Multiply Large Numbers Represented as Strings in C++



Given two numbers in the string formats. We need to multiply them. The idea to solve the problem is to maintain a previous digit multiplication answer and carry. We can use the previous digits multiplication answer and carry to get the next set digits multiplication.

Let's see an example.

Input

15
2

Output

30

Algorithm

  • Initialise the numbers in string.

  • Initialise a string of length number_one_length + number_two_length.

  • Iterate over the first number from the end.

    • Iterate over the second number from the end.

      • Multiply two digits and add the corresponding previous row digit.

      • Update the previous row digit.

      • Store the carry in the previous index of the result string.

  • Convert the char to digits by adding 0 character to every character in the result.

  • Return the result by ignoring the leading zero.

Implementation

Following is the implementation of the above algorithm in C++

Open Compiler
#include <bits/stdc++.h> using namespace std; string multiplyTwoNumbers(string num1, string num2) { if (num1 == "0" || num2 == "0") { return "0"; } string product(num1.size() + num2.size(), 0); for (int i = num1.size() - 1; i >= 0; i--) { for (int j = num2.size() - 1; j >= 0; j--) { int n = (num1[i] - '0') * (num2[j] - '0') + product[i + j + 1]; product[i + j + 1] = n % 10; product[i + j] += n / 10; } } for (int i = 0; i < product.size(); i++) { product[i] += '0'; } if (product[0] == '0') { return product.substr(1); } return product; } int main() { string num1 = "34"; string num2 = "57"; if((num1.at(0) == '-' || num2.at(0) == '-') && (num1.at(0) != '-' || num2.at(0) != '-')) { cout << "-"; } if(num1.at(0) == '-') { num1 = num1.substr(1); } if(num2.at(0) == '-') { num2 = num2.substr(1); } cout << multiplyTwoNumbers(num1, num2) << endl; return 0; }

Output

If you run the above code, then you will get the following result.

1938
Updated on: 2021-10-25T07:10:44+05:30

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