Chewbacca and Number

#1

how do I approach chewbacca and Number problem

#2

hello @s.kumar

read a number as string (say s) .
if left most character is ‘9’

  • then leave the leftmost character as it is and for other characters simpl replace s[i] with min(s[i],‘9’-s[i])

else:

  • for each i replace s[i] with min(s[i],‘9’-s[i])

print s

#include<bits/stdc++.h>

using namespace std;

int main() {

    string n;

    cin>>n;

    int start=0;

    if(n[0]=='9')

       start=1;

    for(int i=start;i<n.length();i++){

            char c='9'-n[i]+'0';

            n[i]=min(n[i],c );

    }

    cout<<n;

    return 0;

}
#3

why is ‘0’ added in the following statement,
char c=‘9’-n[i]+‘0’;

#4

to convert int to char we add ‘0’ to the number.
here we did that for the same purpose

#5

but both ‘9’ and s[i] are char only, isn’t it?

#6

…

no, that will be int

#7

‘9’ is char, right?
4 is int,
then why ‘9’-(4+‘0’) is giving correct o/p

#8

it will give 5 which is int and not char.

u again need to to add ‘0’ to convert 5 in to ‘5’

#9

bro,
n[i]=min(n[i],c); //here n[i] on the right side holds the int value, and c holds char, is it possible to compare char and int?

#10

…

n[i] is char and c is also char.

#11

why is n[i]=min(n[i], ‘9’-n[i]+‘0’) statement not working

for(int i=start;i<n.length();i++){

        //char c='9'-n[i]+'0';

        n[i]=min(n[i],'9'-n[i]+'0' );

}
#12

compiler is treating this -> ‘9’-n[i]+‘0’ as an int .
use char(‘9’-n[i]+‘0’)

or write ur own min function

#13

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