# Concise 3ms Java solution

• The multiplication result will have at most `num1.length() + num2.length()` digits. So we start by allocating a `char` array of this length. The advantage of using a char array is that it can be easily (and efficiently) turned into a `String` by invoking the appropriate constructor.

To compute the final result, we need to

• multiply `num1` with each digit in `num2`, going from the least significant to the most significant digit
• at each iteration `i` starting from `0` multiply the intermediate result by `10^i`
• sum up all the intermediate results

In the solution below we compute a running sum, adding each intermediate result to the sum as it is computed. We accumulate the sum at each position, only taking care of carries at the end so that it has to be done just once with a linear scan. The multiplication by `10^i` at each iteration is achieved by decrementing `startpos` at each iteration.

``````public String multiply(String num1, String num2) {
char[] arr = new char[num1.length() + num2.length()];
char[] num2c = new char[num2.length()];

// Copy num2 into an auxiliary array to avoid repeated subtraction of '0'
num2.getChars(0, num2.length(), num2c, 0);
for (int i = 0; i < num2c.length; i++) {
num2c[i] -= '0';
}

for (int i = num1.length() - 1, startpos = arr.length - 1; i >= 0; i--, startpos--) {
int digit = num1.charAt(i) - '0';
for (int j = num2c.length - 1, pos = startpos; j >= 0; j--, pos--) {
arr[pos] += num2c[j] * digit;
}
}

// Take care of carrys
for (int i = arr.length - 1; i > 0; i--) {
arr[i - 1] += arr[i] / 10;
arr[i] = (char)(arr[i] % 10 + '0');
}
arr[0] += '0';