This seems like a stupidly simple thing to do, but I can't figure out what I'm doing wrong. The goal is to have an array of 8 other arrays, which each contain hands of cards (in the example, the arrays just contain arbitrary numbers). Then, depending on whether passDirection is set to -1 or 1, each array is cycled through and replaced with the one next to it. The desired end result is that the values of playerList essentially shift by 1 either up or down, and this can be repeated several times without issue.
What's actually happening with the code I have below, though, is that all the arrays are just being replaced with what's at index 0, except for the first one. How can I fix this?
var playerList = new Array;
var passDirection = -1;
for(i = 0; i < 8; i++) {
playerList.push([playerList.length,i]); // Fill Arrays with arbitrary data
}
for (i=0; i< playerList.length; i++) {
console.log(i + ": " + playerList[i]); // Check their values before anything is done to them
}
for(q=0; q < 5; q++){ // Repeat the process 5 times, just because.
var bufferArray = playerList[0]; // Put Array Element 0's value in a buffer as it will be replaced first
for(i = 0; i < playerList.length && i > (playerList.length * -1); i += passDirection) {
var catcher = i; // 'catcher' should be the array that gets replaced
var passer = catcher - passDirection; // 'passer' should be the one it gets replaced with
if (catcher < 0) {
catcher = catcher + playerList.length;
}
if (passer < 0) {
passer = passer + playerList.length;
} else if (passer >= playerList.length) {
passer = passer - playerList.length;
}
if (passer == 0) {
playerList[catcher] = bufferArray;
} else {
playerList[catcher] = playerList[passer];
}
}
for (i=0; i< playerList.length; i++) {
console.log(i + ": " + playerList[i]);
}
console.log("...");
}
https://jsfiddle.net/3r1Lhwc5
You have two errors in your code:
if (passer = 0) is performing an assignment. You need if (passer === 0).
The passer index is looking at the wrong side of the value. Currently you are first getting from 1 and putting at 0, then getting from 0 and putting at 7 (i.e. -1). Notice how you are moving the same value in the second iteration. You need to change passer = catcher - passDirection to passer = catcher + passDirection
Note that all this can be done much easier with the splice, shift, unshift, pop and push Array methods (on the main playerList).
You can make your life easier by using Array methods to move elements from the beginning to end of an array or vice versa. Using this in the body of your for loop should do the trick:
if (passDirection === -1) {
const first = playerList.shift();
playerList.push(first);
}
else {
const last = playerList.pop();
playerList.unshift(last);
}
I need to do something like this: Let's say I have an array:
[3, 4, 1, 2]
I need to swap 3 and 4, and 1 and 2, so my array looks like [4, 3, 2, 1]. Now, I can just do the sort(). Here I need to count how many iterations I need, to change the initial array to the final output. Example:
// I can sort one pair per iteration
let array = [3, 4, 1, 2, 5]
let counter = 0;
//swap 3 and 4
counter++;
// swap 1 and 2
counter++;
// 5 goes to first place
counter++
// now counter = 3 <-- what I need
EDIT: Here is what I tried. doesn't work always tho... it is from this question: Bubble sort algorithm JavaScript
let counter = 0;
let swapped;
do {
swapped = false;
for (var i = 0; i < array.length - 1; i++) {
if (array[i] < array[i + 1]) {
const temp = array[i];
array[i] = array[i + 1];
array[i + 1] = temp;
swapped = true;
counter++;
}
}
} while (swapped);
EDIT: It is not correct all the time because I can swap places from last to first, for example. Look at the example code above, it is edited now.
This is most optimal code I have tried so far, also the code is accepted as optimal
answer by hackerrank :
function minimumSwaps(arr) {
var arrLength = arr.length;
// create two new Arrays
// one record value and key separately
// second to keep visited node count (default set false to all)
var newArr = [];
var newArrVisited = [];
for (let i = 0; i < arrLength; i++) {
newArr[i]= [];
newArr[i].value = arr[i];
newArr[i].key = i;
newArrVisited[i] = false;
}
// sort new array by value
newArr.sort(function (a, b) {
return a.value - b.value;
})
var swp = 0;
for (let i = 0; i < arrLength; i++) {
// check if already visited or swapped
if (newArr[i].key == i || newArrVisited[i]) {
continue;
}
var cycle = 0;
var j = i;
while (!newArrVisited[j]) {
// mark as visited
newArrVisited[j] = true;
j = newArr[j].key; //assign next key
cycle++;
}
if (cycle > 0) {
swp += (cycle > 1) ? cycle - 1 : cycle;
}
}
return swp;
}
reference
//You are given an unordered array consisting of consecutive integers [1, 2, 3, ..., n] without any duplicates.
//still not the best
function minimumSwaps(arr) {
let count = 0;
for(let i =0; i< arr.length; i++){
if(arr[i]!=i+1){
let temp = arr[i];
arr[arr.indexOf(i+1)] =temp;
arr[i] = i+1;
count =count+1;
}
}
return count;
}
I assume there are two reasons you're wanting to measure how many iterations a sort takes. So I will supply you with some theory (if the mathematics is too dense, don't worry about it), then some practical application.
There are many sort algorithms, some of them have a predicable number of iterations based on the number of items you are sorting, some of them are luck of the draw simply based on the order of the items to be sorted and which item how you select what is called a pivot. So if optimisation is very important to you, then you'll want to select the right algorithm for the purpose of the sort algorithm. Otherwise go for a general purpose algorithm.
Here are most popular sorting algorithms for the purpose of learning, and each of them have least, worst and average running-cases. Heapsort, Radix and binary-sort are worth looking at if this is more than just an theoretical/learning exercise.
Quicksort
Worst Case: Θ(n 2)
Best case: Θ(n lg n)
Average case: Θ(n lg n)
Here is a Quicksort implementation by Charles Stover
Merge sort
Worst case: Θ(n lg n)
Best case: Θ(n lg n)
Average Case: Θ(n lg n)
(note they're all the same)
Here is a merge sort implementation by Alex Kondov
Insertion sort
Worst case: Θ(n2)
Best case: Θ(n)
Average case:Θ(n2)
(Note that its worst and average case are the same, but its best case is the best of any algorithm)
Here is an insertion sort implementation by Kyle Jensen
Selection sort
Worst case: Θ(n2)
Best case: Θ(n2)
Average case: Θ(n2)
(note they're all the same, like a merge sort).
Here is a selection sort algorithm written by #dbdavid updated by myself for ES6
You can quite easily add an iterator variable to any of these examples to count the number of swaps they make, and play around with them to see which algorithms work best in which circumstance.
If there's a very good chance the items will already be well sorted, insertion sort is your best choice. If you have absolutely no idea, of the four basic sorting algorithms quicksort is your best choice.
function minimumSwaps(arr) {
var counter = 0;
for (var i = arr.length; i > 0; i--) {
var minval = Math.min(...arr); console.log("before", arr);
var minIndex = arr.indexOf(minval);
if (minval != = arr[0]) {
var temp = arr[0];
arr[0] = arr[minIndex];
arr[minIndex] = temp; console.log("after", arr);
arr.splice(0, 1);
counter++;
}
else {
arr.splice(0, 1); console.log("in else case")
}
} return counter;
}
This is how I call my swap function:
minimumSwaps([3, 7, 6, 9, 1, 8, 4, 10, 2, 5]);
It works with Selection Sort. Logic is as follows:
Loop through the array length
Find the minimum element in the array and then swap with the First element in the array, if the 0th Index doesn't have the minimum value founded out.
Now remove the first element.
If step 2 is not present, remove the first element(which is the minimum value present already)
increase counter when we swap the values.
Return the counter value after the for Loop.
It works for all values.
However, it fails due to a timeout for values around 50,000.
The solution to this problem is not very intuitive unless you are already somewhat familiar with computer science or real math wiz, but it all comes down to the number of inversions and the resulting cycles
If you are new to computer science I recommend the following resources to supplement this solution:
GeeksforGeeks Article
Informal Proof Explanation
Graph Theory Explanation
If we define an inversion as:
arr[i]>arr[j]
where "i" is the current index and "j" is the following index --
if there are no inversions the array is already in order and requires no sorting.
For Example:
[1,2,3,4,5]
So the number of swaps is related to the number of inversions, but not directly because each inversion can lead to a series of swaps (as opposed to a singular swap EX: [3,1,2]).
So if one consider's the following array:
[4,5,2,1,3,6,10,9,7,8]
This array is composed of three cycles.
Cycle One- 4,1,3 (Two Swaps)
Cycle Two- 5,2 (One Swap)
Cycle Three- 6 (0 Swaps)
Cycle Four- 10,9,7,8 (3 Swaps)
Now here's where the CS and Math magic really kicks in: each cycle will only require one pass through to properly sort it, and this is always going to be true.
So another way to say this would be-- the minimum number of swaps to sort any cycle is the number of element in that cycle minus one, or more explicitly:
minimum swaps = (cycle length - 1)
So if we sum the minimum swaps from each cycle, that sum will equal the minimum number of swaps for the original array.
Here is my attempt to explain WHY this algorithm works:
If we consider that any sequential set of numbers is just a section of a number line, then any set starting at zero should be equal to its own index should the set be expressed as a Javascript array. This idea becomes the criteria to programmatically determined if in element is already in the correct position based on its own value.
If the current value is not equal to its own index then the program should detect a cycle start and recording its length. Once the while loop reaches the the original value in the cycle it will add the minimum number of swaps in the cycle to a counter variable.
Anyway here is my code-- it is very verbose but should work:
export const minimumSwaps = (arr) => {
//This function returns the lowest value
//from the provided array.
//If one subtracts this value the from
//any value in the array it should equal
//that value's index.
const shift = (function findLowest(arr){
let lowest=arr[0];
arr.forEach((val,i)=>{
if(val<lowest){
lowest=val;
}
})
return lowest;
})(arr);
//Declare a counter variable
//to keep track of the swaps.
let swaps = 0;
//This function returns an array equal
//in size to the original array provided.
//However, this array is composed of
//boolean values with a value of false.
const visited = (function boolArray(n){
const arr=[];
for(let i = 0; i<n;i++){
arr.push(false);
}
return arr;
})(arr.length);
//Iterate through each element of the
//of the provided array.
arr.forEach((val, i) => {
//If the current value being assessed minus
//the lowest value in the original array
//is not equal to the current loop index,
//or, if the corresponding index in
//the visited array is equal to true,
//then the value is already sorted
if (val - shift === i || visited[i]) return;
//Declare a counter variable to record
//cycle length.
let cycleLength = 0;
//Declare a variable for to use for the
//while loop below, one should start with
//the current loop index
let x = i;
//While the corresponding value in the
//corresponding index in the visited array
//is equal to false, then we
while (!visited[x]) {
//Set the value of the current
//corresponding index to true
visited[x] = true;
//Reset the x iteration variable to
//the next potential value in the cycle
x = arr[x] - shift;
//Add one to the cycle length variable
cycleLength++;
};
//Add the minimum number of swaps to
//the swaps counter variable, which
//is equal to the cycle length minus one
swaps += cycleLength - 1;
});
return swaps
}
This solution is simple and fast.
function minimumSwaps(arr) {
let minSwaps = 0;
for (let i = 0; i < arr.length; i++) {
// at this position what is the right number to be here
// for example at position 0 should be 1
// add 1 to i if array starts with 1 (1->n)
const right = i+1;
// is current position does not have the right number
if (arr[i] !== right) {
// find the index of the right number in the array
// only look from the current position up passing i to indexOf
const rightIdx = arr.indexOf(right, i);
// replace the other position with this position value
arr[rightIdx] = arr[i];
// replace this position with the right number
arr[i] = right;
// increment the swap count since a swap was done
++minSwaps;
}
}
return minSwaps;
}
Here is my solution, but it timeouts 3 test cases with very large inputs. With smaller inputs, it works and does not terminate due to timeout.
function minimumSwaps(arr) {
let swaps = 0;
for (let i = 0; i < arr.length; i++) {
if (arr[i] === i + 1) continue;
arr.splice(i, 1, arr.splice(arr.indexOf(i + 1), 1, arr[i])[0]); //swap
swaps++;
}
return swaps;
}
I'm learning how to make it more performant, any help is welcome.
This is my solution to the Main Swaps 2 problem in JavaScript. It passed all the test cases. I hope someone finds it useful.
//this function calls the mainSwaps function..
function minimumSwaps(arr){
let swaps = 0;
for (var i = 0; i < arr.length; i++){
var current = arr[i];
var targetIndex = i + 1;
if (current != targetIndex){
swaps += mainSwaps(arr, i);
}
}
return swaps;
}
//this function is called by the minimumSwaps function
function mainSwaps(arr, index){
let swapCount = 0;
let currentElement = arr[index];
let targetIndex = currentElement - 1;
let targetElement = arr[currentElement - 1];
while (currentElement != targetElement){
//swap the elements
arr[index] = targetElement;
arr[currentElement - 1] = currentElement;
//increase the swapcount
swapCount++;
//store the currentElement, targetElement with their new values..
currentElement = arr[index];
targetElement = arr[currentElement - 1];
}
return swapCount;
}
var myarray = [2,3,4,1,5];
var result = console.log(minimumSwaps(myarray));
you can also do it with a map. But its O(nlogn)
const minSwaps = (arr) =>{
let arrSorted = [...arr].sort((a,b)=>a-b);
let indexMap = new Map();
// fill the indexes
for(let i=0; i<arr.length; i++){
indexMap.set(arr[i],i);
}
let count = 0;
for(let i=0; i<arrSorted.length;i++){
if(arr[i] != arrSorted[i]){
count++;
// swap the index
let newIdx = indexMap.get(arrSorted[i]);
indexMap.set(arr[i],newIdx);
indexMap.set(arrSorted[i],i);
// sawp the values
[arr[i],arr[newIdx]] =[arr[newIdx],arr[i]];
}
}
return count;
}
I am trying to insert a value into an array. I am not changing the size of the array. All I want to do is insert a value and then move all numbers after the insertion to the right using this algorighm:
Go to the last element in the array n = (length-1)
If it is not the passed index (n > index), set it's value to the value of the previous element A(n) = A(n-1)
If it is the passed index (n = index), set the value to the passed value A(n) = value and exit
Move left one element n = n-1
Repeat steps 2, 3, and 4
How do I do this? Also, I can't use any built-in array functions. Here is an example of my Javascript code:
var array = [];
for(var i=1; i<=1000; i++) {
array.push(Math.round(Math.random()*100));
}
function InsertIntoArray(array,index,number){
var numCount = 0
var move = 0 ;
for(var move = array.length - 1; move > index; move--)
{
if (move > index)
{
array[i] = array[i-1];
numCount ++;
}
else (move == index)
{
array[index] = number;
numCount++;
break;
}
}
console.log(move);
console.log(numCount);
console.log(array);
}
console.log(array);
InsertIntoArray(array, 1, 11);
You're pretty close but doing a log more than required. Hopefully the comments are sufficient:
// Just use simple test cases initially
var array = [0,1,2,3];
// The original function had some unused and pointless variables,
// they're removed
function insertIntoArray(array, index, value){
// Don't allow index to be greater than length - 1
if (index > array.length - 1) return;
// Loop until the required index is reached, shifting
// values to the next higher index
for(var move = array.length - 1; move > index; move--) {
array[move] = array[move - 1];
}
// Must now have reached required index and have shifted
// values, so just insert
array[index] = value;
}
// Original array
document.write(array + '<br>');
// Do insert
insertIntoArray(array, 2, 15);
// Modified array
document.write(array);
Note that you can have sparse arrays, the above will create new elements in such arrays so they aren't sparse any more. Also, for large arrays, it will be quite inefficient, Barmar's splice + slice answer is likely better in that regard, though it does change length along the way.
You could separate the arrays into a left and right array at the index, add the item onto the left array, and remove the last item from the right one. Then combine the two arrays:
function InsertIntoArray(array,index,number){
var leftArray = array.slice(0, index);
var rightArray = array.slice(index, array.length - 1);
leftArray.push(number);
return leftArray.concat(rightArray);
}
Fiddle Example. Note using return is to change the value of the array given other than the local array value. Simply changin array in the function will not change the global array variable.
The problem with your loop was that you were using array[i] = array[i-1], but the index variable in your loop was move, not i.
You don't need to do the if inside the loop. Just insert the new element when the loop is done. You also had a syntax error in the else -- you don't put a test after else, it's automatically the opposite of the if.
function InsertIntoArray(array, index, number) {
// Move all the elements after index up by 1
for (var move = array.length - 1; move > index; move--) {
array[move] = array[move - 1];
}
// Insert the new element
array[index] = number;
}
var array = [];
for (var i = 1; i <= 30; i++) {
array.push(i);
}
document.getElementById("before").textContent = JSON.stringify(array);
InsertIntoArray(array, 1, 11);
document.getElementById("results").textContent = JSON.stringify(array);
<b>Before:</b>
<div id="before"></div>
<b>After:</b>
<div id="results"></div>
I want to do something like:
var arr = []
for var(i=0;i<x;i++){
arr.push{ get num(){return this.previousArrayElement.num + randomNumber}}
}
how can I treat "previousArrayElement"?
I think you are just trying to create an array of size x containing numbers in order of size and separated by randomNumber intervals? Something like this would work:
var x = 100;
var arr = [0]
for (i=1; i<x; i++) {
arr.push( arr[i-1] + Math.random() );
}
Note that by starting the array out with an initial value (index 0) and beginning your iteration with the second value (index 1) you don't have to worry about accessing the 0-1 element at the first iteration.
I hope that helps!
Not 100% sure this is what you want. Expected output shown is not valid syntax and details provided are very open to interpretation
var arr = []
for (var i=0; i < x; i++){
var num = i > 0 ? arr[i-1].num : 0;
num= num + randomNumber; // is this an existing variable?
arr.push({ num: num}); // used object with property `num` based on example `previousArrayElement.num `
}
I have:
function getRandomInt(min, max){
return Math.floor(Math.random() * (max - min + 1)) + min;
}
But the problem is I want randomise the population of something with elements in an array (so they do not appear in the same order every time in the thing I am populating) so I need to ensure the number returned is unique compared to the other numbers so far.
So instead of:
for(var i = 0; i < myArray.length; i++) {
}
I have:
var i;
var count = 0;
while(count < myArray.length){
count++;
i = getRandomInt(0, myArray.length); // TODO ensure value is unique
// do stuff with myArray[i];
}
It looks like rather than independent uniform random numbers you rather want a random permutation of the set {1, 2, 3, ..., N}. I think there's a shuffle method for arrays that will do that for you.
As requested, here's the code example:
function shuffle(array) {
var top = array.length;
while (top--) {
var current = Math.floor(Math.random() * top);
var tmp = array[current];
array[current] = array[top - 1];
array[top - 1] = tmp;
}
return array;
}
Sometimes the best way to randomize something (say a card deck) is to not shuffle it before pulling it out, but to shuffle it as you pull it out.
Say you have:
var i,
endNum = 51,
array = new Array(52);
for(i = 0; i <= endNum; i++) {
array[i] = i;
}
Then you can write a function like this:
function drawNumber() {
// set index to draw from
var swap,
drawIndex = Math.floor(Math.random() * (endNum+ 1));
// swap the values at the drawn index and at the "end" of the deck
swap = array[drawIndex];
array[drawIndex] = array[endNum];
array[endNum] = swap;
endNum--;
}
Since I decrement the end counter the drawn items will be "discarded" at the end of the stack and the randomize function will only treat the items from 0 to end as viable.
This is a common pattern I've used, I may have adopted it into js incorrectly since the last time I used it was for writing a simple card game in c#. In fact I just looked at it and I had int ____ instead of var ____ lol
If i understand well, you want an array of integers but sorted randomly.
A way to do it is described here
First create a rand function :
function randOrd(){
return (Math.round(Math.random())-0.5); }
Then, randomize your array. The following example shows how:
anyArray = new Array('1','2','3','4','5');
anyArray.sort( randOrd );
document.write('Random : ' + anyArray + '<br />';);
Hope that will help,
Regards,
Max
You can pass in a function to the Array.Sort method. If this function returns a value that is randomly above or below zero then your array will be randomly sorted.
myarray.sort(function() {return 0.5 - Math.random()})
should do the trick for you without you having to worry about whether or not every random number is unique.
No loops and very simple.