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.


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.

// CPP program to find multiplication
// of two number without use of
// multiplication operator
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)
        m /= 2;
    return ans;

// Driver program 
int main()
    int n = 20 , m = 13;
    cout << multiply(n, m);
    return 0;



Time Complexity : O(log n)

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

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