# Multiplication of two numbers with shift operator

• Difficulty Level : Medium
• Last Updated : 21 Jun, 2022

For any given two numbers n and m, you have to find n*m without using any multiplication operator.
Examples :

```Input: n = 25 , m = 13
Output: 325

Input: n = 50 , m = 16
Output: 800```

Method 1
We can solve this problem with the shift operator. The idea is based on the fact that every number can be represented in binary form. And multiplication with a number is equivalent to multiplication with powers of 2. Powers of 2 can be obtained using left shift operator.
Check for every set bit in the binary representation of m and for every set bit left shift n, count times where count if place value of the set bit of m and add that value to answer.

## C++

 `// CPP program to find multiplication``// of two number without use of``// multiplication operator``#include``using` `namespace` `std;` `// Function for multiplication``int` `multiply(``int` `n, ``int` `m)``{ ``    ``int` `ans = 0, count = 0;``    ``while` `(m)``    ``{``        ``// check for set bit and left``        ``// shift n, count times``        ``if` `(m % 2 == 1)             ``            ``ans += n << count;` `        ``// increment of place value (count)``        ``count++;``        ``m /= 2;``    ``}``    ``return` `ans;``}` `// Driver code``int` `main()``{``    ``int` `n = 20 , m = 13;``    ``cout << multiply(n, m);``    ``return` `0;``}`

## Java

 `// Java program to find multiplication``// of two number without use of``// multiplication operator``class` `GFG``{``    ` `    ``// Function for multiplication``    ``static` `int` `multiply(``int` `n, ``int` `m)``    ``{``        ``int` `ans = ``0``, count = ``0``;``        ``while` `(m > ``0``)``        ``{``            ``// check for set bit and left``            ``// shift n, count times``            ``if` `(m % ``2` `== ``1``)            ``                ``ans += n << count;``    ` `            ``// increment of place``            ``// value (count)``            ``count++;``            ``m /= ``2``;``        ``}``        ` `        ``return` `ans;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `main (String[] args)``    ``{``        ``int` `n = ``20``, m = ``13``;``        ` `        ``System.out.print( multiply(n, m) );``    ``}``}` `// This code is contributed by Anant Agarwal.`

## Python3

 `# python 3 program to find multiplication``# of two number without use of``# multiplication operator` `# Function for multiplication``def` `multiply(n, m):``    ``ans ``=` `0``    ``count ``=` `0``    ``while` `(m):``        ``# check for set bit and left``        ``# shift n, count times``        ``if` `(m ``%` `2` `=``=` `1``):``            ``ans ``+``=` `n << count` `        ``# increment of place value (count)``        ``count ``+``=` `1``        ``m ``=` `int``(m``/``2``)` `    ``return` `ans` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ``n ``=` `20``    ``m ``=` `13``    ``print``(multiply(n, m))``    ` `# This code is contributed by``# Ssanjit_Prasad`

## C#

 `// C# program to find multiplication``// of two number without use of``// multiplication operator``using` `System;` `class` `GFG``{``    ` `    ``// Function for multiplication``    ``static` `int` `multiply(``int` `n, ``int` `m)``    ``{``        ``int` `ans = 0, count = 0;``        ``while` `(m > 0)``        ``{``            ``// check for set bit and left``            ``// shift n, count times``            ``if` `(m % 2 == 1)        ``                ``ans += n << count;``    ` `            ``// increment of place``            ``// value (count)``            ``count++;``            ``m /= 2;``        ``}``        ` `        ``return` `ans;``    ``}``    ` `    ``// Driver Code``    ``public` `static` `void` `Main ()``    ``{``        ``int` `n = 20, m = 13;``        ` `        ``Console.WriteLine( multiply(n, m) );``    ``}``}` `// This code is contributed by vt_m.`

## PHP

 ``

## Javascript

 ``

Output

`260`

Time Complexity : O(log n)

Auxiliary Space: O(1)

Method 2

We can use shift operator in loops.

## C++

 `#include ``using` `namespace` `std;`` ` `int` `multiply(``int` `n, ``int` `m){``    ``bool` `isNegative = ``false``;``    ``if` `(n < 0 && m < 0) {``        ``n = -n, m = -m;``    ``}``    ``if` `(n < 0) {``        ``n = -n, isNegative = ``true``;``    ``}``    ``if` `(m < 0) {``        ``m = -m, isNegative = ``true``;``    ``}``    ``int` `result = 0;``    ``while` `(m){``        ``if` `(m & 1) {``            ``result += n;``        ``}``        ``// multiply a by 2``        ``n = n << 1;``        ``// divide b by 2``        ``m = m >> 1;``    ``}``    ``return` `(isNegative) ? -result : result;``}`` ` `int` `main()``{``    ``int` `n = 20 , m = 13;``    ``cout << multiply(n, m);``    ``return` `0;``}`

Output

`260`
```Related Article:
Russian Peasant (Multiply two numbers using bitwise operators)
This article is contributed by Shivam Pradhan (anuj_charm). If you like GeeksforGeeks 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 GeeksforGeeks 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.
```

My Personal Notes arrow_drop_up