Objective was relatively simple.<\/p>\n
Rule for handshake.<\/p>\n
\n1 = wink\n10 = double blink\n100 = close your eyes\n1000 = jump\n10000 = Reverse the order of the operations in the secret handshake.\n<\/code><\/pre>\n<\/blockquote>\n\n\nExample<\/h2>\n\n\n\n
Let’s say the input is 2 the output should be [“double blink”].<\/p>\n\n\n\n
Version 1.0<\/h2>\n\n\n\n
function binaryOptimized(number) { \n var binaryRep = []; \n do { \n binaryRep.push(parseInt(number%2)); \n number \/= 2; \n } while (parseInt(number) > 0); \n return binaryRep; \n}\n\nfunction getCommand(pseudo, arr) { switch(pseudo) { \ncase 1: arr.push (\"wink\"); break; \ncase 10: arr.push (\"double blink\"); break; \ncase 100: arr.push (\"close your eyes\"); break; \ncase 1000: arr.push (\"jump\"); break; \ncase 10000: arr = arr.reverse(); \n} \nreturn arr; \n} \n \nexport const commands = (input) => { \n var rep = binaryOptimized(input); \n var commands = []; \n rep.map((element, i) => { \n getCommand(element * Math.pow(10, i), commands) \n }); \n return commands; \n};<\/pre>\n\n\n\nThe code is self explanatory<\/p>\n\n\n\n
Defined a binary calculation function, computed the equivalent binary representation. Used the individual element to compute the corresponding command and responded back with the response.<\/p>\n\n\n\n
Can it be optimized?<\/h2>\n\n\n\n
The think I misinterpreted was that, binary calculation of number is an optional thing.<\/p>\n\n\n\n
BITWISE Operator<\/h3>\n\n\n\n
There are two basic bitwise operation ‘&’ and ‘|’. <\/p>\n\n\n\n
Bitwise operators treat their operands as a sequence of 32 bits (zeroes and ones), rather than as decimal, hexadecimal, or octal numbers.<\/p><\/blockquote>\n\n\n\n
Example<\/h4>\n\n\n\n
5 => 101\nWe can check the bits now\n2^0 = 1 => 1\n2^1 = 2 => 10\n2^2 = 4 => 100\n2^3 = 8 => 1000\n2^4 = 16 => 10000\n2^5 = 32 => 100000<\/pre>\n\n\n\nNow instead of converting it to binary I can simply & to compute the on bits.<\/p>\n\n\n\n
5&1 = 101 & 1 => 1\n5&2 = 101 & 10 => 0\n5&4 = 101 & 100 => 4 \n5&8 = 101 & 1000 => 0\n5&16 = 101 & 10000 => 0\n5&32 = 101 & 100000 => 0<\/pre>\n\n\n\nAlso you can define the left shift to compute the bit rep.<\/p>\n\n\n\n
1<<0 = 1\n1<<1 = 10 = 2\n1<<2 = 100 = 4\n1<<3 = 1000 = 8\n1<<4 = 10000 = 16\n1<<5 = 100000 = 32<\/pre>\n\n\n\nSo to compute the secret handshake it is just matter of computing the on bits and add it to the final array.<\/p>\n\n\n\n
I defined the Array of handshakes <\/p>\n\n\n\n
HANDSHAKES = [\"wink\", \"double blink\", \"close your eyes\", \"jump\"];<\/pre>\n\n\n\nTo compute the handshake sequence I just to validate the ON bits.<\/p>\n\n\n\n
Array.filter<\/h3>\n\n\n\n
let newArray = arr.filter(callback(element[, index[, array]])[, thisArg])<\/p><\/blockquote>\n\n\n\n