How to multiply strings without overflow by c++? - c++

Question

I tried to solve Multiply Strings by c++ by this approach, but I cannot avoid integer overflow by change type from int to long long int or double. Python won't overflow, so my code works like below.
Given two non-negative integers num1 and num2 represented as strings, return the product of num1 and num2, also represented as a string.
Python works:
class Solution:
def multiply(self, num1: str, num2: str) -> str:
n = len(num1) # assume n >= m
m = len(num2)
if n < m:
num1, num2 = num2, num1
n, m = m, n
product = 0
for i in range(1, m + 1):
multiplier = int(num2[m - i]) # current character of num2
sum_ = 0
for j in range(0, n): # multiply num1 by multiplier
multiplicand = int(num1[n - j - 1])
num = multiplicand * (10 ** j) * multiplier
sum_ += num
product += sum_ * (10 ** (i - 1))
return str(product)
C++ failed:
string multiply(string num1, string num2) {
int n = num1.size();
int m = num2.size();
if (n < m) {
std::swap(num1, num2);
std::swap(n, m);
}
long long int product = 0;
for (int i = 1; i <= m; ++i) {
int multiChar = num2[m - i] - '0';
long long int sum = 0;
for (int j = 0; j < n ; ++j) {
int charCand = num1[n - j - 1] - '0';
long long int num = charCand * ((pow(10, j))) * multiChar;
sum += num;
}
product += sum * ((pow(10, i - 1)));
}
return std::to_string(product);
}
As far as I have tested, some cases are OK, but overflow seems unavoidable if the number is too big. Is there any way to fix my code?
Testcase:
"12323247989"
"98549324321"
runtime error: 1.05355e+20 is outside the range of representable values of type 'long long' (solution.cpp)
SUMMARY: UndefinedBehaviorSanitizer: undefined-behavior prog_joined.cpp:28:17
Expected:
"1214447762756072040469"

Answer 1

You are not on the right way. Imagine how you would do that by hand:
abc*def
-------
xxxx
xxxx0
xxxx00
-------
You just add single digits as well, don't you? Only those of same significance – possibly considering some carry.
You might rather reproduce the same in code, too. Producing overflow that way is much less likely (I assume that after multiplying single digits summing up the results in a single integer – recommending an unsigned type for – is acceptable; if not, you'd have to build up a std::string again). The sign you calculate independently, just as you'd do by hand as well.
One difference to multiplication by hand we'll have, though: By hand you would create rather large intermediate numbers by multiplying one number with each digit
of the other number. That would require to store these intermediate numbers as strings again, e. g. in a vector. More efficient, though, is identifying those digit pairs of which the multiplication results in the same significance.
These will be 0|0 -> 0; 0|1, 1|0 -> 1; 0|2, 1|1, 2|0 -> 2, and so on. You produce these pairs by:
for(size_t i = 0, max = std::max(num1.length(), num2.length); i < max; ++i)
{
for(size_t j = 0; j < i; ++j)
{
if(j < num1.length() && i - j < num2.length())
{
// iterate backwards for easy carry handling
size_t idx1 = num1.length() - j;
size_t idx2 = num2.length() - (i - j);
// multiply characters at num1[idx1] and num2[idx2] and add result to sum
}
}
// add carry
// calculate last digit and a p p e n d to a result string
// update carry
}
// append '-' sign, if result is negative
std::reverse(result.begin(), result.end());
Building up the string in reverse order is more efficient, as you do not need to move the subsequent characters all the time. (Untested code, if you find a bug, please fix it yourself).
The loops are in my preferred variant; if you feel better with another, feel free to change; just be aware that with signed types you can produce endless loops if you try e. g. for(unsigned i = n; n >= 0; --i /* overflows to UINT_MAX */).
Side note: You should accept the input strings by reference (std::string const& num1, std::string const& num2), that avoids the needless copies arising by accepting by value.