Decode a given string by removing duplicate occurrences

Given encoded string str such that 'a' is written 1 time, 'b' is written 2 times, and so on, until 'z' is written 26 times, the task is to decode the given string str. 

Note: The letter may contain spaces and punctuation marks, so don't ignore those.

Examples:

Input: str = "bbadddd"
Output: bad
Explanation:
As each letter is written corresponding to the number of appearances in the English Alphabets. Therefore, the resultant string is "bad".

Input: str = "abbbbacccdddd"
Output: abbacd

Approach: The given problem can be solved by iterating the given string str, and for each character and push that character in an output string and move that respective position ahead to check for the other character. Follow the steps below to solve the problem:

• Initialize a variable, say output as an empty string that stores the resultant string.
• Define a function findRepetition(char c) and perform the following steps:
  • If the value of c is in the range from a to z, then return the value of c-'a'.
  • Otherwise, if the value of c is in the range from A to Z, then return the value of c-'Z'.
  • Otherwise, return 0.
• Iterate over a range [0, N] using the variable i and perform the following steps:
  • Push the i^th character of the string str into the string output.
  • Call the function findRepetition(str[i]) to count the number of steps i^th index has to move ahead.
• After performing the above steps, print the string output as the resultant string.

Below is the implementation of the above approach.

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to count the appearances
// of each character
int findRepitition(char a)
{
    // If the character is lower case
    if (a <= 'z' && a >= 'a') {
        return a - 'a';
    }

    // If the character is uppercase
    else if (a <= 'Z' && a >= 'A') {
        return a - 'A';
    }

    // If the character is a
    // punctuation mark
    return 0;
}

// Function to decode the given encoded
// string
void decodeString(string str)
{
    string output = "";

    // Iterate the given string str
    for (int i = 0; i < str.length(); i++) {

        output.push_back(str[i]);

        // Find the index of the next
        // character to be printed
        i += findRepitition(str[i]);
    }

    cout << "Decrypted code is {"
         << output << "}" << endl;
}

// Driver Code
int main()
{
    string str = "abbbb acccdddd";
    decodeString(str);

    return 0;
}

// Java program for the above approach
import java.util.*;

class GFG{

// Function to count the appearances
// of each character
static int findRepitition(char a) 
{
    
    // If the character is lower case
    if (a <= 'z' && a >= 'a')
    {
        return a - 'a';
    }

    // If the character is uppercase
    else if (a <= 'Z' && a >= 'A') 
    {
        return a - 'A';
    }

    // If the character is a
    // punctuation mark
    return 0;
}

// Function to decode the given encoded
// String
static void decodeString(String str)
{
    String output = "";

    // Iterate the given String str
    for(int i = 0; i < str.length(); i++) 
    {
        output += (str.charAt(i));

        // Find the index of the next
        // character to be printed
        i += findRepitition(str.charAt(i));
    }
    System.out.print("Decrypted code is {" + 
                     output + "}" + "\n");
}

// Driver Code
public static void main(String[] args) 
{
    String str = "abbbb acccdddd";
    
    decodeString(str);
}
}

// This code is contributed by umadevi9616 

# Python3 program for the above approach

# Function to count the appearances
# of each character
def findRepetition(a):
  
  # If the character is lower case
    if a <= 'z' and a >= 'a':
        return ord(a)-97
      
    # If the character is uppercase
    elif a <= 'Z' and a >= 'A':
        return ord(a)-65
    else:
      
     # If the character is a
    # punctuation mark
        return 0

# Function to decode the given encoded
# string
def decodeString(str_):
    output = ""
    i = 0
    
    # Iterate the given string str
    while(i < len(str_)):
        output += str_[i]
                
        # Find the index of the next
        # character to be printed
        i += findRepetition(str_[i]) + 1
    print("Decrypted code is {" + output + "}")

# Driver code
str_ = "abbbb acccdddd"
decodeString(str_)

# This code is contributed by Parth Manchanda

// C# program for the above approach
using System;
using System.Collections.Generic;

class GFG{

// Function to count the appearances
// of each character
static int findRepitition(char a)
{
    // If the character is lower case
    if (a <= 'z' && a >= 'a') {
        return (int)a - 97;
    }

    // If the character is uppercase
    else if (a <= 'Z' && a >= 'A') {
        return (int)a - 65;
    }

    // If the character is a
    // punctuation mark
    return 0;
}

// Function to decode the given encoded
// string
static void decodeString(string str)
{
    string output = "";

    // Iterate the given string str
    for (int i = 0; i < str.Length; i++) {

        output += str[i];

        // Find the index of the next
        // character to be printed
        i += findRepitition(str[i]);
    }

    Console.Write("Decrypted code is {" + output + "}");
}

// Driver Code
public static void Main()
{
    string str = "abbbb acccdddd";
    decodeString(str);
}
}

// This code is contributed by ipg2016107.

<script>

        // JavaScript program for the above approach

        // Function to count the appearances
        // of each character
        function findRepitition(a) 
        {
        
            // If the character is lower case
            if (a.charCodeAt(0) <= 'z'.charCodeAt(0) && a.charCodeAt(0) >= 'a'.charCodeAt(0)) {
                return a.charCodeAt(0) - 'a'.charCodeAt(0);
            }

            // If the character is uppercase
            else if (a <= 'Z' && a >= 'A') {
                return a.charCodeAt(0) - 'A'.charCodeAt(0);
            }

            // If the character is a
            // punctuation mark
            return 0;
        }

        // Function to decode the given encoded
        // string
        function decodeString(str) {
            let output = "";

            // Iterate the given string str
            for (let i = 0; i < str.length; i++) {

                output += (str[i]);

                // Find the index of the next
                // character to be printed
                i += findRepitition(str[i]);
            }

            document.write("Decrypted code is {"
                + output + "}" + "<br>");
        }

        // Driver Code
        let str = "abbbb acccdddd";
        decodeString(str);

// This code is contributed by Potta Lokesh
    </script>

Decrypted code is {abb acd}

Time Complexity: O(N)
Auxiliary Space: O(N)