Multiplication with a power of 2

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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 
?> 

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 
?> 

Output :

Time complexity : O(1)

My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

C++

Java

Python3

C#

PHP

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