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++
#include<bits/stdc++.h>
using namespace std;
int multiply( int n, int m)
{
int ans = 0, count = 0;
while (m)
{
if (m % 2 == 1)
ans += n << count;
count++;
m /= 2;
}
return ans;
}
int main()
{
int n = 20 , m = 13;
cout << multiply(n, m);
return 0;
}
|
Java
class GFG
{
static int multiply( int n, int m)
{
int ans = 0 , count = 0 ;
while (m > 0 )
{
if (m % 2 == 1 )
ans += n << count;
count++;
m /= 2 ;
}
return ans;
}
public static void main (String[] args)
{
int n = 20 , m = 13 ;
System.out.print( multiply(n, m) );
}
}
|
Python3
def multiply(n, m):
ans = 0
count = 0
while (m):
if (m % 2 = = 1 ):
ans + = n << count
count + = 1
m = int (m / 2 )
return ans
if __name__ = = '__main__' :
n = 20
m = 13
print (multiply(n, m))
|
C#
using System;
class GFG
{
static int multiply( int n, int m)
{
int ans = 0, count = 0;
while (m > 0)
{
if (m % 2 == 1)
ans += n << count;
count++;
m /= 2;
}
return ans;
}
public static void Main ()
{
int n = 20, m = 13;
Console.WriteLine( multiply(n, m) );
}
}
|
PHP
<?php
function multiply( $n , $m )
{
$ans = 0; $count = 0;
while ( $m )
{
if ( $m % 2 == 1)
$ans += $n << $count ;
$count ++;
$m /= 2;
}
return $ans ;
}
$n = 20 ; $m = 13;
echo multiply( $n , $m );
?>
|
Javascript
<script>
function multiply(n, m)
{
let ans = 0, count = 0;
while (m)
{
if (m % 2 == 1)
ans += n << count;
count++;
m = Math.floor(m / 2);
}
return ans;
}
let n = 20 , m = 13;
document.write(multiply(n, m));
</script>
|
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;
}
n = n << 1;
m = m >> 1;
}
return (isNegative) ? -result : result;
}
int main()
{
int n = 20 , m = 13;
cout << multiply(n, m);
return 0;
}
|
Java
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;
}
n = n << 1 ;
m = m >> 1 ;
}
return (isNegative) ? -result : result;
}
public static void main (String[] args) {
int n = 20 , m = 13 ;
System.out.println(multiply(n, m));
}
}
|
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
n = n << 1
m = m >> 1
return - result if is_negative else result
n = 20
m = 13
print (multiply(n, m))
|
C#
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;
}
n = n << 1;
m = m >> 1;
}
return (isNegative) ? -result : result;
}
public static void Main () {
int n = 20 , m = 13;
Console.WriteLine(multiply(n, m));
}
}
|
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;
}
n = n << 1;
m = m >> 1;
}
return (isNegative) ? -result : result;
}
console.log(multiply(20, 13));
|
Time Complexity : O(log(m))
Auxiliary Space: O(1)
Related Article:
Russian Peasant (Multiply two numbers using bitwise operators)
Last Updated :
12 Sep, 2023
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