Multiplication of two numbers with shift operator - GeeksforGeeks

05 Mar, 2021
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++
Java
Python3
C#
PHP
Javascript
// CPP program to find multiplication// of two number without use of// multiplication operator#include<bits/stdc++.h>using namespace std; // Function for multiplicationint 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 codeint main(){    int n = 20 , m = 13;    cout << multiply(n, m);    return 0;}
// Java program to find multiplication// of two number without use of// multiplication operatorclass 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.
# python 3 program to find multiplication# of two number without use of# multiplication operator # Function for multiplicationdef 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 codeif __name__ == '__main__':    n = 20    m = 13    print(multiply(n, m))     # This code is contributed by# Ssanjit_Prasad
// C# program to find multiplication// of two number without use of// multiplication operatorusing 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 program to find multiplication// of two number without use of// multiplication operator // Function for multiplicationfunction 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.?>
<script> // JavaScript program to find multiplication// of two number without use of// multiplication operator // Function for multiplicationfunction 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.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
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