My code doesn't work

#1

#include
#include
#include
using namespace std;
int main() {
string n;
cin>>n;
int currO = (int)n[0];
int alterO = 9 - (int)n[0];
if(alterO!=0 && alterO<currO) {
char nowO = (char)alterO;
n[0] = nowO ;
}

for(int i=1; n[i] ; i++) {
	int curr,alter;
	curr = (int)n[i];
	alter = 9 - (int)n[i];
	if(alterO<currO) {
		char now = (char)alterO;
		n[i] = now ;
	}
}
cout<<n;

}

#2

@priyamthakuria27
There are lot of mistake. I am pointing them out. let say testcase is"4545" ,see (int)n[0] won’t give 4, it will give ASCII value of character ‘4’ i.e 52. Then when you do alterO=9-(int)n[0], it would result in (9-52) which is negative number. and it’ll give wrong result.
To deal with this type of situation, you code should go like this
int currO = n[0]-‘0’;
int alterO = 9 - currO;
if(alterO!=0 && alterO<currO) {
char nowO = (char)(alterO+‘0’);
n[0] = nowO ;
}

#3

Do the same operation inside for loop as well

#4

what does the " -‘0’ " do?

#5

@priyamthakuria27
inorder to get the integral value of ith index in the string we are using ASCII value. So difference of ASCII value of char(a) and ‘0’ will give difference between them. let say char a=‘5’, so if we subtract ‘0’ from it we will get int(5). This is because we are actually subtracting their ASCII values, So ASCII_value(‘5’) - ASCII_value(‘0’) = 53 - 48 = 5.
You can’t directly change char to int.