Medium — Leetcode: https://leetcode.com/problems/decoded-string-at-index/solution/
An encoded string
S is given. To find and write the decoded string to a tape, the encoded string is read one character at a time and the following steps are taken:
- If the character read is a letter, that letter is written onto the tape.
- If the character read is a digit (say
d), the entire current tape is repeatedly written
d-1more times in total.
Now for some encoded string
S, and an index
K, find and return the
K-th letter (1 indexed) in the decoded string.
Input: S = "leet2code3", K = 10
The decoded string is "leetleetcodeleetleetcodeleetleetcode".
The 10th letter in the string is "o".
Input: S = "ha22", K = 5
The decoded string is "hahahaha". The 5th letter is "h".
Input: S = "a2345678999999999999999", K = 1
The decoded string is "a" repeated 8301530446056247680 times. The 1st letter is "a".
2 <= S.length <= 100
Swill only contain lowercase letters and digits
Sstarts with a letter.
1 <= K <= 10^9
- It’s guaranteed that
Kis less than or equal to the length of the decoded string.
- The decoded string is guaranteed to have less than
Intuition :Work Backwards
We dont actually need to constuct the whole string to be able to do this better. ALl we need is that index of the character to get the value from the original string itself.
If we have a decoded string like
appleappleappleappleappleapple and an index like
K = 24, the answer is the same if
K = 4.
Explaination : Here we did apple ( size 5) repeated 6 times resulting in 5*6=30 character long string. But this 30 characters are 6 times the same original string Apple. The index of “l” in apple is 4. And the index of “l” in apple repeated 6 times are [4, 5 * 1 +4 = 9, 5*2 +4 = 14, 5*3 +4 = 19, 5 * 4 + 4 = 24 and 5 * 5 +4 = 29] ,
In general, when a decoded string is equal to some word with
size length repeated some number of times (such as
size = 5 repeated 6 times), the answer is the same for the index
K as it is for the index
K % size.
We can use this insight by working backwards, keeping track of the size of the decoded string. Whenever the decoded string would equal some
d times, we can reduce
K % (word.length).
- First, find the length of the decoded string.
- After, we’ll work backwards, keeping track of
size: the length of the decoded string after parsing symbols
S, S, ..., S[i].
- If we see a digit
S[i], it means the size of the decoded string after parsing
S, S, ..., S[i-1]will be
size / Integer(S[i]).
- If we see a character, it means the size after decoding this character is
size - 1.
def decodeAtIndex(self, S: str, K: int) -> str:
size = 0
# Find size = length of decoded string
for c in S:
size *= int(c)
size += 1
for c in reversed(S):
K %= size
# print(K, size)
if K == 0 and c.isalpha():
return cif c.isdigit():
size /= int(c)
size -= 1