# Multiplication of two numbers with shift operator

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
```

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

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

 ` `

Output :

```260
```

Time Complexity : O(log n)

Related Article:
Russian Peasant (Multiply two numbers using bitwise operators)

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