Multiplication with a power of 2

Difficulty Level : Medium
Last Updated : 07 Apr, 2021

Given two numbers x and n, we need to multiply x with 2ⁿ
Examples :

Input  : x = 25, n = 3
Output : 200
25 multiplied by 2 raised to power 3
is 200.

Input : x = 70, n = 2
Output : 280

A simple solution is to compute n-th power of 2 and then multiply with x.

C++

// Simple C/C++ program
// to compute x * (2^n)
#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
 
// Returns 2 raised to power n
ll power2(ll n)
{
    if (n == 0)
        return 1;
         
    if (n == 1)
        return 2;
         
    return power2(n / 2) *
                    power2(n / 2);
}
 
ll multiply(ll x, ll n)
{
    return x * power2(n);
}
 
// Driven program
int main()
{
    ll x = 70, n = 2;
    cout<<multiply(x, n);
    return 0;
}

Java

// Simple Java program
// to compute x * (2^n)
import java.util.*;
 
class GFG {
     
    // Returns 2 raised to power n
    static long power2(long n)
    {
        if (n == 0)
            return 1;
             
        if (n == 1)
            return 2;
             
        return power2(n / 2)
                          * power2(n / 2);
    }
      
    static long multiply(long x, long n)
    {
        return x * power2(n);
    }
     
    /* Driver program */
    public static void main(String[] args)
    {
        long x = 70, n = 2;
         
        System.out.println(multiply(x, n));
    }
}
     
// This code is contributed by Arnav Kr. Mandal.   

Python3

# Simple Python program
# to compute x * (2^n)
 
# Returns 2 raised to power n
def power2(n):
 
    if (n == 0):
        return 1
    if (n == 1):
        return 2
    return power2(n / 2) *
                  power2(n / 2);
 
 
def multiply(x, n):
    return x * power2(n);
 
 
# Driven program
x = 70
n = 2
print(multiply(x, n))
 
# This code is contributed by Smitha Dinesh Semwal

C#

// Simple C# program
// to compute x * (2^n)
using System;
 
class GFG {
     
    // Returns 2 raised to power n
    static long power2(long n)
    {
        if (n == 0)
            return 1;
             
        if (n == 1)
            return 2;
             
        return power2(n / 2)
                        * power2(n / 2);
    }
     
    static long multiply(long x, long n)
    {
        return x * power2(n);
    }
     
    /* Driver program */
    public static void Main()
    {
        long x = 70, n = 2;
         
        Console.WriteLine(multiply(x, n));
    }
}
     
// This code is contributed by Vt_m.

PHP

<?php
// Simple PHP program
// to compute x * (2^n)
 
// Returns 2 raised to power n
function power2($n)
{
    if ($n == 0)
        return 1;
         
    if ($n == 1)
        return 2;
         
    return power2($n / 2) *
           power2($n / 2);
}
 
function multiply( $x, $n)
{
    return $x * power2($n);
}
 
// Driver Code
$x = 70; $n = 2;
echo multiply($x, $n);
 
// This code is contributed by ajit
?>

Javascript

<script>
 
// Simple JavaScript program
// to compute x * (2^n)
 
// Returns 2 raised to power n
function power2(n)
{
    if (n == 0)
        return 1;
         
    if (n == 1)
        return 2;
         
    return power2(n / 2) *
        power2(n / 2);
}
 
function multiply( x, n)
{
    return x * power2(n);
}
 
// Driver Code
let x = 70
let n = 2;
document.write( multiply(x, n));
 
// This code is contributed by mohan
 
</script>

Output :

Time complexity : O(Log n)
An efficient solution is to use bitwise leftshift operator. We know 1 << n means 2 raised to power n.

C++

// Efficient C/C++ program to compute x * (2^n)
#include <stdio.h>
typedef long long int ll;
 
ll multiply(ll x, ll n)
{
    return x << n;
}
 
// Driven program to check above function
int main()
{
    ll x = 70, n = 2;
    printf("%lld", multiply(x, n));
    return 0;
}

Java

// JAVA Code for Multiplication with a
// power of 2
import java.util.*;
 
class GFG {
     
    static long multiply(long x, long n)
    {
        return x << n;
    }
     
    /* Driver program to test above function */
    public static void main(String[] args)
    {
        long x = 70, n = 2;
        System.out.println(multiply(x, n));
    }
}
 
//This code is contributed by Arnav Kr. Mandal.

Python3

# Efficient Python3 code to compute x * (2^n)
 
def multiply( x , n ):
    return x << n
     
# Driven code to check above function
x = 70
n = 2
print( multiply(x, n))
 
# This code is contributed by "Sharad_Bhardwaj".

C#

// C# Code for Multiplication with a
// power of 2
using System;
 
class GFG {
     
    static int multiply(int x, int n)
    {
        return x << n;
    }
     
    /* Driver program to test above function */
    public static void Main()
    {
        int x = 70, n = 2;
     
        Console.WriteLine(multiply(x, n));
    }
}
 
//This code is contributed by vt_m.

PHP

<?php
// Efficient PHP program to compute x * (2^n)
 
function multiply($x, $n)
{
    return $x << $n;
}
 
    // Driver Code
    $x = 70;
    $n = 2;
    echo multiply($x, $n);
     
// This code is contributed by ajit
?>

Javascript

<script>
 
// Efficient JavaScript program to compute x * (2^n)
 
function multiply(x, n)
{
    return x << n;
}
 
    // Driver Code
    let x = 70;
    let n = 2;
    document.write(multiply(x, n));
 
// This code is contributed by mohan
 
</script>

Output :

Time complexity : O(1)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes