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JAVASCRIPT: EvolveU C5 coding challenge - week #1 - answers
// This answer comes in 2 flavours: the "plays nice with others" version and the "hax0rz rUlez!!" version.
//
//--------------------------------------------------------------------------------------
// PLAYS NICE WITH OTHERS:
//
const arr2num = (arr) => {
return arr
.reverse() // work from least-to-most significant digits
.reduce(
(accumulatedVal, curVal, index) =>
// multiply by 10 to the power of the significant position (index)
(accumulatedVal + Math.abs(curVal) * 10 ** index) *
(curVal < 0 ? -1 : 1),
null, // return null if given arr is empty
);
};
const num2arr = (num) => {
return num === null
? [] // null means arr1 & arr2 were empty arrays
: Math.abs(num)
.toString()
.split('') // array with one element per digit of num
.map((i, idx) =>
// if the first digit and it is negative, make result negative
idx === 0 && num < 0 ? new Number(-i) : new Number(i),
);
};
function addArrays(arr1, arr2) {
let a1 = arr2num(arr1);
let a2 = arr2num(arr2);
let sum;
if (a1 && a2) {
sum = a1 + a2;
} else if (a1) {
sum = a1;
} else if (a2) {
sum = a2;
} else {
sum = null;
}
return num2arr(sum);
}
//
// -------------------------------------------------------------------------------------
//--------------------------------------------------------------------------------------
// Hackers Rulez (...NOT!):
//
const sumOrNull = (v1, v2) => ((v1 || v1 === 0) && (v2 || v2 === 0)) ? v1 + v2 : (v1 || v1 === 0) ? v1 : (v2 || v2 === 0) ? v2 : null;
const reNegate = (idx, num) => (idx === 0 && num < 0 ? -1 : 1);
const arr2num = (arr) => arr.reverse().reduce((acc, curVal, idx) => (acc + Math.abs(curVal) * 10 ** idx) * (curVal < 0 ? -1 : 1), null)
const num2arr = (num) => num === null ? [] : Math.abs(num).toString().split('').map((i, idx) => i * reNegate(idx, num))
function addArrays(arr1, arr2) {
return num2arr(sumOrNull(arr2num(arr1), arr2num(arr2)))
}
//
// -------------------------------------------------------------------------------------
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