Python Code to Count Possible Decoding Of A Given Digit Sequence
Count Possible Decoding Of A Given Digit Sequence in Python
Here we will discuss how to count all the possible decoding of a given digit sequence in Python. Before counting the number of decoding’s first let’s see how the numbers are assumed to be coded. The decoding programs are the most possible questions asked and are largely practiced in Python programming. The program counts the number of possible decoding’s of the entered digit by the user of the given sequence.
1 = A, 2 = B, 3 = C, . . . . , 26 = Z.
The decoding programs are the most possible questions asked and are largely practiced in C programming. The program counts the number of possible decoding’s of the entered digit by the user of the given sequence.
For example :- if the digit sequence is “12” then there are two possible decoding’s for this – One of them is ‘AB’ when we decode 1 as ‘A’ and 2 as ‘B’. Now we can also decode this digit sequence “12” as ‘L’ when we decode digits 1 and 2 taken together as an integer 12.
Now let’s take another example for reference,
Sequence = 131
- Possible decoding (1, 3, 1,) = ACA
- Possible decoding (13, 1) = MA
So, the total possible decoding’s of sequence 131 is 2.
- Initialize a array cnt of size 1 + length of input and store 1 at 0 & 1 position
- Iterate from 2 to a using variable k
- If the last digit not equal to 0, then last digit must added to the number of words if (dig[k-1] > ‘0’)
- In case second last digit is smaller than 2 and last digit is smaller than 7, then last two digits form a valid character
- Print cnt[a]
Rundef cnt_decoding_digits(dig, a): cnt =  * (a + 1) cnt, cnt = 1, 1 for k in range(2, a + 1): cnt[k] = 0 cnt[k] = cnt[k - 1] if dig[k - 2] == '1' or (dig[k - 2] == '2' and dig[k - 1] < '7'): cnt[k] += cnt[k - 2] return cnt[a] dig = input("Enter Number : ") print("Possible count of decoding of the sequence :", cnt_decoding_digits(dig, len(dig)))
OutputEnter the sequence : 141 Possible count of decoding of the sequence : 2