M = 5
Output: 15
Explanation: LCM of 3, 5 is 15.

Approach: The idea is to use the basic elementary method of finding LCM of two numbers. Follow the steps below to solve the problem:

• Define a recursive function LCM() with 3 integer parameters N, M, and K to find LCM of N and M.
• The following base conditions need to be considered:
  • If N or M is equal to 1, return N * M.
  • If N is equal to M, return N.
• If K
  • If K divides both N and M, return K * LCM(N/K, M/K, 2).
  • Otherwise, increment K by 1 and return LCM(N, M, K+1).
• Otherwise, return the product of N and M.
• Finally, print the result of the recursive function as the required LCM.

Below is the implementation of the above approach:

C
// C program for the above approach

#include <stdio.h>

// Function to return the
// minimum of two numbers
int Min(int Num1, int Num2)
{
    return Num1 >= Num2
               ? Num2
               : Num1;
}

// Utility function to calculate LCM
// of two numbers using recursion
int LCMUtil(int Num1, int Num2, int K)
{
    // If either of the two numbers
    // is 1, return their product
    if (Num1 == 1 || Num2 == 1)
        return Num1 * Num2;

    // If both the numbers are equal
    if (Num1 == Num2)
        return Num1;

    // If K is smaller than the
    // minimum of the two numbers
    if (K <= Min(Num1, Num2)) {

        // Checks if both numbers are
        // divisible by K or not
        if (Num1 % K == 0 && Num2 % K == 0) {

            // Recursively call LCM() function
            return K * LCMUtil(
                           Num1 / K, Num2 / K, 2);
        }

        // Otherwise
        else
            return LCMUtil(Num1, Num2, K + 1);
    }

    // If K exceeds minimum
    else
        return Num1 * Num2;
}

// Function to calculate LCM
// of two numbers
void LCM(int N, int M)
{
    // Stores LCM of two number
    int lcm = LCMUtil(N, M, 2);

    // Print LCM
    printf("%d", lcm);
}

// Driver Code
int main()
{
    // Given N & M
    int N = 2, M = 4;

    // Function Call
    LCM(N, M);

    return 0;
}
Output:

4

Time Complexity: O(max(N, M))
Auxiliary Space: O(1)