Problem of the day
'A' -> "1"
'B' -> "2"
...
'Z' -> "26"
Input: βSβ = "11106"
Output: 2
The possible ways are:-
(1 1 10 6),
(11 10 6)
Note that the grouping (1 11 06) is invalid because "06" cannot be mapped into 'F' since "6" is different from "06".
The first line contains βTβ, denoting the number of test cases.
Each test case's first and only line contains the string βSβ.
Return the number of ways to decode the string βSβ.
You don't need to print anything. It has already been taken care of. Just implement the given function.
1 <= T <= 10
1 <= | S | <= 100
Time Limit: 1 sec
2
12
226
2
3
For the first test case:-
"12" could be decoded as "AB" (1 2) or "L" (12).
For the second test case:-
"226" could be decoded as "BZ" (2 26), "VF" (22 6), or "BBF" (2 2 6).
2
1234
333
3
1