Chewbacca and the number

#1

how to approach to the question

#2

@anmolsr755

The logic behind this Problem was pretty simple as we can invert any digit ( 9 - digit) but we need to invert only such digits that will eventually end up giving the smallest number possible.

So, we should invert only digits greater then or equal to 5 as after inverting them, the result gives us smallest number.

For e.g.,

9 - 5 = 4 viz, smaller than the original number that was 5.

9 - 8 = 1 viz, smaller than the original number that was 8.

but 9 - 1 = 8 viz, greater than the original number that was 1.

Important point to consider is, After inverting any digits their should not be trailing zeros that means, if their is 9 at the starting of the number then it must remain the same

Also input can be quite big so, u need to capture the number in long long int in c++ and long in java.

#3

can we do it without using array

#4

You can but then you will have to print the numbers side by side.
Something like this
while(n>0){
if(n%10>=5) cout<<9-(n%10); else cout<<n%10;
n/=10;
}
but you need to do this in reverse order, so you may call it recursively, but then again it uses space in the stack. So use a simple array please.

#5

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