# Karatsuba Algorithm for fast Multiplication of Large Decimal Numbers represented as Strings

• Difficulty Level : Medium
• Last Updated : 30 May, 2022

Given two numeric strings A and B, the task is to find the product of the two numeric strings efficiently.

Example:

Input: A = 5678, B = 1234
Output: 7006652

Input: A = 74638463789, B = 35284567382
Output: 2633585904851937530398

Approach: The given problem can be solved using Karastuba’s Algorithm for Fast Multiplication, the idea is to append zeroes in front of the integers such that both the integers have an equal and even number of digits n. Thereafter, divide the numbers in the following way:

A =  Al * 10n/2 + Ar    [Al and Ar contain leftmost and rightmost n/2 digits of A]
B =  Bl * 10n/2 + Br    [Bl and Br contain leftmost and rightmost n/2 digits of B]

• Therefore, the product A * B can also be represented as follows:

A * B = (Al * 10n/2 + Ar) * (Bl * 10n/2 + Br)
=> A * B = 10n*Al*Bl + 10n/2*(Al*Br + Bl*Ar) + Ar*Br
=> A * B = 10n*Al*Bl + 10n/2*((Al + Ar)*(Bl + Br) – Al*Bl – Ar*Br) + Ar*Br  [since Al*Br + Bl*Ar = (Al + Ar)*(Bl + Br) – Al*Bl – Ar*Br]

Notice that the above expression only requires three multiplications Al*Bl, Ar*Br, and (Al + Ar)*(Bl + Br), instead of the standard four. Hence, the recurrence becomes T(n) = 3T(n/2) + O(n) and solution of this recurrence is O(n1.59). This idea has been discussed more thoroughly in this article. Therefore the above problem can be solved using the steps below:

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ``using` `namespace` `std;` `// Function to find the sum of larger``// numbers represented as a string``string findSum(string str1, string str2)``{``    ``// Before proceeding further, make``    ``// sure length of str2 is larger``    ``if` `(str1.length() > str2.length())``        ``swap(str1, str2);` `    ``// Stores the result``    ``string str = ``""``;` `    ``// Calculate length of both string``    ``int` `n1 = str1.length();``    ``int` `n2 = str2.length();` `    ``// Reverse both of strings``    ``reverse(str1.begin(), str1.end());``    ``reverse(str2.begin(), str2.end());` `    ``int` `carry = 0;``    ``for` `(``int` `i = 0; i < n1; i++) {` `        ``// Find the sum of the current``        ``// digits and carry``        ``int` `sum``            ``= ((str1[i] - ``'0'``)``               ``+ (str2[i] - ``'0'``)``               ``+ carry);``        ``str.push_back(sum % 10 + ``'0'``);` `        ``// Calculate carry for next step``        ``carry = sum / 10;``    ``}` `    ``// Add remaining digits of larger number``    ``for` `(``int` `i = n1; i < n2; i++) {``        ``int` `sum = ((str2[i] - ``'0'``) + carry);``        ``str.push_back(sum % 10 + ``'0'``);``        ``carry = sum / 10;``    ``}` `    ``// Add remaining carry``    ``if` `(carry)``        ``str.push_back(carry + ``'0'``);` `    ``// Reverse resultant string``    ``reverse(str.begin(), str.end());` `    ``return` `str;``}` `// Function to find difference of larger``// numbers represented as strings``string findDiff(string str1, string str2)``{``    ``// Stores the result of difference``    ``string str = ``""``;` `    ``// Calculate length of both string``    ``int` `n1 = str1.length(), n2 = str2.length();` `    ``// Reverse both of strings``    ``reverse(str1.begin(), str1.end());``    ``reverse(str2.begin(), str2.end());` `    ``int` `carry = 0;` `    ``// Run loop till small string length``    ``// and subtract digit of str1 to str2``    ``for` `(``int` `i = 0; i < n2; i++) {` `        ``// Compute difference of the``        ``// current digits``        ``int` `sub``            ``= ((str1[i] - ``'0'``)``               ``- (str2[i] - ``'0'``)``               ``- carry);` `        ``// If subtraction < 0 then add 10``        ``// into sub and take carry as 1``        ``if` `(sub < 0) {``            ``sub = sub + 10;``            ``carry = 1;``        ``}``        ``else``            ``carry = 0;` `        ``str.push_back(sub + ``'0'``);``    ``}` `    ``// Subtract the remaining digits of``    ``// larger number``    ``for` `(``int` `i = n2; i < n1; i++) {``        ``int` `sub = ((str1[i] - ``'0'``) - carry);` `        ``// If the sub value is -ve,``        ``// then make it positive``        ``if` `(sub < 0) {``            ``sub = sub + 10;``            ``carry = 1;``        ``}``        ``else``            ``carry = 0;` `        ``str.push_back(sub + ``'0'``);``    ``}` `    ``// Reverse resultant string``    ``reverse(str.begin(), str.end());` `    ``// Return answer``    ``return` `str;``}` `// Function to remove all leading 0s``// from a given string``string removeLeadingZeros(string str)``{``    ``// Regex to remove leading 0s``    ``// from a string``    ``const` `regex pattern(``"^0+(?!\$)"``);` `    ``// Replaces the matched value``    ``// with given string``    ``str = regex_replace(str, pattern, ``""``);``    ``return` `str;``}` `// Function to multiply two numbers``// using Karatsuba algorithm``string multiply(string A, string B)``{``    ``if` `(A.length() > B.length())``        ``swap(A, B);` `    ``// Make both numbers to have``    ``// same digits``    ``int` `n1 = A.length(), n2 = B.length();``    ``while` `(n2 > n1) {``        ``A = ``"0"` `+ A;``        ``n1++;``    ``}` `    ``// Base case``    ``if` `(n1 == 1) {` `        ``// If the length of strings is 1,``        ``// then return their product``        ``int` `ans = stoi(A) * stoi(B);``        ``return` `to_string(ans);``    ``}` `    ``// Add zeros in the beginning of``    ``// the strings when length is odd``    ``if` `(n1 % 2 == 1) {``        ``n1++;``        ``A = ``"0"` `+ A;``        ``B = ``"0"` `+ B;``    ``}` `    ``string Al, Ar, Bl, Br;` `    ``// Find the values of Al, Ar,``    ``// Bl, and Br.``    ``for` `(``int` `i = 0; i < n1 / 2; ++i) {``        ``Al += A[i];``        ``Bl += B[i];``        ``Ar += A[n1 / 2 + i];``        ``Br += B[n1 / 2 + i];``    ``}` `    ``// Recursively call the function``    ``// to compute smaller product` `    ``// Stores the value of Al * Bl``    ``string p = multiply(Al, Bl);` `    ``// Stores the value of Ar * Br``    ``string q = multiply(Ar, Br);` `    ``// Stores value of ((Al + Ar)*(Bl + Br)``    ``// - Al*Bl - Ar*Br)``    ``string r = findDiff(``        ``multiply(findSum(Al, Ar),``                 ``findSum(Bl, Br)),``        ``findSum(p, q));` `    ``// Multiply p by 10^n``    ``for` `(``int` `i = 0; i < n1; ++i)``        ``p = p + ``"0"``;` `    ``// Multiply s by 10^(n/2)``    ``for` `(``int` `i = 0; i < n1 / 2; ++i)``        ``r = r + ``"0"``;` `    ``// Calculate final answer p + r + s``    ``string ans = findSum(p, findSum(q, r));` `    ``// Remove leading zeroes from ans``    ``ans = removeLeadingZeros(ans);` `    ``// Return Answer``    ``return` `ans;``}` `// Driver Code``int` `main()``{``    ``string A = ``"74638463789"``;``    ``string B = ``"35284567382"``;` `    ``cout << multiply(A, B);` `    ``return` `0;``}`

Output:

`2633585904851937530398`

Time Complexity: O(Nlog 3) or O(N1.59), where N is the maximum among the lengths given strings A and B.
Auxiliary Space: O(N2)

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