Given an encoded string where repetitions of substrings are represented as substring followed by count of substrings. For example, if encrypted string is “ab2cd2” and k=4, so output will be ‘b’ because decrypted string is “ababcdcd” and 4th character is ‘b’.
Note: Frequency of encrypted substring can be of more than one digit. For example, in “ab12c3”, ab is repeated 12 times. No leading 0 is present in frequency of substring.
Input: "a2b2c3", k = 5 Output: c Decrypted string is "aabbccc" Input: "ab4c2ed3", k = 9 Output : c Decrypted string is "ababababccededed"
The solution discussed in previous post requires additional O(n) space. The following post discuss a solution that requires constant space. The stepwise algorithm is:
- Find length of current substring. Use two pointers. Fix one pointer at beginning of substring and move another pointer until a digit is not found.
- Find frequency of repetition by moving the second pointer further until an alphabet is not found.
- Find length of substring if it is repeated by multiplying frequency and its original length.
- If this length is less than k, then required character lies in substring that follows. Subtract this length from k to keep count of number of characters that are required to be covered.
- If length is less than or equal to k, then required character lies in current substring. As k is 1-indexed reduce it by 1 and then take its mod with original substring length. Required character is kth character from starting of substring which is pointed by first pointer.
Below is the implementation of the above approach:
# Python3 program to find K’th
# character in decrypted string
# Function to find K’th character
# in Encoded String
def encodedChar(string, k):
n = len(string)
i = 0
while i < n: j = i length = 0 freq = 0 # Find length of substring by # traversing the string until # no digit is found. while j < n and string[j].isalpha(): j += 1 length += 1 # Find frequency of preceding substring. while j < n and string[j].isdigit(): freq = freq * 10 + int(string[j]) j += 1 # Find the length of the substring # when it is repeated. num = freq * length # If the length of the repeated substring # is less than k then required character # is present in next substring. Subtract # the length of repeated substring from # k to keep account of the number # of characters required to be visited. if k > num:
k -= num
i = j
# If length of repeated substring is
# more or equal to k then required
# character lies in current substring.
k -= 1
k %= length
return string[i + k]
# This is for the case when there are no
# repetition in string. e.g. str=”abced”.
return string[k – 1]
# Driver Code
if __name__ == “__main__”:
string = “abced”
k = 4
# This code is contributed
# by Rituraj Jain
Time Complexity: O(n)
Auxiliary Space: O(1)
- Find k'th character of decrypted string | Set 1
- Find a string such that every character is lexicographically greater than its immediate next character
- Find the character in first string that is present at minimum index in second string
- Queries to find the first non-repeating character in the sub-string of a string
- Queries to find the last non-repeating character in the sub-string of a given string
- Find one extra character in a string
- Find the last non repeating character in string
- Find last index of a character in a string
- Find the first repeated character in a string
- Given a string, find its first non-repeating character
- Find repeated character present first in a string
- Program to find the kth character after decrypting a string
- Find i'th Index character in a binary string obtained after n iterations
- Efficiently find first repeated character in a string without using any additional data structure in one traversal
- Replace every character of string by character whose ASCII value is K times more than it
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Improved By : rituraj_jain