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Find k-th character of decrypted string - Set β 2 in Python
Suppose we have one encoded string where repetitions of substrings are represented as substring followed by count of substrings. As an example, if the string is "pq2rs2" and k=5, so output will be 'r', this is because the decrypted string is "pqpqrsrs" and 5th character is 'r'. We have to keep in mind that the frequency of encrypted substring can be of more than one digit.
So, if the input is like string = "pq4r2ts3" and k = 11, then the output will be i, as the string is pqpqpqpqrrtststs
To solve this, we will follow these steps −
encoded := blank string
occurrence := 0, i := 0
while i < size of str, do
temp := blank string
occurrence := 0
while i < size of str and str[i] is an alphabet, do
temp := temp + str[i]
i := i + 1
while i < size of str and str[i] is a digit, do
occurrence := occurrence * 10 + ASCII of (str[i]) - ASCII of ('0')
i := i + 1
for j in range 1 to occurrence + 1, increase by 1, do
encoded := encoded + temp
if occurrence is same as 0, then
encoded := encoded + temp
return encoded[k - 1]
Example
Let us see the following implementation to get better understanding −
def find_kth_char(str, k):
encoded = ""
occurrence = 0
i = 0
while i < len(str):
temp = ""
occurrence = 0
while (i < len(str) and ord(str[i]) >= ord('a') and ord(str[i]) <= ord('z')):
temp += str[i]
i += 1
while (i < len(str) and ord(str[i]) >= ord('1') and ord(str[i]) <= ord('9')):
occurrence = occurrence * 10 + ord(str[i]) - ord('0')
i += 1
for j in range(1, occurrence + 1, 1):
encoded += temp
if occurrence == 0:
encoded += temp
return encoded[k - 1]
str = "pq4r2ts3"
k = 11
print(find_kth_char(str, k))Input
"pq4r2ts3", 11
Output
t
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