Given a string str consisting of characters and numbers and an integer k, the task is to decrypt the string and the return the kth character in the decrypted string.
In order to decrypt the string, traverse the string character by character and if the current character is an alphabet then append it to the resultant string else if it is a numeric digit then parse the number and repeat the resultant string this parsed number of times and continue with the original string. For example, str = “ab2c3” will be decrypted as “ababcababcababc”.
Examples:
Input: str = “ab2c3”, k = 5
Output: c
Decrypted string will be “ababcababcababc” and ‘c’ is the fifth character.
Input: str = “x2y3”, k = 3
Output: y
Approach:
- Initialize starting index i = 0 and total_len = 0.
- Loop while i is less than the length of the input string and check if current character is an alphabet or not. If yes then increment total_len by 1 and check if total length is less than or equal to k if yes then return the string else increment i.
- Initialize n = 0 again loop while i is less than length of input string and i is not an alphabet and parse the number and increment i and find the next_total_len = total_len * n.
- If k < total_len then get the position of the character by initializing pos = k % total_len.
- If position is not found then update position = total_len and finally return the character at kth position. If not found then return -1.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <cstdlib>#include <iostream>using namespace std;// Function to print kth character of// String s after decrypting itchar findKthChar(string s, int k){ // Get the length of string int len = s.length(); // Initialise pointer to character // of input string to zero int i = 0; // Total length of resultant string int total_len = 0; // Traverse the string from starting // and check if each character is // alphabet then increment total_len while (i < len) { if (isalpha(s[i])) { total_len++; // If total_len equal to k then // return string else increment i if (total_len == k) return s[i]; i++; } else { // Parse the number int n = 0; while (i < len && !isalpha(s[i])) { n = n * 10 + (s[i] - '0'); i++; } // Update next_total_len int next_total_len = total_len * n; // Get the position of kth character if (k <= next_total_len) { int pos = k % total_len; // Position not found then update // position with total_len if (!pos) { pos = total_len; } // Recursively find the kth position return findKthChar(s, pos); } else { // Else update total_len // by next_total_len total_len = next_total_len; } } } // Return -1 if character not found return -1;}// Driver codeint main(){ string s = "ab2c3"; int k = 5; cout << findKthChar(s, k); return 0;} |
Java
// Java implementation of the approach import java.util.*;class GfG { // Function to print kth character of // String s after decrypting it static Character findKthChar(String s, int k) { // Get the length of string int len = s.length(); // Initialise pointer to character // of input string to zero int i = 0; // Total length of resultant string int total_len = 0; // Traverse the string from starting // and check if each character is // alphabet then increment total_len while (i < len) { if (Character.isLetter(s.charAt(i))) { total_len++; // If total_len equal to k then // return string else increment i if (total_len == k) return s.charAt(i); i++; } else { // Parse the number int n = 0; while (i < len && !Character.isLetter(s.charAt(i))) { n = n * 10 + (s.charAt(i) - '0'); i++; } // Update next_total_len int next_total_len = total_len * n; // Get the position of kth character if (k <= next_total_len) { int pos = k % total_len; // Position not found then update // position with total_len if (pos == 0) { pos = total_len; } // Recursively find the kth position return findKthChar(s, pos); } else { // Else update total_len // by next_total_len total_len = next_total_len; } } } // Return -1 if character not found return ' '; } // Driver code public static void main(String[] args) { String s = "ab2c3"; int k = 5; System.out.println(findKthChar(s, k)); }} // This code is contributed by Prerna Saini. |
Python3
# Python 3 implementation of the approach# Function to print kth character of# String s after decrypting itdef findKthChar(s, k): # Get the length of string len1 = len(s) # Initialise pointer to character # of input string to zero i = 0 # Total length of resultant string total_len = 0 # Traverse the string from starting # and check if each character is # alphabet then increment total_len while (i < len1): if (s[i].isalpha()): total_len += 1 # If total_len equal to k then # return string else increment i if (total_len == k): return s[i] i += 1 else: # Parse the number n = 0 while (i < len1 and s[i].isalpha() == False): n = n * 10 + (ord(s[i]) - ord('0')) i += 1 # Update next_total_len next_total_len = total_len * n # Get the position of kth character if (k <= next_total_len): pos = k % total_len # Position not found then update # position with total_len if (pos == 0): pos = total_len # Recursively find the kth position return findKthChar(s, pos) else: # Else update total_len # by next_total_len total_len = next_total_len # Return -1 if character not found return -1# Driver codeif __name__ == '__main__': s = "ab2c3" k = 5 print(findKthChar(s, k)) # This code is contributed by# Surendra_Gangwar |
C#
// C# implementation of the approach using System; class GFG { // Function to print kth character of // String s after decrypting it static char findKthChar(String s, int k) { // Get the length of string int len = s.Length; // Initialise pointer to character // of input string to zero int i = 0; // Total length of resultant string int total_len = 0; // Traverse the string from starting // and check if each character is // alphabet then increment total_len while (i < len) { if (char.IsLetter(s[i])) { total_len++; // If total_len equal to k then // return string else increment i if (total_len == k) return s[i]; i++; } else { // Parse the number int n = 0; while (i < len && !char.IsLetter(s[i])) { n = n * 10 + (s[i] - '0'); i++; } // Update next_total_len int next_total_len = total_len * n; // Get the position of kth character if (k <= next_total_len) { int pos = k % total_len; // Position not found then update // position with total_len if (pos == 0) { pos = total_len; } // Recursively find the kth position return findKthChar(s, pos); } else { // Else update total_len // by next_total_len total_len = next_total_len; } } } // Return -1 if character not found return ' '; } // Driver code public static void Main(String[] args) { String s = "ab2c3"; int k = 5; Console.WriteLine(findKthChar(s, k)); }}// This code is contributed by PrinciRaj1992 |
Javascript
// Javascript implementation of the approach// Function to print kth character of// String s after decrypting itfunction findKthChar(s, k){ // Get the length of string let len = s.length; // Initialise pointer to character // of input string to zero let i = 0; // Total length of resultant string let total_len = 0; // To check a character is an alphabet or not let isAlpha = function(ch){ return typeof ch === "string" && ch.length === 1 && (ch >= "a" && ch <= "z" || ch >= "A" && ch <= "Z"); } // Traverse the string from starting // and check if each character is // alphabet then increment total_len while (i < len) { if (isAlpha(s[i])) { total_len++; // If total_len equal to k then // return string else increment i if (total_len == k) return s[i]; i++; } else { // Parse the number let n = 0; while (i < len && !isAlpha(s[i])) { n = n * 10 + (s[i] - '0'); i++; } // Update next_total_len let next_total_len = total_len * n; // Get the position of kth character if (k <= next_total_len) { let pos = k % total_len; // Position not found then update // position with total_len if (!pos) { pos = total_len; } // Recursively find the kth position return findKthChar(s, pos); } else { // Else update total_len // by next_total_len total_len = next_total_len; } } } // Return -1 if character not found return -1;}// Driver codelet s = "ab2c3";let k = 5;console.log(findKthChar(s, k));// This code is contributed by Samim Hossain Mondal. |
Output:
c
Time Complexity: O(N)
Auxiliary Space: O(1)
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