Given an encrypted string str and the encryption algorithm, the task is to decrypt the string. The encryption algorithm is as follows:
The 1st character of the string will be repeated once in the encrypted string, the 2nd character will be repeated twice, …, nth character will be repeated n times. For example, the string “abcd” will be encrypted as “abbcccdddd”.
Examples:
Input: str = “geeeeekkkksssss”
Output: neveropenInput: str = “abbcccdddd”
Output: abcd
Approach: Initialize i = 0 and print str[i].
Update i = i + 1 and print str[i], then update i = i + 2 and print str[i] and so on while i < length(str).
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the decrypted stringstring decryptString(string str, int n){ // Initial jump will be 1 int i = 0, jump = 1; string decryptedStr = ""; while (i < n) { decryptedStr += str[i]; i += jump; // Increment jump by 1 with every character jump++; } return decryptedStr;}// Driver codeint main(){ string str = "geeeeekkkksssss"; int n = str.length(); cout << decryptString(str, n); return 0;} |
Java
// Java implementation of the approachclass GFG{// Function to return the decrypted stringstatic String decryptString(String str, int n){ // Initial jump will be 1 int i = 0, jump = 1; String decryptedStr = ""; while (i < n) { decryptedStr += str.charAt(i); i += jump; // Increment jump by 1 with every character jump++; } return decryptedStr;}// Driver codepublic static void main(String[] args){ String str = "geeeeekkkksssss"; int n = str.length(); System.out.println(decryptString(str, n));}}// This code is contributed by Code_Mech |
Python3
# Python 3 implementation of the approach# Function to return the decrypted stringdef decryptString(str, n): # Initial jump will be 1 i = 0 jump = 1 decryptedStr = "" while (i < n): decryptedStr += str[i]; i += jump # Increment jump by 1 with # every character jump += 1 return decryptedStr# Driver codeif __name__ == '__main__': str = "geeeeekkkksssss" n = len(str) print(decryptString(str, n))# This code is contributed by# Surendra_Gangwar |
C#
// C# implementation of the approachusing System; class GFG{// Function to return the decrypted stringstatic string decryptString(string str, int n){ // Initial jump will be 1 int i = 0, jump = 1; string decryptedStr = ""; while (i < n) { decryptedStr += str[i]; i += jump; // Increment jump by 1 with every character jump++; } return decryptedStr;}// Driver codepublic static void Main(){ string str = "geeeeekkkksssss"; int n = str.Length; Console.Write(decryptString(str, n));}}// This code is contributed by ita_c |
PHP
<?php// PHP implementation of the approach // Function to return the decrypted string function decryptString($str, $n) { // Initial jump will be 1 $i = 0 ; $jump = 1 ; $decryptedStr = ""; while ($i < $n) { $decryptedStr .= $str[$i]; $i += $jump; // Increment jump by 1 with // every character $jump++; } return $decryptedStr; } // Driver code $str = "geeeeekkkksssss"; $n = strlen($str); echo decryptString($str, $n); // This code is contributed by Ryuga?> |
Javascript
<script>// Javascript implementation of the approach// Function to return the decrypted stringfunction decryptString(str,n){ // Initial jump will be 1 let i = 0, jump = 1; let decryptedStr = ""; while (i < n) { decryptedStr += str[i]; i += jump; // Increment jump by 1 with every character jump++; } return decryptedStr;}// Driver codelet str = "geeeeekkkksssss";let n = str.length;document.write(decryptString(str, n));// This code is contributed by rag2127</script> |
Output
neveropen
Time Complexity: O(?n), where n is the length of the given string.
Auxiliary Space: O(?n), where n is the length of the given string.
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