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C Program to find LCM of two numbers using Recursion

  • Last Updated : 15 Dec, 2020

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 < min(N, M):
    • 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)

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