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It would only accept my version when I did:
when there is no Xth number, return 0 (instead of undefined).
Minimal version using .sort()
although I had to provide a comparison function to it for it to work properly.
const compare = (a, b) => {
if (a < b) { return -1; }
else if (a > b) { return 1; }
else { return 0; }
}
const getX = (x, nums) => {
const result = Array.from(nums).sort(compare)[x - 1];
return result ? result : 0;
}
Here’s another version where I made an iterator (using a generator function) and did a simple sorting algorithm in it (not in-place).
The algorithm I used is: find min, insert it in new array, find min of remaining stuff, insert it, etc. 0(n2 )
my code (long)
I split the tasks into 3 functions, instead of writing more loops inside loops.
function min(nums, bools, changeMinToFalse) {
// finds minimum only for positions where bools has true
let minimum = undefined;
let indexOfMin = undefined;
const length = nums.length;
for (let i = 0; i < length; i++) {
if ((bools == undefined) || bools[i]) {
if ((minimum == undefined) || (nums[i] < minimum)) {
minimum = nums[i];
indexOfMin = i;
}
}
}
if (changeMinToFalse && !(indexOfMin == undefined)) {
bools[indexOfMin] = false;
}
return minimum;
}
function* inOrder(arr) {
// generator function to iterate from least to greatest
const length = arr.length;
const notUsedYet = new Array(length);
notUsedYet.fill(true);
for (let i = 0; i < length; i++) {
yield min(arr, notUsedYet, true);
}
}
function getX(x, nums) {
const length = nums.length;
if (x < 0) { x = length + x; }
const iterator = inOrder(nums);
for (let i = 1; i <= length; i++) {
if (i == x) {
return iterator.next().value;
}
else {
iterator.next();
}
}
return 0;
/* had to return 0 instead of undefined
because of answer checker */
}
Thank you! I returned “Error” instead of 0 for broken inputs and failed testcases and I couldn’t find out why. The instructions missing some points. As someone who learned C, returning 0 when the function is broken is very painful.
Your solution with BubbleSort is radically faster than other
Test: x=500, array with 1000 random integer from -50 to 50 (warming up, 1000 calls of 20 passes)
0.022 Array.sort
1.536 janbazant1107978602’s solution with generator
My idea was not to sort all list because we need only x minimums from the nums array to get the correct number from the list, so we will sort all list only in the worst case when x=nums.length. In the worst case, we got time complexity O(N^2) same as bubble sort, but in other cases, we got an advantage because we don’t need to sort all list.
The outer loop will run x times to find x minimums, each time you find a minimum you push it to the sorted array, record the index of this minimum and remove this minimum from the nums length decrease n-1, and all repeats, we search next minimum until sorted array length become x. I hope my logic didn’t go wrong and there is still some advantage.
function getX(x, nums) {
// Write your code here
const sortedArray = [];
let min = nums[0];
let index = 0;
if (x > 0 && x <= nums.length) {
while (sortedArray.length < x) {
for (let i = 0; i < nums.length; i++) {
if (min > nums[i + 1]) {
min = nums[i + 1];
index = i + 1;
}
}
sortedArray.push(min);
nums.splice(index, 1);
index = 0;
min = nums[0];
}
return sortedArray[x - 1];
} else {
return 0;
}
}
console.log(getX(2, [5, 10, -3, -3, 7, 9]));
// Leave this so we can test your code:
module.exports = getX;
function getX(x, nums) {
//if x is NOT in the range of the amount of numbers available in the array, return 0
if(x < 1 || x > nums.length){
return 0;
}
//if it is, sort the array in order then return the number at X location in the array
else{
nums.sort((a,b)=>a-b);
}
return nums[x-1]
}
module.exports = getX;
Here is my code for this chalenge, It took me sometime to understand that in the test, there were situations where x was outside the range of the array.length.
Therefore i decided to first test if x was valid and only sort the numbers once it passed the first test
-here is a more concise alternative, if we don’t mind sorting the array before checking the validity of x
function getX(x, nums) {
//sort nums in ascending order
nums.sort((a,b)=>a-b);
//return 0 if x not in the range of the array, else return number at x-1 position
return (x < 1 || x > nums.length)? 0: nums[x-1]
}
//test code
module.exports = getX;
function getX(x, nums) {
nums.sort((a,b) => a - b)
const result = nums[x - 1]
return result ? result : 0;
}
function getX(x, nums) {
nums.sort((a,b) => a - b)
const result = nums[x - 1]
return result ? result : 0;
}
console.log(getX(4, [5, 10, -3 , -3, 7, 9]));
console.log(getX(2, [5, 10, -3, -3, 7, 9]));
// Leave this so we can test your code:
module.exports = getX;
function getX(x, nums) {
if(nums.length===0 || x < 1 || x > nums.length){
return 0
}
let swapping=true;
while (swapping){
swapping=false;
for(let i=0; i < nums.length-1; i++){
if(nums[i] > nums[i+1]){
swap(nums, i, i+1);
swapping=true;
}
}
}
return nums[x-1];
}
const swap = (arr, index1, index2) =>{
let temp= arr[index2];
arr[index2]=arr[index1];
arr[index1]=temp;
}; // Write your code here
console.log(getX(1, [5, 10, -3, -3, 7, 9]));
console.log(getX(4, [5, 10, -3 , -3, 7, 9]));
// Leave this so we can test your code:
module.exports = getX;
function getX(x, nums) {
if (x > nums.length || nums.length == 0) return 0;
let numsSorted = nums.sort((a,b) => a - b);
return numsSorted[x-1];
}
console.log(getX(2, [5, 10, -3, -3, 7, 9]));
// Leave this so we can test your code:
module.exports = getX;
) below.
) to up-vote the contribution!