# Codingame Solution: Chuck Norris

Original problem

## The Goal

Binary with 0 and 1 is good, but binary with only 0, or almost, is even better! Originally, this is a concept designed by Chuck Norris to send so called unary messages.

Write a program that takes an incoming message as input and displays as output the message encoded using Chuck Norris’ method.

## Rules

Here is the encoding principle:

• The input message consists of ASCII characters (7-bit)
• The encoded output message consists of blocks of 0
• A block is separated from another block by a space
• Two consecutive blocks are used to produce a series of same value bits (only1 or0 values):
- First block: it is always 0 or 00. If it is 0, then the series contains 1, if not, it contains 0
- Second block: the number of 0 in this block is the number of bits in the series

## Example

Let’s take a simple example with a message which consists of only one character: Capital C. C in binary is represented as 1000011, so with Chuck Norris’ technique this gives:

• 0 0 (the first series consists of only a single 1)
• 00 0000 (the second series consists of four 0)
• 0 00 (the third consists of two 1)

So C is coded as: 0 0 00 0000 0 00

Second example, we want to encode the message CC (i.e. the 14 bits 10000111000011) :

• 0 0 (one single 1)
• 00 0000 (four 0)
• 0 000 (three 1)
• 00 0000 (four 0)
• 0 00 (two 1)

So CC is coded as: 0 0 00 0000 0 000 00 0000 0 00

## Game Input

Input
Line 1: the message consisting of N ASCII characters (without carriage return)
Output
The encoded message
Constraints
0 < N < 100

## JavaScript Solution

A solution can be found quite easily by walking over the string and for each character decide if a new block must be created. An implementation can then look as follows:

```var MESSAGE = readline();

var res = "";
var pre = -1;

for (var i = 0; i < MESSAGE.length; i++) {

for (var j = 6; j >= 0; j--) {

var bit = MESSAGE.charCodeAt(i) >> j & 1;
if (bit !== pre) {

if (-1 !== pre) {
res+= " ";
}
res+= 1 == bit ? "0 " : "00 ";
pre = bit;
}
res+= "0";
}
}
print(res);```

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