# Decoded String at Index in Python

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Suppose one encoded string S is given. We have to find and write the decoded string to a tape, here the encoded string is read one character at a time and the following steps are performed −

• If the character read is a letter, that letter is simply written onto the tape.
• If the character read is a digit, the entire current tape is repeatedly written digit – 1 more times in total.

Now if some encoded string S, and an index K is given, find and return the K-th letter (starting indices from 1) in the decoded string.

So if the string is “hello2World3” and k = 10, then the output will be “o”. This is because the decoded string will be “hellohelloWorldhellohelloWorldhellohelloWorld”, so 10th character is “o”.

To solve this, we will follow these steps −

• size := 0
• for i in string s
• if i is a numeric character, then size := size * integer from of i, otherwise size := size + 1
• for i in range length of s – 1 down to 0
• k := k mod size
• if s[i] is numeric and k = 0, then return s[i]
• if s[i] is numeric, then decrease size by 1, otherwise size := size / integer of s[i]
• return empty string

Let us see the following implementation to get better understanding −

## Example

Live Demo

class Solution(object):
def decodeAtIndex(self, s, k):
"""
:type S: str
:type K: int
:rtype: str
"""
size = 0
for i in s:
if i.isdigit():
size *= int(i)
else:
size += 1
#print(size)
for i in range(len(s) - 1, -1, -1):
k %= size
if s[i].isalpha() and k == 0:
return s[i]
if s[i].isalpha():
size -=1
else:
size /= int(s[i])
return ""
ob = Solution()
print(ob.decodeAtIndex("hello2World3", 10))

## Input

"hello2World3"
10
ob.decodeAtIndex("hello2World3", 10)

## Output

o
Updated on 30-Apr-2020 06:09:03