Find k-th character of decrypted string | Set – 2

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.

Examples:

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:

  1. 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.
  2. Find frequency of repetition by moving the second pointer further until an alphabet is not found.
  3. Find length of substring if it is repeated by multiplying frequency and its original length.
  4. 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.
  5. 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:

C++

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// C++ program to find K'th character in
// decrypted string
#include <bits/stdc++.h>
using namespace std;
  
// Function to find K'th character in 
// Encoded String
char encodedChar(string str, int k)
{
    int i, j;
  
    int n = str.length();
  
    // To store length of substring
    int len;
  
    // To store length of substring when
    // it is repeated
    int num;
  
    // To store frequency of substring
    int freq;
  
    i = 0;
  
    while (i < n) {
        j = i;
        len = 0;
        freq = 0;
  
        // Find length of substring by
        // traversing the string until 
        // no digit is found.
        while (j < n && isalpha(str[j])) {
            j++;
            len++;
        }
  
        // Find frequency of preceding substring.
        while (j < n && isdigit(str[j])) {
            freq = freq * 10 + (str[j] - '0');
            j++;
        }
  
        // Find length of substring when
        // it is repeated.
        num = freq * len;
  
        // If length of repeated substring is less than
        // k then required character is present in next
        // substring. Subtract length of repeated
        // substring from k to keep account of 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.
        else {
            k--;
            k %= len;
            return str[i + k];
        }
    }
  
    // This is for the case when there
    // are no repetition in string.
    // e.g. str="abced".
    return str[k - 1];
}
  
// Driver Code
int main()
{
    string str = "abced";
    int k = 4;
  
    cout << encodedChar(str, k) << endl;
  
    return 0;
}

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Python3

# 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.
else:
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

print(encodedChar(string, k))

# This code is contributed
# by Rituraj Jain

Output:

e

Time Complexity: O(n)
Auxiliary Space: O(1)



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Improved By : rituraj_jain