# Karatsuba algorithm for fast multiplication using Divide and Conquer algorithm

Given two binary strings that represent value of two integers, find the product of two strings. For example, if the first bit string is “1100” and second bit string is “1010”, output should be 120.

For simplicity, let the length of two strings be same and be n.

A Naive Approach is to follow the process we study in school. One by one take all bits of second number and multiply it with all bits of first number. Finally add all multiplications. This algorithm takes O(n^2) time.

Using Divide and Conquer, we can multiply two integers in less time complexity. We divide the given numbers in two halves. Let the given numbers be X and Y.

For simplicity let us assume that n is even

```X =  Xl*2n/2 + Xr    [Xl and Xr contain leftmost and rightmost n/2 bits of X]
Y =  Yl*2n/2 + Yr    [Yl and Yr contain leftmost and rightmost n/2 bits of Y]```

The product XY can be written as follows.

```XY = (Xl*2n/2 + Xr)(Yl*2n/2 + Yr)
= 2n XlYl + 2n/2(XlYr + XrYl) + XrYr```

If we take a look at the above formula, there are four multiplications of size n/2, so we basically divided the problem of size n into four sub-problems of size n/2. But that doesn’t help because the solution of recurrence T(n) = 4T(n/2) + O(n) is O(n^2). The tricky part of this algorithm is to change the middle two terms to some other form so that only one extra multiplication would be sufficient. The following is tricky expression for middle two terms.

`XlYr + XrYl = (Xl + Xr)(Yl + Yr) - XlYl- XrYr`

So the final value of XY becomes

`XY = 2n XlYl + 2n/2 * [(Xl + Xr)(Yl + Yr) - XlYl - XrYr] + XrYr`

With above trick, the recurrence becomes T(n) = 3T(n/2) + O(n) and solution of this recurrence is O(n1.59).

What if the lengths of input strings are different and are not even? To handle the different length case, we append 0’s in the beginning. To handle odd length, we put floor(n/2) bits in left half and ceil(n/2) bits in right half. So the expression for XY changes to following.

`XY = 22ceil(n/2) XlYl + 2ceil(n/2) * [(Xl + Xr)(Yl + Yr) - XlYl - XrYr] + XrYr`

The above algorithm is called Karatsuba algorithm and it can be used for any base.

Recommended Practice

Following is C++ implementation of above algorithm.

## C++

 `// C++ implementation of Karatsuba algorithm for bit string multiplication.` `#include` `#include`   `using` `namespace` `std;`   `// FOLLOWING TWO FUNCTIONS ARE COPIED FROM http://goo.gl/q0OhZ` `// Helper method: given two unequal sized bit strings, converts them to` `// same length by adding leading 0s in the smaller string. Returns the` `// the new length` `int` `makeEqualLength(string &str1, string &str2)` `{` `    ``int` `len1 = str1.size();` `    ``int` `len2 = str2.size();` `    ``if` `(len1 < len2)` `    ``{` `        ``for` `(``int` `i = 0 ; i < len2 - len1 ; i++)` `            ``str1 = ``'0'` `+ str1;` `        ``return` `len2;` `    ``}` `    ``else` `if` `(len1 > len2)` `    ``{` `        ``for` `(``int` `i = 0 ; i < len1 - len2 ; i++)` `            ``str2 = ``'0'` `+ str2;` `    ``}` `    ``return` `len1; ``// If len1 >= len2` `}`   `// The main function that adds two bit sequences and returns the addition` `string addBitStrings( string first, string second )` `{` `    ``string result;  ``// To store the sum bits`   `    ``// make the lengths same before adding` `    ``int` `length = makeEqualLength(first, second);` `    ``int` `carry = 0;  ``// Initialize carry`   `    ``// Add all bits one by one` `    ``for` `(``int` `i = length-1 ; i >= 0 ; i--)` `    ``{` `        ``int` `firstBit = first.at(i) - ``'0'``;` `        ``int` `secondBit = second.at(i) - ``'0'``;`   `        ``// boolean expression for sum of 3 bits` `        ``int` `sum = (firstBit ^ secondBit ^ carry)+``'0'``;`   `        ``result = (``char``)sum + result;`   `        ``// boolean expression for 3-bit addition` `        ``carry = (firstBit&secondBit) | (secondBit&carry) | (firstBit&carry);` `    ``}`   `    ``// if overflow, then add a leading 1` `    ``if` `(carry)  result = ``'1'` `+ result;`   `    ``return` `result;` `}`   `// A utility function to multiply single bits of strings a and b` `int` `multiplyiSingleBit(string a, string b)` `{  ``return` `(a[0] - ``'0'``)*(b[0] - ``'0'``);  }`   `// The main function that multiplies two bit strings X and Y and returns` `// result as long integer` `long` `int` `multiply(string X, string Y)` `{` `    ``// Find the maximum of lengths of x and Y and make length` `    ``// of smaller string same as that of larger string` `    ``int` `n = makeEqualLength(X, Y);`   `    ``// Base cases` `    ``if` `(n == 0) ``return` `0;` `    ``if` `(n == 1) ``return` `multiplyiSingleBit(X, Y);`   `    ``int` `fh = n/2;   ``// First half of string, floor(n/2)` `    ``int` `sh = (n-fh); ``// Second half of string, ceil(n/2)`   `    ``// Find the first half and second half of first string.` `    ``// Refer http://goo.gl/lLmgn for substr method` `    ``string Xl = X.substr(0, fh);` `    ``string Xr = X.substr(fh, sh);`   `    ``// Find the first half and second half of second string` `    ``string Yl = Y.substr(0, fh);` `    ``string Yr = Y.substr(fh, sh);`   `    ``// Recursively calculate the three products of inputs of size n/2` `    ``long` `int` `P1 = multiply(Xl, Yl);` `    ``long` `int` `P2 = multiply(Xr, Yr);` `    ``long` `int` `P3 = multiply(addBitStrings(Xl, Xr), addBitStrings(Yl, Yr));`   `    ``// Combine the three products to get the final result.` `    ``return` `P1*(1<<(2*sh)) + (P3 - P1 - P2)*(1<

## Python3

 `# Python implementation of Karatsuba algorithm for bit string multiplication.`   `# Helper method: given two unequal sized bit strings, converts them to` `# same length by adding leading 0s in the smaller string. Returns the` `# the new length` `def` `make_equal_length(str1, str2):` `    ``len1 ``=` `len``(str1)` `    ``len2 ``=` `len``(str2)` `    ``if` `len1 < len2:` `        ``for` `i ``in` `range``(len2 ``-` `len1):` `            ``str1 ``=` `'0'` `+` `str1` `        ``return` `len2` `    ``elif` `len1 > len2:` `        ``for` `i ``in` `range``(len1 ``-` `len2):` `            ``str2 ``=` `'0'` `+` `str2` `    ``return` `len1 ``# If len1 >= len2`   `# The main function that adds two bit sequences and returns the addition` `def` `add_bit_strings(first, second):` `    ``result ``=` `""  ``# To store the sum bits`   `    ``# make the lengths same before adding` `    ``length ``=` `make_equal_length(first, second)` `    ``carry ``=` `0`  `# Initialize carry`   `    ``# Add all bits one by one` `    ``for` `i ``in` `range``(length``-``1``, ``-``1``, ``-``1``):` `        ``first_bit ``=` `int``(first[i])` `        ``second_bit ``=` `int``(second[i])`   `        ``# boolean expression for sum of 3 bits` `        ``sum` `=` `(first_bit ^ second_bit ^ carry) ``+` `ord``(``'0'``)`   `        ``result ``=` `chr``(``sum``) ``+` `result`   `        ``# boolean expression for 3-bit addition` `        ``carry ``=` `(first_bit & second_bit) | (second_bit & carry) | (first_bit & carry)`   `    ``# if overflow, then add a leading 1` `    ``if` `carry:` `        ``result ``=` `'1'` `+` `result`   `    ``return` `result`   `# A utility function to multiply single bits of strings a and b` `def` `multiply_single_bit(a, b):` `    ``return` `int``(a[``0``]) ``*` `int``(b[``0``])`   `# The main function that multiplies two bit strings X and Y and returns` `# result as long integer` `def` `multiply(X, Y):` `    ``# Find the maximum of lengths of x and Y and make length` `    ``# of smaller string same as that of larger string` `    ``n ``=` `max``(``len``(X), ``len``(Y))` `    ``X ``=` `X.zfill(n)` `    ``Y ``=` `Y.zfill(n)`   `    ``# Base cases` `    ``if` `n ``=``=` `0``: ``return` `0` `    ``if` `n ``=``=` `1``: ``return` `int``(X[``0``])``*``int``(Y[``0``])`   `    ``fh ``=` `n``/``/``2`  `# First half of string` `    ``sh ``=` `n ``-` `fh  ``# Second half of string`   `    ``# Find the first half and second half of first string.` `    ``Xl ``=` `X[:fh]` `    ``Xr ``=` `X[fh:]`   `    ``# Find the first half and second half of second string` `    ``Yl ``=` `Y[:fh]` `    ``Yr ``=` `Y[fh:]`   `    ``# Recursively calculate the three products of inputs of size n/2` `    ``P1 ``=` `multiply(Xl, Yl)` `    ``P2 ``=` `multiply(Xr, Yr)` `    ``P3 ``=` `multiply(``str``(``int``(Xl, ``2``) ``+` `int``(Xr, ``2``)), ``str``(``int``(Yl, ``2``) ``+` `int``(Yr, ``2``)))`   `    ``# Combine the three products to get the final result.` `    ``return` `P1``*``(``1``<<(``2``*``sh)) ``+` `(P3 ``-` `P1 ``-` `P2)``*``(``1``<

## C#

 `// C# implementation of Karatsuba algorithm for bit string` `// multiplication.` `using` `System;` `using` `System.Collections.Generic;`   `class` `GFG {`   `  ``// Convert bit strings to same length by adding leading` `  ``// 0s in the smaller string.` `  ``// Returns the new length.` `  ``private` `static` `int` `MakeEqualLength(``ref` `string` `str1,` `                                     ``ref` `string` `str2)` `  ``{` `    ``int` `len1 = str1.Length;` `    ``int` `len2 = str2.Length;` `    ``if` `(len1 < len2) {` `      ``str1 = str1.PadLeft(len2 - len1 + len1, ``'0'``);` `      ``return` `len2;` `    ``}` `    ``else` `if` `(len1 > len2) {` `      ``str2 = str2.PadLeft(len1 - len2 + len2, ``'0'``);` `    ``}` `    ``return` `len1; ``// If len1 >= len2` `  ``}`   `  ``// Adds two bit sequences and returns the addition` `  ``private` `static` `string` `AddBitStrings(``string` `first,` `                                      ``string` `second)` `  ``{` `    ``string` `result = ``""``; ``// To store the sum bits`   `    ``// make the lengths same before adding` `    ``int` `length = MakeEqualLength(``ref` `first, ``ref` `second);` `    ``int` `carry = 0; ``// Initialize carry`   `    ``// Add all bits one by one` `    ``for` `(``int` `i = length - 1; i >= 0; i--) {` `      ``int` `first_bit = ``int``.Parse(first[i].ToString());` `      ``int` `second_bit` `        ``= ``int``.Parse(second[i].ToString());`   `      ``// boolean expression for sum of 3 bits` `      ``int` `sum` `        ``= (first_bit ^ second_bit ^ carry) + ``'0'``;`   `      ``result = (``char``)sum + result;`   `      ``// boolean expression for 3-bit addition` `      ``carry = (first_bit & second_bit)` `        ``| (second_bit & carry)` `        ``| (first_bit & carry);` `    ``}`   `    ``// if overflow, then add a leading 1` `    ``if` `(carry != 0) {` `      ``result = ``'1'` `+ result;` `    ``}`   `    ``return` `result;` `  ``}`   `  ``// Multiplies single bits of strings a and b` `  ``private` `static` `int` `MultiplySingleBit(``char` `a, ``char` `b)` `  ``{` `    ``return` `int``.Parse(a.ToString())` `      ``* ``int``.Parse(b.ToString());` `  ``}`   `  ``// Multiplies two bit strings X and Y and returns result` `  ``// as long integer` `  ``private` `static` `long` `Multiply(``string` `X, ``string` `Y)` `  ``{` `    ``// Find the maximum of lengths of x and Y and make` `    ``// length of smaller string same as that of larger` `    ``// string` `    ``int` `n = Math.Max(X.Length, Y.Length);` `    ``X = X.PadLeft(n, ``'0'``);` `    ``Y = Y.PadLeft(n, ``'0'``);`   `    ``// Base cases` `    ``if` `(n == 0)` `      ``return` `0;` `    ``if` `(n == 1)` `      ``return` `MultiplySingleBit(X[0], Y[0]);`   `    ``int` `fh = n / 2; ``// First half of string` `    ``int` `sh = n - fh; ``// Second half of string`   `    ``// Find the first half and second half of first` `    ``// string.` `    ``string` `Xl = X.Substring(0, fh);` `    ``string` `Xr = X.Substring(fh);`   `    ``// Find the first half and second half of second` `    ``// string` `    ``string` `Yl = Y.Substring(0, fh);` `    ``string` `Yr = Y.Substring(fh);`   `    ``// Recursively calculate the three products of` `    ``// inputs of size n/2` `    ``long` `P1 = Multiply(Xl, Yl);` `    ``long` `P2 = Multiply(Xr, Yr);` `    ``long` `P3 = Multiply(AddBitStrings(Xl, Xr),` `                       ``AddBitStrings(Yl, Yr));`   `    ``// Combine the three products to get the final` `    ``// result.` `    ``return` `P1 * (1L << (2 * sh))` `      ``+ (P3 - P1 - P2) * (1L << sh) + P2;` `  ``}`   `  ``// Test the implementation` `  ``public` `static` `void` `Main(``string``[] args)` `  ``{` `    ``Console.WriteLine(Multiply(``"1100"``, ``"1010"``));` `    ``Console.WriteLine(Multiply(``"110"``, ``"1010"``));` `    ``Console.WriteLine(Multiply(``"11"``, ``"1010"``));` `    ``Console.WriteLine(Multiply(``"1"``, ``"1010"``));` `    ``Console.WriteLine(Multiply(``"0"``, ``"1010"``));` `    ``Console.WriteLine(Multiply(``"111"``, ``"111"``));` `    ``Console.WriteLine(Multiply(``"11"``, ``"11"``));` `  ``}` `}`

## Javascript

 `// JavaScript implementation of Karatsuba algorithm for bit string multiplication.`   `// Helper method: given two unequal sized bit strings, converts them to` `// same length by adding leading 0s in the smaller string. Returns the` `// the new length` `function` `make_equal_length(str1, str2) {` `    ``let len1 = str1.length;` `    ``let len2 = str2.length;` `    ``if` `(len1 < len2) {` `        ``for` `(let i = 0; i < len2 - len1; i++) {` `            ``str1 = ``'0'` `+ str1;` `        ``}` `        ``return` `len2;` `    ``} ``else` `if` `(len1 > len2) {` `        ``for` `(let i = 0; i < len1 - len2; i++) {` `            ``str2 = ``'0'` `+ str2;` `        ``}` `    ``}` `    ``return` `len1; ``// If len1 >= len2` `}`   `// The main function that adds two bit sequences and returns the addition` `function` `add_bit_strings(first, second) {` `    ``let result = ``""``; ``// To store the sum bits`   `    ``// make the lengths same before adding` `    ``let length = make_equal_length(first, second);` `    ``let carry = 0; ``// Initialize carry`   `    ``// Add all bits one by one` `    ``for` `(let i = length - 1; i >= 0; i--) {` `        ``let first_bit = parseInt(first[i]);` `        ``let second_bit = parseInt(second[i]);`   `        ``// boolean expression for sum of 3 bits` `        ``let sum = (first_bit ^ second_bit ^ carry) + ``'0'``.charCodeAt(0);`   `        ``result = String.fromCharCode(sum) + result;`   `        ``// boolean expression for 3-bit addition` `        ``carry = (first_bit & second_bit) | (second_bit & carry) | (first_bit & carry);` `    ``}`   `    ``// if overflow, then add a leading 1` `    ``if` `(carry) {` `        ``result = ``'1'` `+ result;` `    ``}`   `    ``return` `result;` `}`   `// A utility function to multiply single bits of strings a and b` `function` `multiply_single_bit(a, b) {` `    ``return` `parseInt(a[0]) * parseInt(b[0]);` `}`   `// The main function that multiplies two bit strings X and Y and returns` `// result as long integer` `function` `multiply(X, Y) {` `    ``// Find the maximum of lengths of x and Y and make length` `    ``// of smaller string same as that of larger string` `    ``let n = Math.max(X.length, Y.length);` `    ``X = X.padStart(n, ``'0'``);` `    ``Y = Y.padStart(n, ``'0'``);`     `    ``// Base cases` `    ``if` `(n == 0) ``return` `0;` `    ``if` `(n == 1) ``return` `parseInt(X[0]) * parseInt(Y[0]);`   `    ``let fh = Math.floor(n / 2); ``// First half of string` `    ``let sh = n - fh; ``// Second half of string`   `    ``// Find the first half and second half of first string.` `    ``let Xl = X.slice(0, fh);` `    ``let Xr = X.slice(fh);`   `    ``// Find the first half and second half of second string` `    ``let Yl = Y.slice(0, fh);` `    ``let Yr = Y.slice(fh);`   `    ``// Recursively calculate the three products of inputs of size n/2` `    ``let P1 = multiply(Xl, Yl);` `    ``let P2 = multiply(Xr, Yr);` `    ``let P3 = multiply((parseInt(Xl, 2) + parseInt(Xr, 2)).toString(2), (parseInt(Yl, 2) + parseInt(Yr, 2)).toString(2));`   `    ``// Combine the three products to get the final result.` `    ``return` `P1 * (1 << (2 * sh)) + (P3 - P1 - P2) * (1 << sh) + P2` `}`   `console.log(multiply(``"1100"``, ``"1010"``))` `console.log(multiply(``"110"``, ``"1010"``))` `console.log(multiply(``"11"``, ``"1010"``))` `console.log(multiply(``"1"``, ``"1010"``))` `console.log(multiply(``"0"``, ``"1010"``))` `console.log(multiply(``"111"``, ``"111"``))` `console.log(multiply(``"11"``, ``"11"``))`

## Java

 `// Java implementation of Karatsuba algorithm for bit string multiplication.`   `public` `class` `GFG {` `      ``// Driver Code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``System.out.println(multiply(``"1100"``, ``"1010"``));` `        ``System.out.println(multiply(``"110"``, ``"1010"``));` `        ``System.out.println(multiply(``"11"``, ``"1010"``));` `        ``System.out.println(multiply(``"1"``, ``"1010"``));` `        ``System.out.println(multiply(``"0"``, ``"1010"``));` `        ``System.out.println(multiply(``"111"``, ``"111"``));` `        ``System.out.println(multiply(``"11"``, ``"11"``));` `    ``}` `  `  `    ``// Helper method: given two unequal sized bit strings,` `    ``// converts them to same length by adding leading 0s in` `    ``// the smaller string. Returns the new length` `    ``private` `static` `int` `makeEqualLength(StringBuilder str1,` `                                       ``StringBuilder str2)` `    ``{` `        ``int` `len1 = str1.length();` `        ``int` `len2 = str2.length();` `        ``if` `(len1 < len2) {` `            ``for` `(``int` `i = ``0``; i < len2 - len1; i++) {` `                ``str1.insert(``0``, ``'0'``);` `            ``}` `            ``return` `len2;` `        ``}` `        ``else` `if` `(len1 > len2) {` `            ``for` `(``int` `i = ``0``; i < len1 - len2; i++) {` `                ``str2.insert(``0``, ``'0'``);` `            ``}` `        ``}` `        ``return` `len1; ``// If len1 >= len2` `    ``}`   `    ``// The main function that adds two bit sequences and` `    ``// returns the addition` `    ``private` `static` `StringBuilder` `    ``addBitStrings(StringBuilder first, StringBuilder second)` `    ``{` `        ``StringBuilder result = ``new` `StringBuilder();` `        ``int` `length = makeEqualLength(first, second);` `        ``int` `carry = ``0``;`   `        ``// Add all bits one by one` `        ``for` `(``int` `i = length - ``1``; i >= ``0``; i--) {` `            ``int` `firstBit = first.charAt(i) - ``'0'``;` `            ``int` `secondBit = second.charAt(i) - ``'0'``;`   `            ``// boolean expression for sum of 3 bits` `            ``int` `sum = (firstBit ^ secondBit ^ carry) + ``'0'``;` `            ``result.insert(``0``, (``char``)sum);`   `            ``// boolean expression for 3-bit addition` `            ``carry = (firstBit & secondBit)` `                    ``| (secondBit & carry)` `                    ``| (firstBit & carry);` `        ``}`   `        ``// if overflow, then add a leading 1` `        ``if` `(carry == ``1``) {` `            ``result.insert(``0``, ``'1'``);` `        ``}`   `        ``return` `result;` `    ``}`   `    ``// A utility function to multiply single bits of strings` `    ``// a and b` `    ``private` `static` `int` `multiplySingleBit(``int` `a, ``int` `b)` `    ``{` `        ``return` `a * b;` `    ``}`   `    ``// The main function that multiplies two bit strings X` `    ``// and Y and returns result as long integer` `    ``public` `static` `long` `multiply(String X, String Y)` `    ``{` `        ``// Find the maximum of lengths of X and Y and make` `        ``// length of smaller string same as that of larger` `        ``// string` `        ``int` `n = Math.max(X.length(), Y.length());` `        ``X = String.format(``"%"` `+ n + ``"s"``, X)` `                ``.replace(``' '``, ``'0'``);` `        ``Y = String.format(``"%"` `+ n + ``"s"``, Y)` `                ``.replace(``' '``, ``'0'``);`   `        ``// Base cases` `        ``if` `(n == ``0``)` `            ``return` `0``;` `        ``if` `(n == ``1``)` `            ``return` `Integer.parseInt(X)` `                ``* Integer.parseInt(Y);`   `        ``int` `fh = n / ``2``; ``// First half of string` `        ``int` `sh = n - fh; ``// Second half of string`   `        ``// Find the first half and second half of first` `        ``// string.` `        ``String Xl = X.substring(``0``, fh);` `        ``String Xr = X.substring(fh);`   `        ``// Find the first half and second half of second` `        ``// string` `        ``String Yl = Y.substring(``0``, fh);` `        ``String Yr = Y.substring(fh);`   `        ``// Recursively calculate the three products of` `        ``// inputs of size n/2` `        ``long` `P1 = multiply(Xl, Yl);` `        ``long` `P2 = multiply(Xr, Yr);` `        ``long` `P3 = multiply(Integer.toBinaryString(` `                               ``Integer.parseInt(Xl, ``2``)` `                               ``+ Integer.parseInt(Xr, ``2``)),` `                           ``Integer.toBinaryString(` `                               ``Integer.parseInt(Yl, ``2``)` `                               ``+ Integer.parseInt(Yr, ``2``)));`   `        ``// Combine the three products to get the final` `        ``// result.` `        ``return` `P1 * (1L << (``2` `* sh))` `            ``+ (P3 - P1 - P2) * (1L << sh) + P2;` `    ``}` `}`

Output

```120
60
30
10
0
49
9```

Time Complexity: Time complexity of the above solution is O(nlog23) = O(n1.59).
Time complexity of multiplication can be further improved using another Divide and Conquer algorithm, fast Fourier transform. We will soon be discussing fast Fourier transform as a separate post.

Auxiliary Space: O(n)

Exercise:
The above program returns a long int value and will not work for big strings. Extend the above program to return a string instead of a long int value.

Solution:
Multiplication process for large numbers is an important problem in Computer Science. Given approach uses Divide and Conquer methodology.
Run the code to see the time complexity comparison for normal Binary Multiplication and Karatsuba Algorithm.
You can see the full code in this repository

Examples:

```First Binary Input : 101001010101010010101001010100101010010101010010101
Second Binary Input : 101001010101010010101001010100101010010101010010101
Decimal Output : Not Representable
Output : 2.1148846e+30```
```First Binary Input : 1011
Second Binary Input : 1000
Decimal Output : 88
Output : 5e-05```

## C++

 `#include ` `#include ` `#include ` `#include ` `#include ` `#include `   `using` `namespace` `std;`   `// classical method class` `class` `BinaryMultiplier` `{` `public``:` `    ``string MakeMultiplication(string,string);      ` `    ``string MakeShifting(string,``int``);               ` `    ``string addBinary(string,string);` `    ``void` `BinaryStringToDecimal(string);` `};`   `// karatsuba method class` `class` `Karatsuba` `{` `public``:` `    ``int` `lengthController(string &,string &);` `    ``string addStrings(string,string);` `    ``string multiply(string,string);` `    ``string DecimalToBinary(``long` `long` `int``);` `    ``string Subtraction(string,string);` `    ``string MakeShifting(string,``int``);` `};`   `// this function get strings and go over str2 bit ` `// if it sees 1  it calculates the shifted version according to position bit` `// Makes add operation for binary strings` `// returns result string` `string BinaryMultiplier::MakeMultiplication(string str1, string str2)` `{` `    ``string allSum = ``""``;` `    ``for` `(``int` `j = 0 ; j= 0 || j >= 0 || s == 1)` `    ``{` `        ``s += ((i >= 0)? a[i] - ``'0'``: 0);` `        ``s += ((j >= 0)? b[j] - ``'0'``: 0);` `        `  `        ``result = ``char``(s % 2 + ``'0'``) + result;` `        `  `        ``s /= 2;` `    `  `        ``i--;` `        ``j--;` `    ``}` `    ``return` `result;` `}`   `// this function shifts the given string according to given number` `// returns shifted version ` `string BinaryMultiplier::MakeShifting(string str, ``int` `stepnum)` `{` `    ``string shifted = str;` `    ``for` `(``int` `i = 0 ; i < stepnum ; i++)` `        ``shifted = shifted + ``'0'``;` `    ``return` `shifted;` `}`   `// this function converts Binary String Number to Decimal Number` `// After 32 bits it gives 0 because it overflows the size of int` `void` `BinaryMultiplier::BinaryStringToDecimal(string result)` `{` `    ``cout<<``"Binary Result : "``<= 0; i--)` `    ``{` `        ``if` `(result[i] == ``'1'``)` `        ``{` `            ``val += ``pow``(2,(result.length()-1)-i);` `        ``}` `    ``}` `    ``cout<<``"Decimal Result (Not proper for Large Binary Numbers):"` `< len2)` `    ``{` `        ``for` `(``int` `i = 0 ; i < len1 - len2 ; i++)` `            ``str2 = ``'0'` `+ str2;` `    ``}` `    ``return` `len1;` `}`   `// this function add strings with carry ` `// uses one by one bit addition methodology` `// returns result string` `string Karatsuba::addStrings(string first, string second)` `{` `    ``string result;  ``// To store the sum bits` `    `  `    ``// make the lengths same before adding` `    ``int` `length = lengthController(first, second);` `    ``int` `carry = 0;  ``// Initialize carry` `    `  `    ``// Add all bits one by one` `    ``for` `(``int` `i = length-1 ; i >= 0 ; i--)` `    ``{` `        ``int` `firstBit = first.at(i) - ``'0'``;` `        ``int` `secondBit = second.at(i) - ``'0'``;` `        `  `        ``// boolean expression for sum of 3 bits` `        ``int` `sum = (firstBit ^ secondBit ^ carry)+``'0'``;` `        `  `        ``result = (``char``)sum + result;` `        `  `        ``// Boolean expression for 3-bit addition` `        ``carry = (firstBit&secondBit) | (secondBit&carry) | (firstBit&carry);` `    ``}` `    `  `    ``// if overflow, then add a leading 1` `    ``if` `(carry)` `    ``{` `        ``result = ``'1'` `+ result;` `    ``}` `    `  `    ``return` `result;` `}`   `// this function converts decimal number to binary string` `string Karatsuba::DecimalToBinary(``long` `long` `int` `number)` `{` `    ``string result = ``""``;` `    ``if` `(number <= 0)` `    ``{` `        ``return` `"0"``;` `    ``}` `    ``else` `    ``{` `        ``int` `i = 0;` `        ``while` `(number > 0)` `        ``{` `            `  `            ``long` `long` `int` `num= number % 2;` `            ``stringstream ss;` `            ``ss<= 0; i--)` `    ``{` `        ``diff = (lhs[i]-``'0'``) - (rhs[i]-``'0'``);` `        ``if` `(diff >= 0)` `        ``{` `            ``result = DecimalToBinary(diff) + result;` `        ``}` `        ``else` `        ``{` `            ``for` `(``int` `j = i-1; j>=0; j--)` `            ``{` `                ``lhs[j] = ((lhs[j]-``'0'``) - 1) % 10 + ``'0'``;` `                ``if` `(lhs[j] != ``'1'``)` `                ``{` `                    ``break``;` `                ``}` `            ``}` `            ``result = DecimalToBinary(diff+2) + result;` `        ``}` `    ``}` `    ``return` `result;` `}`   `// this function makes shifting` `string Karatsuba::MakeShifting(string str, ``int` `stepnum)` `{` `    ``string shifted = str;` `    ``for` `(``int` `i = 0 ; i < stepnum ; i++)` `        ``shifted = shifted + ``'0'``;` `    ``return` `shifted;` `}`   `// this function is the core of the Karatsuba ` `// divides problem into 4 subproblems` `// recursively multiplies them` `// returns the result string` `string Karatsuba::multiply(string X,  string Y)` `{` `    ``int` `n = lengthController(X, Y);` `    `  `    ``if` `(n == 1) ``return` `((Y[0]-``'0'` `== 1) && (X[0]-``'0'` `== 1)) ? ``"1"` `: ``"0"``;` `    `  `    ``int` `fh = n/2;   ``// First half of string, floor(n/2)` `    ``int` `sh = (n-fh); ``// Second half of string, ceil(n/2)` `    `  `    ``// Find the first half and second half of first string.` `    ``string Xl = X.substr(0, fh);` `    ``string Xr = X.substr(fh, sh);` `    `  `    ``// Find the first half and second half of second string` `    ``string Yl = Y.substr(0, fh);` `    ``string Yr = Y.substr(fh, sh);` `    `  `    ``// Recursively calculate the three products of inputs of size n/2` `    ``string P1 = multiply(Xl, Yl);` `    ``string P2 = multiply(Xr, Yr);` `    ``string P3 = multiply(addStrings(Xl, Xr), addStrings(Yl, Yr));` `    `  `    ``// return added string version` `    ``return` `addStrings(addStrings(MakeShifting(P1, 2*(n-n/2)),P2),MakeShifting(Subtraction(P3,addStrings(P1,P2)), n-(n/2)));` `}`     `int` `main(``int` `argc, ``const` `char` `* argv[])` `{` `    ``// get the binary numbers as strings` `    ``string firstNumber,secondNumber;` `  `  `    ``cout<<``"Please give the First Binary number : "``;` `    ``cin>>firstNumber;` `    ``cout<>secondNumber;` `    ``cout << endl;` `    `    `    ``// make the initial lengths equal by adding zeros` `    ``int` `len1 = firstNumber.size();` `    ``int` `len2 = secondNumber.size();` `    ``int` `general_len = firstNumber.size();` `    `  `    ``if` `(len1 < len2)` `    ``{` `        ``for` `(``int` `i = 0 ; i < len2 - len1 ; i++)` `            ``firstNumber = ``'0'` `+ firstNumber;` `        ``general_len = firstNumber.size();` `    ``}` `    ``else` `if` `(len1 > len2)` `    ``{` `        ``for` `(``int` `i = 0 ; i < len1 - len2 ; i++)` `            ``secondNumber = ``'0'` `+ secondNumber;` `        ``general_len = secondNumber.size();` `    ``}` `    `  `    ``// In classical methodology Binary String Multiplication` `    ``cout<<``"Classical Algorithm : "``<

Time Complexity:
The time complexity of both Classical and Karatsuba methods of binary string multiplication is O(n^2).

In the classical method, the time complexity is O(n^2) because the loop is iterated n times. The time complexity of the addBinary() method is constant because the loop runs with a maximum of two iterations.

In the Karatsuba method, the time complexity is O(n^2) because the ‘multiply’ method of the Karatsuba class is called recursively for each of the three products. The time complexity of the addStrings() method is constant because the loop runs with a maximum of two iterations.

Auxiliary Space :
The Auxiliary Space of both Classical and Karatsuba methods of binary string multiplication is O(n).

In the classical method, the Auxiliary Space is O(n) because the loop is iterated n times and a single string is used to store the result. The space complexity of the addBinary() method is constant because the loop runs with a maximum of two iterations.

In the Karatsuba method, the auxiliary Space is O(n) because the ‘multiply’ method of the Karatsuba class is called recursively for each of the three products.

Related Article :
Multiply Large Numbers Represented as Strings

Feeling lost in the world of random DSA topics, wasting time without progress? It's time for a change! Join our DSA course, where we'll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 geeks!