As we know that every number can be represented as sum(or difference) of powers of 2, therefore what we can do is represent the constant as a sum of powers of 2.
For this purpose we can use the bitwise left shift operator. When a number is bitwise left shifted it is multiplied by 2 for every bit shift.
For example, suppose we want to multiply a variable say “a” by 10 then what we can do is
a = a << 3 + a << 1;
The expression a << 3 multiplies a by 8 ans expression a<<1 multiplies it by 2.
So basically what we have here is a = a*8 + a*2 = a*10
Similarly for multiplying with 7 what we can do is
a = a<<3 - a; or a = a<<2 + a<<1 + a;
Both these statements multiply a by 7.
C++
#include<iostream>using namespace std;// Returns n * 7int multiplyBySeven(int n){ // OR (n << 2) + (n << 1) + n return (n << 3) - n;}// Returns n * 12int multiplyByTwelve(int n){ return (n << 3) + (n << 2);}int main(){ cout << multiplyBySeven(5) << endl; cout << multiplyByTwelve(5) << endl; return 0;} |
Java
class GFG { // Returns n * 7 static int multiplyBySeven(int n) { // OR (n << 2) + (n << 1) + n return (n << 3) - n; } // Returns n * 12 static int multiplyByTwelve(int n) { return (n << 3) + (n << 2); } // Driver code public static void main(String[] args) { System.out.println(multiplyBySeven(5)); System.out.println(multiplyByTwelve(5)); }}// This code is contributed by Anant Agarwal. |
Python3
# Python3 program to Multiplying a# variable with a constant# Returns n * 7def multiplyBySeven(n): # OR (n << 2) + (n << 1) + n return (n << 3) - n# Returns n * 12def multiplyByTwelve(n): return (n << 3) + (n << 2) # Driver codeprint(multiplyBySeven(5))print(multiplyByTwelve(5))# This code is contributed by Anant Agarwal. |
C#
// C# program to Multiplying a// variable with a constantusing System;class GFG{ // Returns n * 7 static int multiplyBySeven(int n) { // OR (n << 2) + (n << 1) + n return (n << 3) - n; } // Returns n * 12 static int multiplyByTwelve(int n) { return (n << 3) + (n << 2); } // Driver code public static void Main() { Console.WriteLine(multiplyBySeven(5)); Console.WriteLine(multiplyByTwelve(5)); }}// This code is contributed by Anant Agarwal. |
PHP
<?php// PHP program of multiply operator // Returns n * 7function multiplyBySeven($n){ return ($n << 3) - $n;}// Returns n * 12function multiplyByTwelve($n){ return ($n << 3) + ($n << 2);}// Driver Codeecho multiplyBySeven(5), "\n";echo multiplyByTwelve(5), "\n";// This code is contributed by Ajit?> |
Javascript
<script>// Python3 program to Multiplying a // variable with a constant// Returns n * 7 function multiplyBySeven(n) { // OR (n << 2) + (n << 1) + n return (n << 3) - n; } // Returns n * 12 function multiplyByTwelve(n) { return (n << 3) + (n << 2); } // Driver code document.write(multiplyBySeven(5) + "<br>"); document.write(multiplyByTwelve(5) + "<br>"); // This code is contributed by Surbhi Tyagi.</script> |
Output :
35 60
Time Complexity: O(1)
Auxiliary Space: O(1)
We just need to find the combination of powers of 2. Also, this comes really handy when we have a very large dataset and each one of them requires multiplication with the same constant as bitwise operators are faster as compared to mathematical operators.
This article is contributed by kp93. If you like neveropen and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
