# Problem Statement#

We have to implement an encode function that takes a list of string values as input and returns a single encoded string that could be transmitted over a network. On the other side of the network, the decode function will take the encoded string as input and return the original list of string values as output.

A delimiter could be used to differentiate between words. For example, ["Hello", "World"] could be encoded with comma delimiter (,) to "Hello,World".

# Brute Force Solution#

In the brute-force approach to solving this problem, we can execute a loop over the input list appending each string with some special character as a delimiter.

If the special character is a part of the string itself (for example "Hello,") then we can add an escape character before it, such that it is easier to differentiate from the delimiter during decoding.

## Psuedo-code for the Brute Force Solution#

func encode(array)
delimiter = ","
encodedString = ""

loop index on array
string = array[index]
new_string = ""

loop index2 on string

if string[index2] is delimiter or "\"
new_string.append("\")

new_string.append(string[index2])

if encodedString is not empty
encodedString.append(delimiter)

encodedString.append(new_string)

return encodedString

func decode(encodedString)
delimiter=","
decodedList = []
string = ""

loop index on encodedString

if encodedString[index] is "\"
string.append(encodedString[index+1])
index++

else if encodedString[index] is delimiter
decodedList.append(string)
string = ""

else
string.append(encodedString[index])

decodedList.append(string)

return decodedList


## Time Complexity Analysis#

### Best Case Scenario#

The best-case input for the brute force solution would be a string full of delimiters since the decoding process will finish earlier (the index is incremented by 2 while encountering the escape character \).

The time complexity of encoding and decoding in the best-case scenario would be $O(k \times n)$ where $k$ is the average size of the string and $n$ is the size of the input array.

### Worst Case Scenario#

For the worst-case time complexity of the brute-force solution, the input should not contain the delimiter or escape characters. The time complexity of encoding and decoding would still be $O(k \times n)$.

## Space Complexity Analysis#

The additional space required to store the encodedString and decodedList will be $O(kn)$.

## Code for the Brute Force Solution#

package main

import "fmt"

func encode(inputArray []string)(string){
delimiter := ","
encodedString := ""

for index:=0;index<len(inputArray);index++{
str := string(inputArray[index])
updatedStr := ""

for index2:=0;index2<len(str);index2++{
char := string(str[index2])

// If the delimiter or "\" is within the string
// add an extra escape character (\))
if char==delimiter || char=="\\"{
updatedStr += "\\"
}

updatedStr += char
}

if len(encodedString)!=0{
encodedString += delimiter
}
encodedString += updatedStr
}

return encodedString
}

func decode(encodedString string)([]string){
delimiter := ","
decodedList := []string{}
str := ""

for index:=0;index<len(encodedString);index++{
char := string(encodedString[index])
if char == "\\"{

// If we encounter an escape character in the encoded string
// Add the next character to the string
// and increment the index by 1
str += string(encodedString[index+1])
index += 1

} else if char==delimiter{

// If we encounter the delimiter
// Add the current string to decodedList
// and reset its values
decodedList = append(decodedList, str)
str = ""

} else {

// Add the string to decodedList
str += char

}
}

decodedList = append(decodedList, str)

return decodedList
}

func main(){
inputArray := []string{"Hello", "World"}
encodedString := encode(inputArray)
fmt.Println("Input Array", inputArray)
fmt.Println("Encoded String:", encodedString)
fmt.Println("Decoded String:", decode(encodedString))

inputArray = []string{"Hel,lo", "World"}
encodedString = encode(inputArray)
fmt.Println("Input Array", inputArray)
fmt.Println("Encoded String:", encodedString)
fmt.Println("Decoded String:", decode(encodedString))

inputArray = []string{"Hel\\,\\lo", "World"}
encodedString = encode(inputArray)
fmt.Println("Input Array", inputArray)
fmt.Println("Encoded String:", encodedString)
fmt.Println("Decoded String:", decode(encodedString))
}

// Output
// Input Array [Hello World]
// Encoded String: Hello,World
// Decoded String: [Hello World]
// Input Array [Hel,lo World]
// Encoded String: Hel\,lo,World
// Decoded String: [Hel,lo World]
// Input Array [Hel\,\lo World]
// Encoded String: Hel\\\,\\lo,World
// Decoded String: [Hel\,\lo World]


# Optimized Solution#

We can improve the time complexity of the brute-force solution if we can somehow omit the loop over each string.

If the length of the string could be used as a delimiter then we can avoid iteration over individual strings because the length function len(string) has $O(1)$ time complexity. But, this presents two more challenges

• if the length of a string is greater than 9 (double digits)
• if the string contains numbers, they could be confused with the length of the strings

To improve on this we can use a special character with the length of the string as delimiters, for example, ["Hello", "World"] could be encoded to 5;Hello5;World.

## Psuedo code for the Optimized Solution#

func encode(array)
delimiter = ","
encodedString = ""
loop each element in array
encodedString.append(len(element), delimiter, element)

return encodedString

func decode(encodedString)
delimiter = ","
decodedList = []

index = 0
while index<len(encodedString)
index2 = index
while encodedString[index2] is not delimiter
index2++

stringLen = toInt(encodedString[index:index2])

string_start = index2+1
string_end = index2+stringLen+1
string = encodedString[string_start:string_end]
decodedList.append(string)

index += string_end

return decodedList


## Time Complexity Analysis#

### Best Case Scenario#

The best-case input for the optimized solution will contain strings with length < 9.

The time complexity of the encoding loop will be $O(n)$ because it will iterate over all the elements in the array.

For the decoding loop, the time complexity appears to be $O(kn)$ but we are incrementing the index by the length of string value ($k$) on every iteration. Thus, the time complexity of decoding is also $O({kn \over k}) = O(n)$.

### Worst Case Scenario#

The worst-case time complexity of encoding is the same as the best-case i.e. $O(n)$.

The time complexity of the decoding function will depend on the number of digits (in the length of the largest string). For example: If the length of the largest individual string is 199990 then the time complexity of decoding the string will be $O(6 \times n)$ (because 199990 has 6 digits).

## Space Complexity Analysis#

The space complexity of the optimized solution will be the same as the brute force solution ($O(kn)$) as no additional data structures are used.

## Code for the Optimized Solution#

package main

import (
"fmt"
"strconv"
)

func encode(inputArray []string)(string){
encodedString := ""
for index:=0;index<len(inputArray);index++{

// For each string append
// the string length
encodedString += strconv.Itoa(len(inputArray[index]))

// "#" delimiter
encodedString += "#"

// the string itself
encodedString += inputArray[index]
}
return encodedString
}

func decode(encodedString string)([]string){
decodedList := []string{}
index := 0
for ;index<len(encodedString);{
index2:=index
for ;string(encodedString[index2])!="#";{
// Exit loop upon encountering the delimiter
index2+=1
}

// The length of individual string would be parsed
lenString, _ := strconv.Atoi(encodedString[index:index2])

// Extract the string from the encodedString
str := string(encodedString[index2+1:index2+lenString+1])

// Append the string to decodedList
decodedList = append(decodedList, str)

// Increment index, moving it to the next string length
index += index2+lenString+1
}

return decodedList
}

func main(){
inputArray := []string{"Hello", "World"}
encodedString := encode(inputArray)
fmt.Println("Input Array", inputArray)
fmt.Println("Encoded String:", encodedString)
fmt.Println("Decoded String:", decode(encodedString))

inputArray = []string{"Hel,lo", "World"}
encodedString = encode(inputArray)
fmt.Println("Input Array", inputArray)
fmt.Println("Encoded String:", encodedString)
fmt.Println("Decoded String:", decode(encodedString))

inputArray = []string{"Hel\\,\\lo", "World"}
encodedString = encode(inputArray)
fmt.Println("Input Array", inputArray)
fmt.Println("Encoded String:", encodedString)
fmt.Println("Decoded String:", decode(encodedString))
}

// Output
// Input Array [Hello World]
// Encoded String: 5#Hello5#World
// Decoded String: [Hello World]
// Input Array [Hel,lo World]
// Encoded String: 6#Hel,lo5#World
// Decoded String: [Hel,lo World]
// Input Array [Hel\,\lo World]
// Encoded String: 8#Hel\,\lo5#World
// Decoded String: [Hel\,\lo World]


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