Multiplication of two numbers with shift operator

For any given two numbers n and m, you have to find n*m without using any multiplication operator. 
Examples :

Input: n = 25 , m = 13
Output: 325

Input: n = 50 , m = 16
Output: 800

Method 1
We can solve this problem with the shift operator. The idea is based on the fact that every number can be represented in binary form. And multiplication with a number is equivalent to multiplication with powers of 2. Powers of 2 can be obtained using left shift operator.
Check for every set bit in the binary representation of m and for every set bit left shift n, count times where count if place value of the set bit of m and add that value to answer.

C++

// CPP program to find multiplication
// of two number without use of
// multiplication operator
#include<bits/stdc++.h>
using namespace std;

// Function for multiplication
int multiply(int n, int m)
{
    int ans = 0, count = 0;
    while (m)
    {
        // check for set bit and left
        // shift n, count times
        if (m % 2 == 1)
            ans += n << count;

        // increment of place value (count)
        count++;
        m /= 2;
    }
    return ans;
}

// Driver code
int main()
{
    int n = 20 , m = 13;
    cout << multiply(n, m);
    return 0;
}

Java

// Java program to find multiplication
// of two number without use of
// multiplication operator
class GFG
{

    // Function for multiplication
    static int multiply(int n, int m)
    {
        int ans = 0, count = 0;
        while (m > 0)
        {
            // check for set bit and left
            // shift n, count times
            if (m % 2 == 1)
                ans += n << count;

            // increment of place
            // value (count)
            count++;
            m /= 2;
        }

        return ans;
    }

    // Driver code
    public static void main (String[] args)
    {
        int n = 20, m = 13;

        System.out.print( multiply(n, m) );
    }
}

// This code is contributed by Anant Agarwal.

Python3

# python 3 program to find multiplication
# of two number without use of
# multiplication operator

# Function for multiplication
def multiply(n, m):
    ans = 0
    count = 0
    while (m):
        # check for set bit and left
        # shift n, count times
        if (m % 2 == 1):
            ans += n << count

        # increment of place value (count)
        count += 1
        m = int(m/2)

    return ans

# Driver code
if __name__ == '__main__':
    n = 20
    m = 13
    print(multiply(n, m))

# This code is contributed by
# Ssanjit_Prasad

C#

// C# program to find multiplication
// of two number without use of
// multiplication operator
using System;

class GFG
{

    // Function for multiplication
    static int multiply(int n, int m)
    {
        int ans = 0, count = 0;
        while (m > 0)
        {
            // check for set bit and left
            // shift n, count times
            if (m % 2 == 1)
                ans += n << count;

            // increment of place
            // value (count)
            count++;
            m /= 2;
        }

        return ans;
    }

    // Driver Code
    public static void Main ()
    {
        int n = 20, m = 13;

        Console.WriteLine( multiply(n, m) );
    }
}

// This code is contributed by vt_m.

PHP

<?php
// PHP program to find multiplication
// of two number without use of
// multiplication operator

// Function for multiplication
function multiply( $n, $m)
{
    $ans = 0; $count = 0;
    while ($m)
    {
        // check for set bit and left
        // shift n, count times
        if ($m % 2 == 1)
            $ans += $n << $count;

        // increment of place value (count)
        $count++;
        $m /= 2;
    }
    return $ans;
}

// Driver code
$n = 20 ; $m = 13;
echo multiply($n, $m);

// This code is contributed by anuj_67.
?>

Javascript

<script>

// JavaScript program to find multiplication
// of two number without use of
// multiplication operator

// Function for multiplication
function multiply(n, m)
{
    let ans = 0, count = 0;
    while (m)
    {
        // check for set bit and left
        // shift n, count times
        if (m % 2 == 1)
            ans += n << count;

        // increment of place value (count)
        count++;
        m = Math.floor(m / 2);
    }
    return ans;
}

// Driver code

    let n = 20 , m = 13;
    document.write(multiply(n, m));

// This code is contributed by Surbhi Tyagi.

</script>

Output

260

Time Complexity : O(log n)

Auxiliary Space: O(1)

Method 2

We can use shift operator in loops.

C++

#include <iostream>
using namespace std;

int multiply(int n, int m){
    bool isNegative = false;
    if (n < 0 && m < 0) {
        n = -n, m = -m;
    }
    if (n < 0) {
        n = -n, isNegative = true;
    }
    if (m < 0) {
        m = -m, isNegative = true;
    }
    int result = 0;
    while (m){
        if (m & 1) {
            result += n;
        }
        // multiply a by 2
        n = n << 1;
        // divide b by 2
        m = m >> 1;
    }
    return (isNegative) ? -result : result;
}

int main()
{
    int n = 20 , m = 13;
    cout << multiply(n, m);
    return 0;
}

Java

// Java program for the above approach

import java.io.*;

class GFG {

    public static int multiply(int n, int m){
    boolean isNegative = false;
    if (n < 0 && m < 0) {
        n = -n;
        m = -m;
    }
    if (n < 0) {
        n = -n;
        isNegative = true;
    }
    if (m < 0) {
        m = -m;
        isNegative = true;
    }
    int result = 0;
    while (m>0){
        if ((m & 1)!=0) {
            result += n;
        }
        // multiply a by 2
        n = n << 1;
        // divide b by 2
        m = m >> 1;
    }
    return (isNegative) ? -result : result;
}

    public static void main (String[] args) {
            int n = 20 , m = 13;
            System.out.println(multiply(n, m));
    }
}

// This code is contributed by Pushpesh Raj.

Python3

def multiply(n, m):
    is_negative = False
    if n < 0 and m < 0:
        n, m = -n, -m
    if n < 0:
        n, is_negative = -n, True
    if m < 0:
        m, is_negative = -m, True
    result = 0
    while m:
        if m & 1:
            result += n
        # multiply a by 2
        n = n << 1
        # divide b by 2
        m = m >> 1
    return -result if is_negative else result


n = 20
m = 13
print(multiply(n, m))

C#

// C# program for the above approach
using System;

class GFG {

    public static int multiply(int n, int m){
    bool isNegative = false;
    if (n < 0 && m < 0) {
        n = -n;
        m = -m;
    }
    if (n < 0) {
        n = -n;
        isNegative = true;
    }
    if (m < 0) {
        m = -m;
        isNegative = true;
    }
    int result = 0;
    while (m>0){
        if ((m & 1)!=0) {
            result += n;
        }
        // multiply a by 2
        n = n << 1;
        // divide b by 2
        m = m >> 1;
    }
    return (isNegative) ? -result : result;
}

    public static void Main () {
            int n = 20 , m = 13;
            Console.WriteLine(multiply(n, m));
    }
}

// This code is contributed by Utkarsh

Javascript

function multiply(n, m) {
    let isNegative = false;
    if (n < 0 && m < 0) {
        n = -n, m = -m;
    }
    if (n < 0) {
        n = -n, isNegative = true;
    }
    if (m < 0) {
        m = -m, isNegative = true;
    }
    let result = 0;
    while (m) {
        if (m & 1) {
            result += n;
        }
        // multiply a by 2
        n = n << 1;
        // divide b by 2
        m = m >> 1;
    }
    return (isNegative) ? -result : result;
}

console.log(multiply(20, 13));

Output

260

Time Complexity : O(log(m))

Auxiliary Space: O(1)

This article is contributed by Shivam Pradhan. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.