class Solution {

// Calculate the sum of all of the results from multiplyOneDigit.

string sumResults(vector<vector<int>>& results) {

// Initialize answer as a number from results.

vector<int> answer = results.back();

vector<int> newAnswer;

results.pop_back();

// Sum each digit from answer and result

for (vector<int> result : results) {

newAnswer.clear();

int carry = 0;

for (int i = 0; i < answer.size() || i < result.size(); ++i) {

// If answer is shorter than result or vice versa, use 0 as the current digit.

int digit1 = i < result.size() ? result[i] : 0;

int digit2 = i < answer.size() ? answer[i] : 0;

// Add current digits of both numbers.

int sum = digit1 + digit2 + carry;

// Set carry equal to the tens place digit of sum.

carry = sum / 10;

// Append the ones place digit of sum to answer.

newAnswer.push_back(sum % 10);

}

if (carry) {

newAnswer.push_back(carry);

}

answer = newAnswer;

}

// Convert answer to a string.

string finalAnswer;

for (int digit : answer) {

finalAnswer.push_back(digit + '0');

}

return finalAnswer;

}

// Multiply the current digit of secondNumber with firstNumber.

vector<int> multiplyOneDigit(string& firstNumber, char& secondNumberDigit, int numZeros) {

// Insert zeros at the beginning based on the current digit's place.

vector<int> currentResult(numZeros, 0);

int carry = 0;

// Multiply firstNumber with the current digit of secondNumber.

for (char firstNumberDigit : firstNumber) {

int multiplication = (secondNumberDigit - '0') * (firstNumberDigit - '0') + carry;

// Set carry equal to the tens place digit of multiplication.

carry = multiplication / 10;

// Append last digit to the current result.

currentResult.push_back(multiplication % 10);

}

if (carry) {

currentResult.push_back(carry);

}

return currentResult;

}

string multiply(string firstNumber, string secondNumber) {

if (firstNumber == "0" || secondNumber == "0") {

return "0";

}

// Reverse both numbers.

reverse(firstNumber.begin(), firstNumber.end());

reverse(secondNumber.begin(), secondNumber.end());

// For each digit in secondNumber, multipy the digit by firstNumber and

// store the multiplication result (reversed) in results.

vector<vector<int>> results;

for (int i = 0; i < secondNumber.size(); ++i) {

results.push_back(multiplyOneDigit(firstNumber, secondNumber[i], i));

}

// Add all the results in the results array, and store the sum in the answer string.

string answer = sumResults(results);

// answer is reversed, so reverse it to get the final answer.

reverse(answer.begin(), answer.end());

return answer;

}

};