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.

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

Output

260

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