#include <iostream>

int main() {
    std::string data;

    std::cout << "Enter string: ";
    std::cin >> data;

    char last = NULL;
    int count = 0;
    for(int i = 0; i <= data.length(); i++) {

        if(i !=0 && last != data[i]) {
            std::cout << last << count;
            count = 0;
        }

        last = data[i];
        count++;
    }
    return 0;
}

input = "wwwwaaadexxxxxx"
list = [0] * 26
for i in input:
    list[ord(i)-97]+=1
i=0
while i < len(input):
    print input[i], list[ord(input[i])-97]
    i += list[ord(input[i])-97]

silento:
30 MINUTES

c pointer is great and bad at same time


http://ideone.com/fzTQgH



<script src="http://ideone.com/e.js/fzTQgH" type="text/javascript" ></script>

pcguru1:


Am curious why did you go through the route of pointers, because i understand you're allocating the same size as the char, you could easily still use a char , yeah Pointers is cool but sometimes complex. Nice one guys

orimion:
https://ideone.com/DwUQSa

pcguru1:


My God what language is this lol

pcguru1:
Run Length Encoding
Given an input string, write a function that returns the Run Length Encoded string for the input string.

For example, if the input string is “wwwwaaadexxxxxx”, then the function should return “w4a3d1e1x6”.

You can use this to compile your code

https://ideone.com

Usage of this Algorithm

Compression of String if the need of character length is highly emphasized.

orimion:

Haskell
might be my inexperience that makes it look cryptic

CryptoCoinr:


Er, is the challenge language specific?

function charcurrence (text)
{
  let counter = 0;
  let char = text[0];
  let charcurredText = "";

  for (let i = 0; i < text.length; i++)
  {
    if (char === text[i])
    {
      counter++;
    }
    else
    {
      charcurredText += char + counter;
      char = text[i];
      counter = 1;
    }
  }

  return charcurredText += char + counter;
}

CryptoCoinr:


pcguru1, did you ignore this or you just didn't see it? Either way, here's mine (in JS as usual):

function charcurrence (text)
{
  let counter = 0;
  let char = text[0];
  let charcurredText = "";

  for (let i = 0; i < text.length; i++)
  {
    if (char === text[i])
    {
      counter++;
    }
    else
    {
      charcurredText += char + counter;
      char = text[i];
      counter = 1;
    }
  }

  return charcurredText += char + counter;
}

pcguru1:


4got to reply

CryptoCoinr:


Okie dokie.

I'm glad I googled what RLE was. I actually completed the challenge within ~15mins - 20mins of posting that question but started wondering what we were meant to do if an already tagged character, say "w", occurred after a whole bunch of other characters. Long story short: many hours spent needlessly reiterating.

segxyhanxy:
a solution of it in java.

public static String rle(String data) {
String compressedData = "";

int counter = 1;
char lastChar = data.charAt(0);
for (int i = 1; i < data.length(); i++) {
char c = data.charAt(i);
if (c != lastChar) {
compressedData += data.substring(i - 1, i) + counter;
counter = 1;
lastChar = c;
} else {
counter++;
lastChar = c;
}
if (i == data.length() - 1) {
compressedData += data.substring(i - 1, i) + counter ;
}
}
return compressedData;
}

deedat205:
Nice challenge. But who can come up with the RLE string decoding solution.
Eg W4E2H3 should be WWWWEEHHH and A12T3Y4 should be AAAAAAAAAAAATTTYYYY

deedat205:
Nice challenge. But who can come up with the RLE string decoding solution.
Eg W4E2H3 should be WWWWEEHHH and A12T3Y4 should be AAAAAAAAAAAATTTYYYY

import Data.Char
 
main = do
  putStrLn $ decode "W4E2H3"
  putStrLn $ decode "A12T3Y4"
 
decode :: String -> String
decode = concat . map expand . split
 
expand :: String -> String
expand "" = ""
expand (x:xs) = take (read xs) $ repeat x
 
split :: String -> [String]
split = words . tail . foldr fn ""
  where fn x acc = if (ord x < 48 || ord x > 57) then (' ':x:acc) else (x:acc)