Smallest string without any multiplication sign that represents the product of two given numbers

Given two numbers A and B, the task is to print the string of the smallest possible length which evaluates to the product of the two given numbers i.e., A*B, without using the multiplication sign.

Any perfect power of 2 can be expressed in the form of a left-shift operator. 
For Example:
N = 2 = 1<<1 = “<<1” 
N = 4 = 1<<2 = “<<2” 
Using the above idea, any number can be expressed using a left-shift operator instead of a multiplication sign. 
For Example:
N = 24 = 6*4 = 6(1<<2) = “6<<2”

Examples:

Input: A = 6, B = 10 
Output: 6<<3+6<<1
Explanation:
The product of the 2 numbers = 6 × 10 = 60. 
The above-given expression evaluates to 6 × (2 × 2 × 2) + 6 × 2 = 60. 
The string “10<<2+10<<1” also evaluates to 60. 
But “6<<3+6<<1” is the required output as its length is smaller.

Input: A = 5, B = 5 
Output: 5<<2+5
Explanation:
The product of the 2 numbers = 5 × 5 = 25. 
The above-given expression evaluates to 5 × (2 × 2) + 5 = 25.

Approach: The idea is to use Left Shift Operator to find the product. Below are the steps: 

• Represent B as powers of 2.

 Let B = 2^k₁ + 2^k₂ + … + 2^k_n, where k₁ > k₂ > .. > k_n 

• Therefore, the product of A and B can be written as

A * B = A * (2^k₁ + 2^k₂+ … + 2^k_n )

• Use the “<<“ (left shift operator) to multiply a number by any power of two.
• Thus A x B = A << k₁ + A << k₂ + … + A << k_n
• To find k_i we use the log() function and continue the process with the remainder B – 2^k_i until the remainder becomes 0 or the log of the remainder becomes zero.
• Similarly, represent A*B = B<< k₁ + B<< k₂ + … + B<< k_n by representing A as the power of 2.
• Compare the two representations and print the string with a smaller length.

Below is the implementation of the above approach:

C++

// C++ program for
// the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the string
// which evaluates to the
// product of A and B
string len(long A, long B)
{
  // Stores the result
  string res = "";
  long Log = 0;

  do
  {
    // 2 ^ log <= B &&
    // 2 ^ (log + 1) > B
    Log = (long)(log(B) / log(2));

    // Update res to
    // res+=A X 2^log
    if (Log != 0)
    {
      res = res + to_string(A) +
            "<<" + to_string(Log);
    }
    else
    {
      // Update res to
      // res+=A X 2^0
      res += A;
      break;
    }

    // Find the remainder
    B = B - (long)pow(2, Log);

    // If remainder is
    // not equal to 0
    if (B != 0)
    {
      res += "+";
    }
    else
      break;
  } while (Log != 0);

  // Return the
  // resultant string
  return res;
}

// Function to find the minimum
// length representation of A*B
void minimumString(long A, long B)
{
  // Find representation of form
  // A << k1 + A << k2 + ... + A << kn
  string res1 = len(A, B);

  // Find representation of form
  // B << k1 + B << k2 + ... + B << kn
  string res2 = len(B, A);

  // Compare the length of
  // the representations
  if (res1.length() > res2.length())
  {
    cout << res2 << endl;
  }
  else
  {
    cout << res1 << endl;
  }
}

// Driver code
int main()
{
  // Product A X B
  long A = 6;
  long B = 10;

  // Function Call
  minimumString(A, B);

  return 0;
}

// This code is contributed by divyeshrabadiya07

Java

// Java program for the above approach

class GFG {

    // Function to find the string
    // which evaluates to the
    // product of A and B
    public static String len(long A,
long B)
    {
        // Stores the result
        String res = "";
        long log = 0;

        do {

            // 2 ^ log <= B &&
            // 2 ^ (log + 1) > B
            log = (long)(Math.log(B)
                         / Math.log(2));

            // Update res to
            // res+=A X 2^log
            if (log != 0) {

                res += A + "<<" + log;
            }
            else {

                // Update res to res+=A X 2^0
                res += A;
                break;
            }

            // Find the remainder
            B = B - (long)Math.pow(2, log);

            // If remainder is not equal
            // to 0
            if (B != 0) {
                res += "+";
            }
            else
                break;
        } while (log != 0);

        // Return the resultant string
        return res;
    }

    // Function to find the minimum
    // length representation of A*B
    public static void
    minimumString(long A, long B)
    {
        // Find representation of form
        // A << k1 + A << k2 + ... + A << kn
        String res1 = len(A, B);

        // Find representation of form
        // B << k1 + B << k2 + ... + B << kn
        String res2 = len(B, A);

        // Compare the length of
        // the representations
        if (res1.length() > res2.length()) {
            System.out.println(res2);
        }
        else {
            System.out.println(res1);
        }
    }

    // Driver Code
    public static void main(String args[])
    {
        // Product A X B
        long A = 6;
        long B = 10;

        // Function Call
        minimumString(A, B);
    }
}

Python3

# Python3 program for the above approach
from math import log

# Function to find the string
# which evaluates to the
# product of A and B
def lenn(A, B):

    # Stores the result
    res = ""
    logg = 0

    while True:

        # 2 ^ logg <= B &&
        # 2 ^ (logg + 1) > B
        logg =log(B) // log(2)

        # Update res to
        # res+=A X 2^logg
        if (logg != 0):
            res += (str(A) + "<<" +
                    str(int(logg)))
        else:

            # Update res to res+=A X 2^0
            res += A
            break

        # Find the remainder
        B = B - pow(2, logg)

        # If remainder is not equal
        # to 0
        if (B != 0):
            res += "+"
        else:
            break

        if logg == 0:
            break

    # Return the resultant string
    return res

# Function to find the minimum
# length representation of A*B
def minimumString(A, B):

    # Find representation of form
    # A << k1 + A << k2 + ... + A << kn
    res1 = lenn(A, B)

    # Find representation of form
    # B << k1 + B << k2 + ... + B << kn
    res2 = lenn(B, A)

    # Compare the length of
    # the representations
    if (len(res1) > len(res2)):
        print(res2)
    else:
        print(res1)

# Driver Code
if __name__ == '__main__':

    # Product A X B
    A = 6
    B = 10

    # Function call
    minimumString(A, B)

# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;

class GFG{

// Function to find the string
// which evaluates to the
// product of A and B
public static string len(long A, long B)
{

    // Stores the result
    string res = "";
    long log = 0;

    do
    {

        // 2 ^ log <= B &&
        // 2 ^ (log + 1) > B
        log = (long)(Math.Log(B) /
                     Math.Log(2));

        // Update res to
        // res+=A X 2^log
        if (log != 0)
        {
            res += A + "<<" + log;
        }
        else
        {

            // Update res to res+=A X 2^0
            res += A;
            break;
        }

        // Find the remainder
        B = B - (long)Math.Pow(2, log);

        // If remainder is not equal
        // to 0
        if (B != 0)
        {
            res += "+";
        }
        else
            break;
    } while (log != 0);

    // Return the resultant string
    return res;
}

// Function to find the minimum
// length representation of A*B
public static void minimumString(long A,
                                 long B)
{

    // Find representation of form
    // A << k1 + A << k2 + ... + A << kn
    string res1 = len(A, B);

    // Find representation of form
    // B << k1 + B << k2 + ... + B << kn
    string res2 = len(B, A);

    // Compare the length of
    // the representations
    if (res1.Length > res2.Length)
    {
        Console.WriteLine(res2);
    }
    else
    {
        Console.WriteLine(res1);
    }
}

// Driver Code
public static void Main()
{

    // Product A X B
    long A = 6;
    long B = 10;

    // Function call
    minimumString(A, B);
}
}

// This code is contributed by code_hunt

Javascript

<script>
// Javascript program for
// the above approach


// Function to find the string
// which evaluates to the
// product of A and B
function len(A, B) {
    // Stores the result
    let res = "";
    let Log = 0;

    do {
        // 2 ^ log <= B &&
        // 2 ^ (log + 1) > B
        Log = Math.floor(Math.log(B) / Math.log(2));

        // Update res to
        // res+=A X 2^log
        if (Log != 0) {
            res = res + String(A) + "<<" + String(Log);
        }
        else {
            // Update res to
            // res+=A X 2^0
            res += A;
            break;
        }

        // Find the remainder
        B = B - Math.pow(2, Log);

        // If remainder is
        // not equal to 0
        if (B != 0) {
            res += "+";
        }
        else
            break;
    } while (Log != 0);

    // Return the
    // resultant string
    return res;
}

// Function to find the minimum
// length representation of A*B
function minimumString(A, B) {
    // Find representation of form
    // A << k1 + A << k2 + ... + A << kn
    let res1 = len(A, B);

    // Find representation of form
    // B << k1 + B << k2 + ... + B << kn
    let res2 = len(B, A);

    // Compare the length of
    // the representations
    if (res1.length > res2.length) {
        document.write(res2 + "<br>");
    }
    else {
        document.write(res1 + "<br>");
    }
}

// Driver code


// Product A X B
let A = 6;
let B = 10;

// Function Call
minimumString(A, B);


// This code is contributed by gfgking
</script>

Output:

6<<3+6<<1

Time Complexity: O(log N)
Auxiliary Space: O(1)