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
          



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

C++

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// 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;
}

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Java

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// 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.    

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Python3

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# 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

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C#

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// 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. 

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PHP

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<?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
?>

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Output :

280

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++

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// 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;
}

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Java

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// 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.

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Python3

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# 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".

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C#

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// 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.

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PHP

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<?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
?>

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Output :

280

Time complexity : O(1)



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