Given a string str of length of N that is in the encoded form with alphabets and * . The task is to find the string from which it was generated. The required string can be generated from the encoded string by replacing all the * with the prefix values of the encoded string.
Examples:
Input: str = ab*c*d
Output: "ababcababcd"
Explanation: For the first occurrence of "*", "ab" is the prefix. So '*' is replaced by "ab" resulting the string to be "ababc*d" and for the next '*' the prefix is "ababc". So the string will now change from "ababc*d" to "ababcababcd".Input : str = "z*z*z"
Output: zzzzzzz
Approach: The solution is based on greedy approach. Follow the steps mentioned below to solve the problem:
- Consider an empty string result.
- Iterate over the given coded string.
- if the current character in the string is not "*" then add the current character to the result.
- Otherwise, if the current character is "*", add the result string formed till now with itself.
- Return the result.
Below is the implementation of the given approach.
// C++ code to implement above approach
#include <bits/stdc++.h>
using namespace std;
// Function to return a string
// found from the coded string
string findstring(string str)
{
// Declaring string to store result
string result = "";
// Loop to generate the original string
for (int i = 0; str[i] != '\0'; i++) {
// If current character in string
// is '*' add result to itself
if (str[i] == '*')
result += result;
// Else add current element only
else
result += str[i];
}
// Return the result
return result;
}
// Driver code
int main()
{
string str = "ab*c*d";
cout << findstring(str);
return 0;
}
// Java code to implement above approach
class GFG {
// Function to return a string
// found from the coded string
static String findstring(String str)
{
// Declaring string to store result
String result = "";
// Loop to generate the original string
for (int i = 0; i < str.length(); i++) {
// If current character in string
// is '*' add result to itself
if (str.charAt(i) == '*')
result += result;
// Else add current element only
else
result += str.charAt(i);
}
// Return the result
return result;
}
// Driver code
public static void main(String[] args)
{
String str = "ab*c*d";
System.out.println(findstring(str));
}
}
// This code is contributed by ukasp.
# Python code for the above approach
# Function to return a string
# found from the coded string
def findstring(str):
# Declaring string to store result
result = ""
# Loop to generate the original string
for i in range(len(str)):
# If current character in string
# is '*' add result to itself
if (str[i] == '*'):
result += result
# Else add current element only
else:
result += str[i]
# Return the result
return result
# Driver code
str = "ab*c*d"
print(findstring(str))
# This code is contributed by Saurabh Jaiswal
// C# code to implement above approach
using System;
class GFG
{
// Function to return a string
// found from the coded string
static string findstring(string str)
{
// Declaring string to store result
string result = "";
// Loop to generate the original string
for (int i = 0; i < str.Length; i++)
{
// If current character in string
// is '*' add result to itself
if (str[i] == '*')
result += result;
// Else add current element only
else
result += str[i];
}
// Return the result
return result;
}
// Driver code
public static void Main()
{
string str = "ab*c*d";
Console.Write(findstring(str));
}
}
// This code is contributed by Samim Hossain Mondal.
<script>
// JavaScript code for the above approach
// Function to return a string
// found from the coded string
function findstring(str) {
// Declaring string to store result
let result = "";
// Loop to generate the original string
for (let i = 0; i < str.length; i++) {
// If current character in string
// is '*' add result to itself
if (str[i] == '*')
result += result;
// Else add current element only
else
result += str[i];
}
// Return the result
return result;
}
// Driver code
let str = "ab*c*d";
document.write(findstring(str));
// This code is contributed by Potta Lokesh
</script>
Output
ababcababcd
Time Complexity: O(N)
Auxiliary Space: O(N)