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{ Bit Shifting }

The Question

A bit shift moves each digit in a set of bits left or right. The last bit in the direction of the shift is lost, and a bit is inserted on the other end.

0010 << // 0100
1011 >> // 0101

1010110 << 2 // 1011000
1011010 >> 3 // 0001011

Bit shifts can optionally take a number of times to shift

Note that each shift to the left multiplies a number by 2 to that power, and each shift to the right divides a number by two to that power (ex. a left shift of 3 multiplies the number by 8)

What you need to do:

Write a function with the following signature:

function (binaryString, direction, numShifts) {}

• binaryString will be a string representing a binary number (ex. "1010"). It will always contain a valid string with 0s and 1s
• direction will be a string - either "left" or "right"
• numShifts will be a number representing the number of times to shift. If left empty, the shift should be 1. It will either contain a valid whole number greater than 0 or be null

Approach

• Your first instinct is probably to start looping...

• You can simply slice and dice the string - run through several examples and you'll see the pattern
• Did you know that ES6 introduces a handy new repeat function on String.prototype...?

Possible Solution

function bitShift (binString, direction, numShifts) {
    numShifts = numShifts || 1;
    
    let l = binString.length,
        zeroes = "0".repeat(numShifts);
    if (direction === 'left') return binString.slice(numShifts, l).concat(zeroes);
    else return zeroes.concat(binString.slice(0, l - numShifts));
}

Conclusion  

REPL.IT solutions

• Looking for patterns in your expected output can lead to simple, elegant solutions 
• String.prototype.repeat is quite handy! Check out some more new String.prototype methods:

startsWith, endsWith, includes