## DEV Community

Akhil

Posted on • Updated on

# String Compression. Facebook interview question.

Question: Design a system that will compress the string and then decode it. The input will consist only of alphabet characters.

Eg: if the string is: aaabbcccccd
1> it's compressed form ie encoded form will be a3b3c5d1.
2> then we have to take the compressed string and decode it.

###### note: this is a string manipulation question and is meant to test your string manipulation skills, don't confuse it with compression algorithms since there are much better compression algorithms.

Imagine a system similar to this :

String "Richard is great but you know" is compressed to RIGBY which both the client and server know what that string represents.

One can use it for their projects:

## Enconding the string / compression

The algorithm is pretty straight forward, keep a pointer and move ahead. Count the number of times a character repeats and append the character and it's count to a result string.

``````var encode = function(string){
let chars = string.split("");
let res = "";
for(let i=0;i<chars.length;i++){
let count = 1;              //default count = 1 since at least 1 character
let char = chars[i];        //the character at that pointer
while(i<chars.length-1 && chars[i] == chars[i+1]){
count++;                  // increment the count
i++;                      // increment the pointer
}
res += char + count;        // append to resultant string.
}
return res;
}
``````

## Decoding the string

Decoding the string is also similar, we keep a pointer, get the character, get its count, generate character and append it to the resultant string.

``````var decode = function(string){
let res = "";
for(let i=0;i<string.length;){
let char = string[i];
let count = 0;
i++;
while(i<string.length && parseInt(string[i])>=0 && parseInt(string[i])<=9){
count = parseInt(count * 10) + parseInt(string[i]);
i++;
}
while(count>0){
res+= char;
count--;
}
}
return res;
}
``````

Putting them together:

``````var encode = function(string){
let chars = string.split("");
let res = "";
for(let i=0;i<chars.length;i++){
let count = 1;
let char = chars[i];
while(i<chars.length-1 && chars[i] == chars[i+1]){
count++;
i++;
}
res += char + count;
}
return res;
}

var decode = function(string){
let res = "";
for(let i=0;i<string.length;){
let char = string[i];
let count = 0;
i++;
while(i<string.length && parseInt(string[i])>=1 && parseInt(string[i])<=9){
count = parseInt(count * 10) + parseInt(string[i]);
i++;
}
while(count>0){
res+= char;
count--;
}
}
return res;
}

let compress = encode("aaabbbccccccccccccd");
console.log(compress);
let res = decode(compress);
console.log(res);

``````

That's it, now you know how to solve a Facebook interview question and a bit of introduction towards compression.

Special thanks to the community for pointing out bugs and edge cases !! ðŸ¥°

Michiel Hendriks

For plain text a Lempel-Ziv (LZ) approach is better than a Run Length Encoding (RLE).

In normal text you rarely have concecutive characters, but really often repeating segments. This is where LZ shines.

RLE is easier and cheaper to implement, as you don't need to keep a dictionary around. So less memory needed.

Akhil

True that ! this question is meant for testing string manipulation skills and not to test compression algorithm skills.

I shall look into LZ approach though ! thanks for the info!

Stefanos Kouroupis

Even if this is not a true compression, the word compression though implies less characters (or the same) so you could make an assumption that for characters repeated less than 3 times there is no need to compress them. so

• asdf will stay asdf
• aasdff will stay aasdff
• aasssdfff will become aas3df3

and when compressing numbers you could use a token approach

Akhil

Yep, we could compare if the "compressed" version is indeed shorter and if it's shorter then use that one.

Akhil

Yep, the function can only take string with characters as a input for encoding, with integers it becomes quite complicated but doable by using some special characters like "\$" or "#" and it must be guaranteed that the said special character won't be used.

Your method is actually a much smarter way of achieving it! Thanks for the tip :)

devguyky

Good post although your decoder assumes that there will never be more than 9 sequential characters. It would throw an error with an input string of a3b2c11d1

Akhil

Thanks for pointing out the bug!

fejese

Akhil

Nice catch !! just fixed it. Thanks a lot for pointing that out !

Tomaz Lovrec • Edited

Using this, encoding a string like 'asdf' will produce a string that is twice the size, since 'asdf' will become, 'a1s1d1f1'.

Akhil

Yep, it's efficient when a string contains multiple repeated characters but still, it won't guarantee the shortest string. Eg: is the string is Mississippi, compressed will be: m1i1s2i1s2i1p2i1 which is actually longer, even though some of the characters are repeating.

This question serves as a general idea towards why we go for compression, I will write about complex topics like Huffman coding in the future.

tareq12345 • Edited

What if i changed the while loop to
let char = chars[i]
If(i < chars.length - 1 && chars[i] == chars[i+1]){
Count++
}
else{
count = 1
res+=char + count
}