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. ?> |
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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