Suppose we have two numbers as a string. We have to multiply them and return the result also in a string. So if the numbers are “26” and “12”, then the result will be “312”
To solve this, we will follow these steps −
- Taking two arguments x and y it indicates x divides y
- if x < -Infinity and y = 1, then return infinity
- a := |x|, b := |y| and ans := 0
- while a – b >= 0
- p := 0
- while a – (left-shifted b (left-shifted 1 p times)) >= 0
- p := p + 1
- a := a – (left shift b, p times)
- ans := ans + left shift 1 p times
- if x > 0 is true and y > 0 is also true, then return ans, otherwise return (– ans)
Example(C++)
Let us see the following implementation to get a better understanding −
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
string multiply(string num1, string num2);
};
string Solution::multiply(string nums1, string nums2) {
int n = nums1.size();
int m = nums2.size();
string ans(n + m, '0');
for(int i = n - 1; i>=0; i--){
for(int j = m - 1; j >= 0; j--){
int p = (nums1[i] - '0') * (nums2[j] - '0') + (ans[i + j + 1] - '0');
ans[i+j+1] = p % 10 + '0';
ans[i+j] += p / 10 ;
}
}
for(int i = 0; i < m + n; i++){
if(ans[i] !='0')return ans.substr(i);
}
return "0";
}
main(){
Solution ob;
cout << ob.multiply("28", "25");
}Input
"26" "12"
Output
"312"