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Find k’th character of decrypted string | Set 1

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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"

Input: "ab4c12ed3", k = 21
Output: e
Decrypted string is "ababababccccccccccccededed"

Approach: 

  1. The idea is simple, Initially take empty decrypted string. 
  2. decompress the string by reading substring and it’s frequency one by one and append current substring in decrypted string by it’s frequency. 
  3. Repeat the process till the end of string  
  4. print the Kth character from decrypted string. 

Algorithm:

  • Step 1: Initialize a string with the given encoded string and the value of k.
  • Step 2: Find the kth character in the encoded string.
  • Step 3: Check if last character of the string is alphabet or numeric value, if it’s alphabet i.e. frequency is zero, append it to the string and return (k-1)th index.
  • Step 4: Print the output.

Implementation:

C++




// 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)
{
    // expand string variable is used to
    // store final string after decompressing string str
    string expand = "";
 
    string temp;  // Current substring
    int freq = 0; // Count of current substring
 
    for (int i=0; str[i]!='\0'; )
    {
        temp = ""; // Current substring
        freq = 0; // count frequency of current substring
 
        // read characters until you find a number
        // or end of string
        while (str[i]>='a' && str[i]<='z')
        {
            // push character in temp
            temp.push_back(str[i]);
            i++;
        }
 
        // read number for how many times string temp
        // will be repeated in decompressed string
        while (str[i]>='1' && str[i]<='9')
        {
            // generating frequency of temp
            freq = freq*10 + str[i] - '0';
            i++;
        }
 
        // now append string temp into expand
        // equal to its frequency
        for (int j=1; j<=freq; j++)
            expand.append(temp);
    }
 
    // this condition is to handle the case
    // when string str is ended with alphabets
    // not with numeric value
    if (freq==0)
        expand.append(temp);
 
    return expand[k-1];
}
 
// Driver program to test the string
int main()
{
    string str = "ab4c12ed3";
    int k = 21;
    cout << encodedChar(str, k) << endl;
    return 0;
}


Java




// Java program to find K'th character in
// decrypted string
public class GFG {
      
    // Function to find K'th character in
    // Encoded String
    static char encodedChar(String str,int k)
    {
        // expand string variable is used to
        // store final string after decompressing
        // string str
        String expand = "";
      
        String temp = ""// Current substring
        int freq = 0; // Count of current substring
      
        for (int i=0; i < str.length() ; )
        {
            temp = ""; // Current substring
            freq = 0; // count frequency of current
                      // substring
      
            // read characters until you find a number
            // or end of string
            while (i < str.length() && str.charAt(i)>='a'
                                && str.charAt(i)<='z')
            {
                // push character in temp
                temp += str.charAt(i);
                i++;
            }
      
            // read number for how many times string temp
            // will be repeated in decompressed string
            while (i < str.length() && str.charAt(i)>='1'
                                && str.charAt(i)<='9')
            {
                // generating frequency of temp
                freq = freq*10 + str.charAt(i) - '0';
                i++;
            }
      
            // now append string temp into expand
            // equal to its frequency
            for (int j=1; j<=freq; j++)
                 expand += temp;
        }
      
        // this condition is to handle the case
        // when string str is ended with alphabets
        // not with numeric value
        if (freq==0)
            expand += temp;
      
        return expand.charAt(k-1);
    }
      
    // Driver program to test the string
    public static void main(String args[])
    {
        String str = "ab4c12ed3";
        int k = 21;
        System.out.println(encodedChar(str, k));
    }
}
// This code is contributed by Sumit Ghosh


Python3




# Python 3 program to find K'th character
# in decrypted string
 
# Function to find K'th character
# in Encoded String
def encodedChar(str, k):
     
    # expand string variable is used
    # to store final string after
    # decompressing string str
    expand = ""
 
    # Current substring
    freq = 0 # Count of current substring
    i = 0
    while(i < len(str)):
        temp = "" # Current substring
        freq = 0 # count frequency of current substring
 
        # read characters until you find
        # a number or end of string
        while (i < len(str) and
               ord(str[i]) >= ord('a') and
               ord(str[i]) <= ord('z')):
                    
            # push character in temp
            temp += str[i]
            i += 1
 
        # read number for how many times string temp
        # will be repeated in decompressed string
        while (i < len(str) and
               ord(str[i]) >= ord('1') and
               ord(str[i]) <= ord('9')):
                    
            # generating frequency of temp
            freq = freq * 10 + ord(str[i]) - ord('0')
            i += 1
 
        # now append string temp into expand
        # equal to its frequency
        for j in range(1, freq + 1, 1):
            expand += temp
 
    # this condition is to handle the case
    # when string str is ended with alphabets
    # not with numeric value
    if (freq == 0):
        expand += temp
 
    return expand[k - 1]
 
# Driver Code
if __name__ == '__main__':
    str = "ab4c12ed3"
    k = 21
    print(encodedChar(str, k))
 
# This code is contributed by
# Shashank_Sharma


C#




// C# program to find K'th
// character in decrypted string
using System;
 
class GFG
{
     
    // Function to find K'th
    // character in Encoded String
    static char encodedChar(string str, int k)
    {
        // expand string variable is
        // used to store final string
        // after decompressing string str
        String expand = "";
     
        String temp = ""; // Current substring
        int freq = 0; // Count of current substring
     
        for (int i = 0; i < str.Length ; )
        {
            temp = ""; // Current substring
            freq = 0; // count frequency of current
                      // substring
     
            // read characters until you
            // find a number or end of string
            while (i < str.Length && str[i]>='a'
                                  && str[i]<='z')
            {
                // push character in temp
                temp += str[i];
                i++;
            }
     
            // read number for how many times
            // string temp will be repeated
            // in decompressed string
            while (i < str.Length && str[i] >= '1'
                                  && str[i] <= '9')
            {
                // generating frequency of temp
                freq = freq * 10 + str[i] - '0';
                i++;
            }
     
            // now append string temp into
            // expand equal to its frequency
            for (int j = 1; j <= freq; j++)
                expand += temp;
        }
     
        // this condition is to handle
        // the case when string str is
        // ended with alphabets not
        // with numeric value
        if (freq == 0)
            expand += temp;
     
        return expand[k - 1];
    }
     
    // Driver Code
    public static void Main()
    {
        string str = "ab4c12ed3";
        int k = 21;
        Console.Write(encodedChar(str, k));
    }
}
 
// This code is contributed
// by ChitraNayal


Javascript




<script>
// Javascript program to find K'th character in
// decrypted string
     
    // Function to find K'th character in
    // Encoded String
    function encodedChar(str, k)
    {
     
        // expand string variable is used to
        // store final string after decompressing
        // string str
        let expand = "";
        let temp = ""// Current substring
        let freq = 0; // Count of current substring
        for (let i=0; i < str.length ; )
        {
            temp = ""; // Current substring
            freq = 0; // count frequency of current
                      // substring
        
            // read characters until you find a number
            // or end of string
            while (i < str.length && str[i].charCodeAt(0)>='a'.charCodeAt(0)
                                && str[i].charCodeAt(0)<='z'.charCodeAt(0))
            {
                // push character in temp
                temp += str[i];
                i++;
            }
        
            // read number for how many times string temp
            // will be repeated in decompressed string
            while (i < str.length && str[i].charCodeAt(0)>='1'.charCodeAt(0)
                                && str[i].charCodeAt(0)<='9'.charCodeAt(0))
            {
                // generating frequency of temp
                freq = freq*10 + str[i].charCodeAt(0) - '0'.charCodeAt(0);
                i++;
            }
        
            // now append string temp into expand
            // equal to its frequency
            for (let j=1; j<=freq; j++)
                 expand += temp;
        }
        
        // this condition is to handle the case
        // when string str is ended with alphabets
        // not with numeric value
        if (freq==0)
            expand += temp;
        
        return expand[k-1];
    }
     
    // Driver program to test the string
    let str = "ab4c12ed3";
    let k = 21;
    document.write(encodedChar(str, k));
     
    // This code is contributed by avanitrachhadiya2155
</script>


Output

e

Time Complexity: O(N) where N is length of string.
Auxiliary Space: O(N)
 

Exercise : The above solution builds the decoded string to find k’th character. Extend the solution to work in O(1) extra space.
Find k-th character of decrypted string | Set – 2



Last Updated : 14 Dec, 2022
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