# 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

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<bits/stdc++.h>` `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

`<?php` `// PHP program to find multiplication` `// of two number without use of` `// multiplication operator` `// Function for multiplication` `function` `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` `++;` ` ` `$m` `/= 2;` ` ` `}` ` ` `return` `$ans` `;` `}` `// Driver code` `$n` `= 20 ; ` `$m` `= 13;` `echo` `multiply(` `$n` `, ` `$m` `);` `// This code is contributed by anuj_67.` `?>` |

## Javascript

`<script>` `// JavaScript program to find multiplication` `// of two number without use of` `// multiplication operator` `// Function for multiplication` `function` `multiply(n, m)` `{` ` ` `let 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 = Math.floor(m / 2);` ` ` `}` ` ` `return` `ans;` `}` `// Driver code` ` ` `let n = 20 , m = 13;` ` ` `document.write(multiply(n, m));` ` ` `// This code is contributed by Surbhi Tyagi.` `</script>` |

**Output :**

260

**Time Complexity :** O(log n)**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 contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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