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# Find k-th character of decrypted string - Set – 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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