Here is the quicksort code I wrote. The function doesn't work because it can't reach the base case. If I log the pivot, r and l to the console, they remain the same no matter how many times the sort function is called. So I wonder if the argument l, r are not really passed into the function as data. Why did it happen?
function sort(data){
if(data.length < 2){
return data;
}
else{
var l = [];
var r = [];
var pivot = parseInt(data.length/2);
for(i=0; i<data.length; i++){
if(data[i] > data[pivot]){
r.push(data[i]);
}
else{
l.push(data[i]);
}
}
return sort(l).concat(sort(r));
}
}
I think that the issue here is that your partitioning step does not necessarily shrink the input array. For example, let's trace what happens if you try sorting [1, 2]. In this case, your pivot element will be the element 2. Since 1 > 2 is false, 1 is added to the list l. Since 2 > 2 is false, 2 is added to the list l. As a result, your recursive call on the list l will have exactly the same arguments as your original call, causing infinite recursion.
To fix this, try splitting the input into three lists - one of smaller values, one of equal values, and one of greater values. This code is shown here:
function sort(data){
if (data.length < 2){
return data;
} else {
var l = [];
var r = [];
var e = [];
var i = 0;
var pivot = (data.length / 2) | 0;
for(i = 0; i < data.length; i++) {
if (data[i] > data[pivot]) {
r.push(data[i]);
} else if (data[i] < data[pivot]) {
l.push(data[i]);
} else {
e.push(data[i]);
}
}
return sort(l).concat(e, sort(r));
}
}
This new version explicitly groups the equal elements into their own list, so they aren't recursively sorted by either of the recursive calls. It also gracefully handles duplicate elements.
If you pick the largest value of the array as the pivot element, then all values of data will end up in the array l and none in r. Thus will make the recursion never stop (and keep l, r and pivot at the same values).
Unless this is a brain excercise, using data.sort() should do a better job. ;)
JavaScript passes objects by reference (arrays are objects too). If you want to pass them by value, you need to use the splice function as explained here.
Note that this will create a lot of copies of your data. You probably want to use the native sort() function.
Related
Since the MIN() function in Sheets only returns a single value and there is no way to make it work with ARRAYFORMULA, I wanted to make a custom function that would take two arrays and compare the values at each entry, and return an array of the minimums. (I know there's a workaround that uses QUERY, but it wasn't going to work for my purposes)
What I have right now will take two arrays with one row and work perfectly. Unfortunately, it breaks when more than one row is introduced. I'm not sure why, so I'm lost on how to move forward. How can I make it work for any size arrays?
When I feed it any two dimensional range, it throws an error:
"TypeError: Cannot set property '0' of undefined"
on this line finalarray[x][y] = Math.min(arr1[x][y], arr2[x][y]);
The current ""working"" code:
function MINARRAY(arr1, arr2) {
if (arr1.length == arr2.length && arr1[0].length == arr2[0].length)
{
var finalarray = [[]];
for (x = 0; x < arr1.length; x++)
{
for(y = 0; y < arr1[x].length; y++)
{
finalarray[x][y] = Math.min(arr1[x][y], arr2[x][y]);
}
}
return finalarray;
}
else
{
throw new Error("These arrays are different sizes");
}
}
finalarray is a 2D array. [[]] sets only the first element of finalarray to a array. It is needed to set all elements of finalarray to a array. Inside the loop, add
finalarray[x]=[]//set `x`th element as a array or finalarray[x]= finalarray[x] || []
finalarray[x][y] = Math.min(arr1[x][y], arr2[x][y]);
Alternatively,
finalarray[x] = [Math.min(arr1[x][y], arr2[x][y])]
I want to partition an array (eg [1,2,3,4,5,6,7,8]), first partition should keep even values, second odd values (example result: [2,4,6,8,1,3,5,7]).
I managed to resolve this problem twice with built-in Array.prototype methods. First solution uses map and sort, second only sort.
I would like to make a third solution which uses a sorting algorithm, but I don't know what algorithms are used to partition lists. I'm thinking about bubble sort, but I think it is used in my second solution (array.sort((el1, el2)=>(el1 % 2 - el2 % 2)))... I looked at quicksort, but I don't know where to apply a check if an integer is even or odd...
What is the best (linear scaling with array grow) algorithm to perform such task in-place with keeping order of elements?
You can do this in-place in O(n) time pretty easily. Start the even index at the front, and the odd index at the back. Then, go through the array, skipping over the first block of even numbers.
When you hit an odd number, move backwards from the end to find the first even number. Then swap the even and odd numbers.
The code looks something like this:
var i;
var odd = n-1;
for(i = 0; i < odd; i++)
{
if(arr[i] % 2 == 1)
{
// move the odd index backwards until you find the first even number.
while (odd > i && arr[odd] % 2 == 1)
{
odd--;
}
if (odd > i)
{
var temp = arr[i];
arr[i] = arr[odd];
arr[odd] = temp;
}
}
}
Pardon any syntax errors. Javascript isn't my strong suit.
Note that this won't keep the same relative order. That is, if you gave it the array [1,2,7,3,6,8], then the result would be [8,2,6,3,7,1]. The array is partitioned, but the odd numbers aren't in the same relative order as in the original array.
If you are insisting on an in-place approach instead of the trivial standard return [arr.filter(predicate), arr.filter(notPredicate)] approach, that can be easily and efficiently achieved using two indices, running from both sides of the array and swapping where necessary:
function partitionInplace(arr, predicate) {
var i=0, j=arr.length;
while (i<j) {
while (predicate(arr[i]) && ++i<j);
if (i==j) break;
while (i<--j && !predicate(arr[j]));
if (i==j) break;
[arr[i], arr[j]] = [arr[j], arr[i]];
i++;
}
return i; // the index of the first element not to fulfil the predicate
}
let evens = arr.filter(i=> i%2==0);
let odds = arr.filter(i=> i%2==1);
let result = evens.concat(odds);
I believe that's O(n). Have fun.
EDIT:
Or if you really care about efficiency:
let evens, odds = []
arr.forEach(i=> {
if(i%2==0) evens.push(i); else odds.push(i);
});
let result = evens.concat(odds);
Array.prototype.getEvenOdd= function (arr) {
var result = {even:[],odd:[]};
if(arr.length){
for(var i = 0; i < arr.length; i++){
if(arr[i] % 2 = 0)
result.odd.push(arr[i]);
else
result.even.push(arr[i]);
}
}
return result ;
};
I'm studying for a technical interview right now, and writing quick javascript implementations of different sorts. The random-array benchmark results for most of the elementary sorts makes sense but the selection sort is freakishly fast. And I don't know why.
Here is my implementation of the Selection Sort:
Array.prototype.selectionSort = function () {
for (var target = 0; target < this.length - 1; target++) {
var min = target;
for (var j = target + 1; j < this.length - 1; j++) {
if (this[min] > this[j]) {
min = j;
}
}
if (min !== target) {
this.swap(min, target);
}
}
}
Here are the results of the same randomly generated array with 10000 elements:
BubbleSort => 148ms
InsertionSort => 94ms
SelectionSort => 91ms
MergeSort => 45ms
All the sorts are using the same swap method. So why is Selection Sort faster? My only guess is that Javascript is really fast at array traversal but slow at value mutation, since SelectionSort uses the least in value mutation, it's faster.
** For Reference **
Here is my Bubble Sort implementation
Array.prototype.bubbleSort = function () {
for (var i = this.length - 1; i > 1; i--) {
var swapped = false;
for (var j = 0; j < i; j++) {
if (this[j + 1] < this[j]) {
this.swap(j, j+1);
swapped = true;
}
}
if ( ! swapped ) {
return;
}
}
}
Here is the swap Implementation
Array.prototype.swap = function (index1, index2) {
var val1 = this[index1],
val2 = this[index2];
this[index1] = val2;
this[index2] = val1;
};
First let me point out two flaws:
The code for your selection sort is faulty. The inner loop needs to be
for (var j = target + 1; j < this.length; j++) {
otherwise the last element is never selected.
Your jsperf tests sort, as you say, the "same randomly generated array" every time. That means that the successive runs in each test loop will try to sort an already sorted array, which would favour algorithms like bubblesort that have a linear best-case performance.
Luckily, your test array is so freakishly large that jsperf runs only a single iteration of its test loop at once, calling the setup code that initialises the array before every run. This would haunt you for smaller arrays, though. You need to shuffle the array inside the "timed code" itself.
Why is Selection Sort faster? My only guess is that Javascript is really fast at array traversal but slow at value mutation.
Yes. Writes are always slower than reads, and have negative effects on cached values as well.
SelectionSort uses the least in value mutation
Yes, and that is quite significant. Both selection and bubble sort do have an O(n²) runtime, which means that both execute about 100000000 loop condition checks, index increments, and comparisons of two array elements.
However, while selection sort does only O(n) swaps, bubble sort does O(n²) of them. That means not only mutating the array, but also the overhead of a method call. And that much much more often than the selection sort does it. Here are some example logs:
> swaps in .selectionSort() of 10000 element arrays
9989
9986
9992
9990
9987
9995
9989
9990
9988
9991
> swaps in .bubbleSort() of 10000 element arrays
24994720
25246566
24759007
24912175
24937357
25078458
24918266
24789670
25209063
24894328
Ooops.
I have been reading through Chapter 5 last night and throughout the morning and can't seem to get the higher order functions concepts to stick. Here are the examples:
//I understand this first function, I am including it because it is used in the next function.
function forEach(array, action) {
for (vari = 0; i < array.length; i++)
action(array[i]);
}
forEach(["Wampeter", "Foma", "Granfalloon"], print);
function sum(numbers) {
var total = 0;
forEach(numbers, function(number) {
total += number;
});
return total;
}
To my understanding the function sum is taking the argument numbers, which I believe comes in as an array? Now, when the forEach function is called (within sum), it takes the array numbers passed to sum and then it also takes an anonymous function?
I am really confused on what this anonymous function is actually doing. It is taking the parameter number but what else is it doing? Does this anonymous function imply that in that parameter, a function like print or show will be passed the parameter number? In other words it would look something like this
function([10,12,11]) {
var total = 0
forEach([10,12,11]), show(???)
//at this point it would iterate over the array, and use the action passed to display `//the pointer in the array. What I think is happening is that it is taking this pointer value and adding it to the total.` //
I have been trying to wrap my head around this example for a while, if anyone knows of a good explanation or any other documentation to read over I would greatly appreciate it, thanks!
The anonymous function is applied to every currently selected element. You can see better how this works if you unroll (execute stepwise) the loop (pseudocode, * means current element):
var total = 0;
forEach([*1, 2, 3]), fun(1)) => total = 0 + 1 = 1
forEach([1, *2, 3]), fun(2)) => total = 1 + 2 = 3
forEach([1, 2, *3]), fun(3)) => total = 3 + 3 = 6
You can rewrite the sum function like this:
// because there is no "pass by reference" in JavaScript for
// "simple" types, total must be wrapped in an object
// in order to return the sum through the parameter for the showcase
var result = { total: 0 }
function sum(numbers_array) {
for (var i = 0; i < numbers_array.length; i++) {
accumulate(result, numbers_array[i]);
}
}
function accumulate(acc, number) {
acc.total += number;
}
In this case the accumulate function does the same as the anonymous function. When the accumulate function is declared within the scope of the sum function, then the total variable is like global (it is known) to the accumulate function and then there is no need of the first parameter, so the function becomes like the one you already know:
var total = 0;
function sum(numbers_array) {
function accumulate(number) {
total += number;
}
for (var i = 0; i < numbers_array.length; i++) {
accumulate(numbers_array[i]);
}
}
Next step would be to extract and pass the accumulate function as parameter:
var total = 0;
function accumulate(number) {
total += number;
}
// notice, that JavaScript knows how many parameters your function expects
function sum(numbers_array, action) {
for (var i = 0; i < numbers_array.length; i++) {
action(numbers_array[i]);
}
}
What left is to extract the iteration and the code will look like this one in the book.
Let me see if I can explain this easily for you:
The forEach() function accepts two parameters, the first one called array is obviously an array or an array-like object, the second parameter called action is actually a function.
forEach() visits each element in the array passed to it and applies to each element in the array the function passed to it as the second parameter.
So forEach() calls the function passed to it named action for each element in the array and it gives the function the array element as a parameter.
The function sum(numbers) accepts an array as you have though, and it uses forEach() inside itself to calculate the sum of numbers in that array using the anonymous function.
Remeber that the anonymous function is called once for each element in the array passed to sum() so it actually sums the elements in the array.
In simple words : to make your code more generic and concise.
Ex:
Lets say we want to find the max element in an Array :
That's pretty easy and cool :
In java script we will write :
var array = [10,20,30,40,50,60]
function maxEle(array){
var max = array[0];
for(var i=0;i< array.length;i++){
if(max < array[i]){
max = array[i];
}
}
console.log(max);
}
So this will give me the maximum element in an array.
Now after few days, some one asked me that your max is working pretty cool, I want a function which will print the minimum in an array.
Again I will redo the same thing, which i was doing in finding Max.
function minEle(array){
var min = array[0];
for(var i=0;i< array.length;i++){
if(min > array[i]){
min = array[i];
}
}
console.log(min);
}
Now this is also working pretty cool.
After sometime, another requirement comes up : I want a function which will print the sum of all the elements of the array.
Again the code will be similar to what we have written till now, except now it will perform summation.
function sumArr(array){
var sum = 0;
for(var i=0;i< array.length;i++){
sum = sum + array[i];
}
}
console.log(sum);
}
Observation :
After writing these bunch of codes, I m rewriting almost the same thing in every function, iterating over the Array and then performing some action.
Now writing the repetitive code is not a cool stuff.
Therefore we will try to encapsulate the varying part i.e action viz min, max, summation.
Since its feasible to pass functions as arguments to a function in FPL.
therefore we will re-factor our previously written code and now write a more generic function.
var taskOnArr = function(array, task){
for(var i=0;i<array.length;i++){
task(array[i]);
}
}
Now this will be our generic function, which can perform task on each element of Array.
Now our tasks will be :
var maxEle = array[0];
var taskMaxEle = function(ele){
if(maxEle < ele){
maxEle = ele;
}
}
Similarly for min element :
var minEle = array[0];
var taskMinEle = function(ele){
if(minEle > ele){
minEle = ele;
}
}
Also for summation of Array :
var sum = 0;
var taskSumArr = function(ele){
sum = sum + ele;
}
Now we need to pass functions to taskOnArr function :
taskOnArr(array,taskSumArr);
console.log(sum);
taskOnArr(array,taskMinEle);
console.log(minEle);
taskOnArr(array,taskMaxEle);
console.log(maxEle);
The following is a function for calculating all the possible combinations of a given array:
function combinations(arr, k) {
var i, subI, sub, combinationsArray = [], next;
for (i = 0; i < arr.length; i++) {
if (k === 1) {
combinationsArray.push([arr[i]]);
} else {
sub = combinations(arr.slice(i + 1, arr.length), k - 1);
for (subI = 0; subI < sub.length; subI++) {
next = sub[subI];
next.unshift(arr[i]);
combinationsArray.push(next);
}
}
}
return combinationsArray;
};
For example:
combinations([1,2,3],2);
returns:
[[1,2],[1,3],[2,3]]
I have a nested for loop which modifies a copy of an array of 12 objects (splicing certain elements,depending on the iteration of the loop), before using it as a parameter to the combinations function and storing certain elements of the array returned.
var resultArray = [];
var paramArray = [obj1,obj2,obj3,obj4,obj5,obj6,obj7,obj8,obj9,obj10,obj11,obj12];
for(i=0;i<length1;i++){
for(n=0;n<length2;n++){
paramArray.splice(...);//modifying array
resultArray[n] = combinations(paramArray,2)[i].slice();//storing an element, there are multiples of each element in the resultArray obviously
}
}
The browser crashes with the above type of code
(firefox returns the messege: "A script on this page may be busy, or it may have stopped responding. You can stop the script now, open the script in the debugger, or let the script continue.") The breakpoint is always the part where the combinations function thats being called.
Because the array parameter is different in each iteration, I cant assign the combinations function call to a variable to optimize the code. Is there a more efficient way of writing this?