Multiply two integers without using multiplication, division and bitwise operators, and no loops

By making use of recursion, we can multiply two integers with the given constraints.

To multiply x and y, recursively add x y times.

C

#include<stdio.h>
/* function to multiply two numbers x and y*/
int multiply(int x, int y)
{
   /* 0  multiplied with anything gives 0 */
   if(y == 0)
     return 0;

   /* Add x one by one */ 
   if(y > 0 )
     return (x + multiply(x, y-1));
 
  /* the case where y is negative */ 
   if(y < 0 )
     return -multiply(x, -y);
}

int main()
{
  printf("\n %d", multiply(5, -11));
  getchar();
  return 0;
}

Java

class GFG {
    
    /* function to multiply two numbers x and y*/
    static int multiply(int x, int y) {
        
        /* 0 multiplied with anything gives 0 */
        if (y == 0)
            return 0;
    
        /* Add x one by one */
        if (y > 0)
            return (x + multiply(x, y - 1));
    
        /* the case where y is negative */
        if (y < 0)
            return -multiply(x, -y);
            
        return -1;
    }
    
    // Driver code
    public static void main(String[] args) {
        
        System.out.print("\n" + multiply(5, -11));
    }
}

// This code is contributed by Anant Agarwal.

Python3

# Function to multiply two numbers
# x and y
def multiply(x,y):

    # 0 multiplied with anything
    # gives 0 
    if(y == 0):
        return 0

    # Add x one by one 
    if(y > 0 ):
        return (x + multiply(x, y - 1))

    # The case where y is negative
    if(y < 0 ):
        return -multiply(x, -y)
    
# Driver code
print(multiply(5, -11))

# This code is contributed by Anant Agarwal.

C#


// Multiply two integers without
// using multiplication, division
// and bitwise operators, and no
// loops
using System;

class GFG {
    
    // function to multiply two numbers
    // x and y
    static int multiply(int x, int y) {
        
        // 0 multiplied with anything gives 0
        if (y == 0)
            return 0;
    
        // Add x one by one
        if (y > 0)
            return (x + multiply(x, y - 1));
    
        // the case where y is negative
        if (y < 0)
            return -multiply(x, -y);
            
        return -1;
    }
    
    // Driver code
    public static void Main() {
        
        Console.WriteLine(multiply(5, -11));
    }
}

// This code is contributed by vt_m.


Output:

-55

Time Complexity: O(y) where y is the second argument to function multiply().

Russian Peasant (Multiply two numbers using bitwise operators)



Please write comments if you find any of the above code/algorithm incorrect, or find better ways to solve the same problem.




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