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 nll 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 ndef 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 = 70n = 2print(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 nfunction 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, $n);// This code is contributed by ajit?> |
Javascript
<script>// Simple JavaScript program// to compute x * (2^n)// Returns 2 raised to power nfunction power2(n){ if (n == 0) return 1; if (n == 1) return 2; return power2(n / 2) * power2(n / 2);}function multiply( x, n){ return x * power2(n);}// Driver Codelet x = 70let n = 2;document.write( multiply(x, n));// This code is contributed by mohan</script> |
Output :
280
Time complexity : O(logn)
Auxiliary Space : O(logn)
An efficient solution is to use bitwise leftshift operator. We know 1 << n means 2 raised to power n.
C++
// Efficient C/C++ program to compute x * (2^n)#include <stdio.h>typedef long long int ll;ll multiply(ll x, ll n){ return x << n;}// Driven program to check above functionint main(){ ll x = 70, n = 2; printf("%lld", multiply(x, n)); return 0;} |
Java
// JAVA Code for Multiplication with a // power of 2 import java.util.*;class GFG { static long multiply(long x, long n) { return x << n; } /* Driver program to test above function */ public static void main(String[] args) { long x = 70, n = 2; System.out.println(multiply(x, n)); }}//This code is contributed by Arnav Kr. Mandal. |
Python3
# Efficient Python3 code to compute x * (2^n)def multiply( x , n ): return x << n # Driven code to check above functionx = 70n = 2print( multiply(x, n))# This code is contributed by "Sharad_Bhardwaj". |
C#
// C# Code for Multiplication with a // power of 2 using System;class GFG { static int multiply(int x, int n) { return x << n; } /* Driver program to test above function */ public static void Main() { int x = 70, n = 2; Console.WriteLine(multiply(x, n)); }}//This code is contributed by vt_m. |
PHP
<?php// Efficient PHP program to compute x * (2^n)function multiply($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 :
280
Time Complexity : O(1)
Auxiliary Space: O(1)
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