Codes of the string

#1

The TAs explain way better than this guy. Can you please explain the approach to this problem

#2

@aiman.mumtaz
see here you have to make all the possible string by mapping the digits to the chars.
Taking the sample input:- 123
ABC using 1-A,2-B,3-C separately
AW using 1-A, (23)-W
LC using (12)-L, 3-C

I hope you got how the above answers are calculated.
Now moving on how recursive code will work.
Here we need to do only two things.

  1. Take the single digit and convert it into mapped characters.
  2. Check it is possible to combine that single digit with next digit.
    For eg:- if current digit is 2 and next is 5, then we can do option 2.
    but if they are 2 and 7, then option 2 is not possible.

So my code is totally based upon these 2 things.

void solve(string s, int i, int n, string curr){
    if(i==n){
        cout<<curr<<endl;
        return;
    }
    else{
        int digit = s[i]-'0';
        solve(s, i+1, n, curr + key[digit]); //calling recursion because step  1 is done

        if(n-i>=2){  //check if i is n-1 then there is no next digit
            int digit1 = s[i]-'0';
            int digit2 = s[i+1]-'0';
            int num = digit1*10 + digit2;
            if(num<=26 && num>9)  //now checking if the number formed is satisfying step 2
                solve(s, i+2, n, curr+key[num]); //if yes then call recursion and change i to i+2
        }
    }
}

I hope you have understood it now.
Let me know if there is some doubt left.
If it is helpful then please mark this doubt as resolved.

#3

Yes. I got the logic and implemented the code. I need one tiny help. How to remove the comma from the last permutation??

#4
#5

@aiman.mumtaz
make a global count variable, print ‘,’ just before printing the cout<<out , with the help of global count variable don’t print ‘,’ for count=0.
this is the code

#6

@aiman.mumtaz
please also remember to mark this doubt as resolved if it was helpful and able to clear your doubt.