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:
- The idea is simple, Initially take empty decrypted string.
- decompress the string by reading substring and it’s frequency one by one and append current substring in decrypted string by it’s frequency.
- Repeat the process till the end of string
- 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++ 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 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 |
# 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# 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 |
<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> |
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