I think I am on the verge of solving this but Im not sure why my code is not executing correctly. Can someone provide some feedback for me and show me where I messed up?
var insert = function(array, rightIndex, value) {
for(var j = rightIndex;
j >= 0 && array[j] > value;
j--) {
array[j + 1] = array[j];
}
array[j + 1] = value;
};
var insertionSort = function(array) {
for(var i = 1; i < array.length; i++){
insert(array, array.length -1, i);
}
};
var array = [22, 11, 99, 88, 9, 7, 42];
insertionSort(array);
println("Array after sorting: " + array);
//Program.assertEqual(array, [7, 9, 11, 22, 42, 88, 99]);
if I do this insert(array, array[i], i);,I get the following output:
Array after sorting: 22,11,12,100,89,10,8,43,5,,4,,1,,
I got here another solution for this insertion sort:
var insert = function(array, rightIndex, value) {
for(var j = rightIndex; j >= 0 && array[j] > value; j--) {
array[j + 1] = array[j];
}
array[j + 1] = value;
};
var insertionSort = function(array) {
for(var i = 0; i < array.length-1; i++){
insert(array, i, array[i+1]);
}
};
var array = [22, 11, 99, 88, 9, 7, 42];
insertionSort(array);
I think you have a probleme here:
in insert(array, array.length -1, i); it should be insert(array, array.length -1, array[i]);
you were inserting array index instead of the value
also you have an array out of bound in array[j + 1] = array[j]; because j start from array.length -1, it should be array[j] = array[j-1]; while j>0
last thing: your rightIndex should be i at each iteration not array.length -1.
Complete code :
var insert = function(array, rightIndex, value) {
for(var j = rightIndex;
j > 0 && array[j-1] > value;
j--) {
array[j] = array[j-1];
}
array[j] = value;
};
var insertionSort = function(array) {
for(var i = 0; i < array.length; i++){
insert(array, i, array[i]);
}
};
var array = [22, 11, 99, 88, 9, 7, 42];
insertionSort(array);
In insertion sort, we divide the initial unsorted array into two parts; sorted part and unsorted part. Initially the sorted part just has one element (Array of only 1 element is a sorted array). We then pick up element one by one from unsorted part; insert into the sorted part at the correct position and expand sorted part one element at a time.
var a = [34, 203, 3, 746, 200, 984, 198, 764, 9];
function insertionSort(values) {
var length = values.length;
for(var i = 1; i < length; ++i) {
var temp = values[i];
var j = i - 1;
for(; j >= 0 && values[j] > temp; --j) {
values[j+1] = values[j];
}
values[j+1] = temp;
}
};
console.log(a);
insertionSort(a);
console.log(a);
I know I am too late at the party. As you are aware there are several ways to do this but the yellow creature on KA apparently wants us to do it in a particular way. Here's the solution that made it happy:
var insert = function(array, rightIndex, value) {
for(var i=rightIndex; i >= 0 && array[i] > value ; i--){
array[i+1] = array[i];
}
array[i+1] = value;
};
Related
I have this code to do selection sort.
function selectionSort(array) {
for(let j = 0; j < array.length; j++) {
let smallest = array[j];
for(let i = j; i >= 0; i--) {
if(array[i] > smallest) {
let temp1 = array[i];
let temp2 = array[j];
array[i] = temp2;
array[j] = temp1;
}
}
}
return array;
}
selectionSort([8, 5, 2, 9, 5, 6, 3]).forEach(element => {
console.log(element);
});
I have added one test case, that fails. My idea here is that j is element selector, like a pointer to it, and next loop iterates backwards checking elements before it, and swapping the smallest element. But it kind of works? Some test cases like [1, 3, 2] work just fine, but some like this one does not
Here is a guide to the selection sort algorithum.
Your outer loop should be tracking the first unsorted element. Technically yours is, but you are calling the first unsorted element smallest which it isn't.
Then your inner loop should be searching for the smallest unsorted element and swapping it with the first unsorted element from the outer loop. You are swapping every larger element (instead of the single smallest) element after the one at i.
Let's fix it. As the algorithm says,
j is running from the start to end
i is running from j+1 to the end looking for smallest
after this, we swap the smallest (at smallest_index) with the original[j]
advanced to next j
function selectionSort(array) {
for (let j = 0; j < array.length - 1; j++) {
let smallest = array[j];
let smallest_index = -1;
for (let i = j + 1; i < array.length; i++) {
if (array[i] <= smallest) {
smallest = array[i]
smallest_index = i;
}
}
let temp1 = array[smallest_index];
let temp2 = array[j];
array[smallest_index] = temp2;
array[j] = temp1;
}
return array;
}
console.log("" + selectionSort([8, 5, 2, 9, 5, 6, 3]))
There is something off about your alogithm, it should look something like the below.
function selectionSort(array) {
for (var i = 0; i < array.length - 1; i++) {
let min = i;
for(var j=i+1;j<array.length;j++){
if(array[j] < array[min])
min = j;
}
const tmp1 = array[min]
const tmp2 = array[i]
array[i] = tmp1;
array[min] = tmp2;
}
return array;
}
selectionSort([8, 5, 2, 9, 5, 6, 3]).forEach(element => {
console.log(element);
});
Example: [1, 4, 9, 78, 42, 4, 11, 56]
Here the duplicate value is 4 and the gap is 3.
I used the array for each array element but I want this query to be optimized.
In Javascript, the following code will do the same for you.
var temp = [1, 4, 9, 78, 42, 4, 11, 56];
var encountered = [];
//This function gets you all the indexes of `val` inside `arr` array.
function getAllIndexes(arr, val) {
var indexes = [], i;
for(i = 0; i < arr.length; i++)
if (arr[i] === val)
indexes.push(i);
return indexes;
}
for(var i=0;i<temp.length;i++) {
if(encountered[temp[i]]) continue;
else {
var indexes = getAllIndexes(temp, temp[i]);
encountered[temp[i]] = true;
if(indexes.length>1) {
var steps = indexes[1]-indexes[0]-1;
$('.container').append('Duplicate item: '+temp[i]+' steps: '+ steps+'<br/>');
}
}
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.7.2/jquery.min.js"></script>
<div class="container"></div>
The below solution will give you if multiple duplicates.
http://jsfiddle.net/saz1d6re - Console Output
http://jsfiddle.net/saz1d6re/2/ - HTML Output
var a = [1, 4, 9, 78, 42, 4, 11, 4, 9];
result = [];
verified = [];
for(var i =0; i<a.length; i++){
b = a[i];
temp = {};
temp.value = a[i];
temp.hasDuplicate = false;
temp.positions = [];
temp.differenceFromFirstOccurence = [];
if(verified.indexOf(b) === -1){
temp.positions = [i+1];
for(var j = 0; j <a.length; j++){
c = a[j];
if( i !== j && b === c){
temp.hasDuplicate = true;
temp.positions.push(j+1);
}
}
verified.push(b);
result.push(temp);
}
}
for(var i = 0; i < result.length; i++){
if(result[i].hasDuplicate){
firstPosition = result[i]['positions'][0];
for(var j = result[i]['positions'].length-1; j > 0; j--){
diff = result[i]['positions'][j] - firstPosition-1;
result[i].differenceFromFirstOccurence.push(diff);
}
}
result[i].differenceFromFirstOccurence.reverse();
}
console.log(result);
for(var i =0; i < result.length; i++){
if(result[i].hasDuplicate && result[i].differenceFromFirstOccurence.length){
console.log("The first occurence of "+result[i].value+" is at "+ result[i].positions[0]);
for(var j = 1; j < result[i].positions.length; j++){
console.log("The duplicate occurence of "+result[i].value+" is at "+ result[i].positions[j] +" and difference is "+ result[i].differenceFromFirstOccurence[j-1]);
}
}
}
You can try this approach. First map all elements that have duplicates, then filter out empty values.
const data = [1, 4, 9, 78, 42, 4, 11, 56];
let duplIndex;
const res = data.map((el, index) => {
duplIndex = data.indexOf(el, index+1);
if(duplIndex !== -1){
return {el:el, gap: duplIndex - (index + 1)}
}
}).filter((el) => {
return el !== undefined;
});
Swaps the values at the position and at the minimum index. Write selection sort, making use of the swap and indexOfMinimum functions. This is the question, I dont know whats wrong in my logic and why my code does not run the assertion statement correctly.
var swap = function(array, firstIndex, secondIndex) {
var temp = array[firstIndex];
array[firstIndex] = array[secondIndex];
array[secondIndex] = temp;
};
var indexOfMinimum = function(array, startIndex) {
var minValue = array[startIndex];
var minIndex = startIndex;
for(var i = minIndex + 1; i < array.length; i++) {
if(array[i] < minValue) {
minIndex = i;
minValue = array[i];
}
}
return minIndex;
};
var selectionSort = function(array) {
var j;
var smallest;
for(j = 0; j < array.length; j++)
{
smallest = indexOfMinimum(array, 0);
swap(array , j , smallest);
}
};
var array = [22, 11, 99, 88, 9, 7, 42];
selectionSort(array);
println("Array after sorting: " + array);
Program.assertEqual(array, [7, 9, 11, 22, 42, 88, 99]);
There was a bug in your code, try running
var swap = function(array, firstIndex, secondIndex) {
var temp = array[firstIndex];
array[firstIndex] = array[secondIndex];
array[secondIndex] = temp;
};
var indexOfMinimum = function(array, startIndex) {
var minValue = array[startIndex];
var minIndex = startIndex;
for(var i = minIndex + 1; i < array.length; i++) {
if(array[i] < minValue) {
minIndex = i;
minValue = array[i];
}
}
return minIndex;
};
var selectionSort = function(array) {
var j;
var smallest;
for(j = 0; j < array.length; j++)
{
smallest = indexOfMinimum(array, j);
swap(array , j , smallest);
}
};
var array = [22, 11, 99, 88, 9, 7, 42];
selectionSort(array);
console.log("Array after sorting: " + array);
in selectionSort()
smallest = indexOfMinimum(array, 0);
should have been
smallest = indexOfMinimum(array, j);
I am new to JavaScript, and I have an array which contains numbers.
var arr = [2,4,8,1,5,9,3,7,6];
How can I sort it using a native for loop in JavaScript?
I know sort function is available, but I want it through for loop.
The output should be:
var res = [1,2,3,4,5,6,7,8,9];
var Arr = [1, 7, 2, 8, 3, 4, 5, 0, 9];
for (var i = 1; i < Arr.length; i++)
for (var j = 0; j < i; j++)
if (Arr[i] < Arr[j]) {
var x = Arr[i];
Arr[i] = Arr[j];
Arr[j] = x;
}
console.log(Arr);
I would do something like this...
var input = [2,3,8,1,4,5,9,7,6];
var output = [];
var inserted;
for (var i = 0, ii = input.length ; i < ii ; i++){
inserted = false;
for (var j = 0, jj = output.length ; j < jj ; j++){
if (input[i] < output[j]){
inserted = true;
output.splice(j, 0, input[i]);
break;
}
}
if (!inserted)
output.push(input[i])
}
console.log(output);
Maybe there are more efficient ways, but if you want to use the for loop, it's my first idea...
First create an empty array where the sorted numbers will be pushed into.
let sorted = [];
Secondly, create a very large amount of numbers that none of the numbers of the array can match. This number will be used for the very first comparison to determine which number of the array is smaller.
let comparison = 9000000000;
Create a for loop.
This loop will have another loop inside of it. The inner loop will check for the smallest number in a given array, and once the smallest number is gotten, it will be push into the empty array we created. The smallest number will also be removed from the initial array and then the array will run again.
for(a = 0; a < arr.length; a++){
//This inner loop fetches the smallest number.
for(b = 0; b < arr.length; a++){
if(comparison > arr[b]){
comparison = arr[b];
}
}
// The smallest number is assigned to comparison
// Now it being pushed to the empty array
sorted.push(comparison);
// Remove the smallest number from the initial array
let indexOfSmallNumber = arr.indexOf(comparison);
arr.splice(indexOfSmallNumber, 1);
// Set the comparison back to 9000000000;
comparison = 90000000000;
a = -1;
// Here, "a" is our main loop index counter and we are
// setting it to -1 because we don't want it to change
// to 2 by default, doing this will make the loop run
// forever until the initial array is empty.
}
let arr = [4, 2, 5, 1]
let temp;
function converter(arr) {
for(let i=0; i<arr.length; i++) {
for (let j=i+1; j<arr.length; j++) {
if(arr[i] > arr[j]) {
temp = arr[i]
arr[i] = arr[j]
arr[j] = temp
}
}
}
return arr
}
const newArr = converter(arr)
console.log(newArr)
Use:
let s = [4, 6, 3, 1, 2];
for (let i = 0; i < s.length;) {
if (s[i] > s[i + 1]) {
let a = s[i];
s[i] = s[i + 1];
s[i + 1] = a;
i--;
}
else {
i++;
}
}
This is a sorting algorithm which has a best time complexity of O(n) and the worst time of O(n^2).
This code checks for each number, and then compares to all numbers on the left side.
To check the time it takes each code to run, you can also use this code below:
let start = process.hrtime.bigint()
let end = process.hrtime.bigint()
console.log(end - start) // This measures the time used in nano seconds.
Also for microseconds, you can use this performance.now().
Here there is a very simple solution that uses a temporary array to store the values greater than the current one. Then it puts the current value between the lesser and the greater values:
var arr = [2,4,8,1,5,9,3,7,6];
var res = [];
for (const c of arr) {
let tmp = [];
while (c < res[res.length-1]) {
tmp.unshift(res.pop());
}
res = [...res, c, ...tmp];
}
const numberArr = [5, 9, 2, 8, 4, 10, 1, 3, 7, 6];
function sortedFunction(arr) {
let sortedArr = [];
for (let i = 0; i < arr.length; i++) {
for (let j = i + 1; j < arr.length; j++) {
let n = 0;
if (arr[i] > arr[j]) {
n = arr[i];
arr[i] = arr[j];
arr[j] = n;
}
}
sortedArr.push(arr[i]);
}
return sortedArr;
}
sortedFunction(numberArr);
Under the JavaScript array sort section of W3Schools it talks about how to compare a value in an array with the others and then order them based on the values being returned. I updated the code to use a for loop to sort values.
// Ascending points
var points = [5.0, 3.7, 1.0, 2.9, 3.4, 4.5];
var output = [];
var i;
for (i = 0; i < points.length; i++) {
points.sort(function (a, b) {
return a - b
});
output += points[i] + "<br>";
}
console.log(output);
// Descending points
var points = [5.0, 3.7, 1.0, 2.9, 3.4, 4.5];
var output = [];
var i;
for (i = 0; i < points.length; i++) {
points.sort(function (a, b) {
return b - a
});
output += points[i] + "<br>";
}
console.log(output);
const array = [12, 3, 45, 61, 23, 45, 6, 7];
function sortArray(array) {
for (var i = 0; i < array.length; ++i) {
for (var j = 0; j < array.length - 1 - i; ++j) {
if (array[j] > array[j + 1]) {
[array[j], array[j + 1]] = [array[j + 1], array[j]];
}
}
}
return array;
}
console.log(sortArray(array));
Here are the two solutions for the same algorithm:
Solution 1:
We can directly use JavaScript functions:
let arr = [2, 4, 8, 1, 5, 9, 3, 7, 6]
const changeOrder = (arr) => {
return arr.sort((a, b) => a - b)
}
let result = changeOrder(arr);
console.log(result) // [ 1, 2, 3, 4, 5, 6, 7, 8, 9 ]
Solution 2:
We can use a JavaScript for loop for doing the same
let arr = [2, 4, 8, 1, 5, 9, 3, 7, 6]
const changeOrder = (arr) => {
for(let i=1; i< arr.length; i++) {
for(let j=0; j < i; j++) {
if(arr[i] < arr[j]) {
let x = arr[i]
arr[i] = arr[j]
arr[j] = x
}
}
}
return arr;
}
let result = changeOrder(arr);
console.log(result) // [ 1, 2, 3, 4, 5, 6, 7, 8, 9 ]
An improvement to previous answer
for (let i = 0; i < arr.length; i++) {
for (let j = 0; j < arr.length - i - 1; j++) {
if (arr[j] > arr[j + 1]) {
let temp = arr[j];
arr[j] = arr[j + 1];
arr[j + 1] = temp;
}
}
}
I have made a bubble sort algorithm (sorta) using JS. It works sometimes, but the problem is that it only iterates through the array once. Here is my code:
function bubble(arr) {
for (var i = 0; i < arr.length; i++) {
if (arr[i] > arr[i + 1]) {
var a = arr[i]
var b = arr[i + 1]
arr[i] = b
arr[i + 1] = a
}
}
return arr;
}
Another bubble sort implementation:
const bubbleSort = array => {
const arr = Array.from(array); // avoid side effects
for (let i = 1; i < arr.length; i++) {
for (let j = 0; j < arr.length - i; j++) {
if (arr[j] > arr[j + 1]) {
[arr[j], arr[j + 1]] = [arr[j + 1], arr[j]];
}
}
}
return arr;
};
You need an inner loop to complete the sort correctly:
function bubble(arr) {
var len = arr.length;
for (var i = 0; i < len ; i++) {
for(var j = 0 ; j < len - i - 1; j++){ // this was missing
if (arr[j] > arr[j + 1]) {
// swap
var temp = arr[j];
arr[j] = arr[j+1];
arr[j + 1] = temp;
}
}
}
return arr;
}
document.write(bubble([1,9,2,3,7,6,4,5,5]));
Please look at the following sequence:
[5, 4, 3, 2, 1]
Now lets say you need to sort this in the ascending order using bubble sort.
So, you iterate the array and swap adjacent elements which are ordered otherwise.
Here is what you will get after the completion of the iteration
[4, 3, 2, 1, 5]
Now if you do this another time, you will get this:
[3, 2, 1, 4, 5]
Likewise, you need to repeat the iteration enough times to get it sorted fully. This means you need 2 nested loops. The inner loop is to iterate the array and the outer loop is to repeat the iteration.
Please see the step-by-step example of this article.
const bubbleSort = (array)=>{
let sorted = false;
let counter =0;
while(!sorted){
sorted = true;
for(let i =0; i < array.length -1 -counter; i++){
if(array[i] > array[i+1]){
helper(i,i+1,array);
sorted = false;
}
}
counter++;
}
return array;
}
//swap function
function helper(i,j, array){
return [array[i],array[j]] = [array[j],array[i]]
}
let array=[8,5,2,9,5,6,3];
console.log(bubbleSort(array))
var array = [6,2,3,7,5,4,1];
function bubbleSort(arr) {
for(let j=0;j<arr.length;j++) {
for(let i = 0; i < arr.length; i++) {
if(arr[i]>arr[i+1] && (i+1 < arr.length)) {
var temp = arr[i];
arr[i] = arr[i+1];
arr[i+1] = temp;
}
}
}
return arr;
}
console.log(bubbleSort(array));
My bubble sort with just a while loop :
function bubbleSort(arr){
var sorted = false
while (!sorted){
sorted = true;
arr.forEach(function (element, index, array){
if (element > array[index+1]) {
array[index] = array[index+1];
array[index+1] = element;
sorted = false;
}
});
}
}
function bubble(arr) {//You need Two Loops for Bubble sort
for (var i = 0; i < arr.length; i++) {//Outer Loop
for(var j=0; j < arr.length - 1; j++){//Inner Loop
if (arr[j] > arr[j + 1]) {
var a = arr[j]
var b = arr[j + 1]
arr[j] = b
arr[j + 1] = a
}
}
}
return arr;
}
Another form of bubble sort includes starting at the end of the array and placing the smallest element first and going till the largest. This is the code:
function bubbleSort(items) {
var length = items.length;
for (var i = (length - 1); i >= 0; i--) {
//Number of passes
for (var j = (length - i); j > 0; j--) {
//Compare the adjacent positions
if (items[j] < items[j - 1]) {
//Swap the numbers
var tmp = items[j];
items[j] = items[j - 1];
items[j - 1] = tmp;
}
}
}
}
Note Bubble sort is one of the slowest sorting algorithms.
It works for me. I commented the code for more understanding
bubbleSort = (numbersArray) => {
const arrayLenght = numbersArray.length;
for (let i = 0; i < arrayLenght; i++) {
for(let j = 0; j < arrayLenght; j++) {
// Print only to debug
// console.log(`i: ${i} - j: ${j}`);
// console.log(`numbersArray[i]: ${numbersArray[i]} | numbersArray[j]: ${numbersArray[j]}`);
// Check if current number is greater than the next number
if (numbersArray[j] > numbersArray[j + 1]) {
// Store current value to generate swap
const currentNumber = numbersArray[j];
// Now the current position get value of the next position
// And de next position get value of the current position
numbersArray[j] = numbersArray[j + 1];
numbersArray[j + 1] = currentNumber;
}
}
}
// Debug: Print the sorted array
console.log(`sorted array: ${numbersArray.toString()}`);
}
const numbers = [
[3, 10, 5, 7],
[8, 5, 2, 9, 5, 6, 3],
[4, 50, 28, 47, 9, 2097, 30, 41, 11, 3, 68],
[3, 10, 5, 7, 8, 5, 2, 9, 5, 6, 3]
];
numbers.forEach(element => {
bubbleSort(element);
});
Output:
sorted array: 3,5,7,10
sorted array: 2,3,5,5,6,8,9
sorted array: 3,4,9,11,28,30,41,47,50,68,2097
sorted array: 2,3,3,5,5,5,6,7,8,9,10
var arr = [5, 3, 4, 1, 2, 6];
function sort (arr) {
for(let i=0; i < arr.length - 1; i++) {
if(arr[i] > arr[i+1]) {
let b = arr[i+1];
arr[i+1] = arr[i];
arr[i] = b;
i = -1; // Resets the loop
}
}
return arr;
}
console.log(sort(arr));
Try this (performance upgrade):
function bubbleSort(inputArr, reverse = false) {
const len = inputArr.length;
for (let i = 0; i < len; i++) {
for (let j = i + 1; j < len; j++) {
let a = inputArr[i];
let b = inputArr[j];
if (reverse ? a < b : a > b) {
const tmp = inputArr[j];
inputArr[j] = inputArr[i];
inputArr[i] = tmp;
}
}
}
return inputArr;
}
Use:
arr = [234,2,4,100, 1,12,5,23,12];
console.log(bubbleSort(arr)); // or console.log(bubbleSort(arr, true));
You need another loop:
var arr = [2, 1]
for(let i = 0;i<arr.length;i++){
for(let b = 0; b<arr.length;i++){
if(arr[b] > arr[b+1]){
var first = arr[b]
var second = arr[b + 1]
arr[b] = second
arr[b + 1] = first
}
}
}
Hope this helps I would recommend using quick sort if you want a high efficiency though.
const bubbleSort = (inputArr) => {
const len = inputArr.length;
for (let i = 0; i < len; i++) {
for (let j = 0; j < len; j++) {
if (inputArr[j] > inputArr[j + 1]) {
let tmp = inputArr[j];
inputArr[j] = inputArr[j + 1];
inputArr[j + 1] = tmp;
}
}
}
return inputArr;
};
const numbers = [50, 30, 10, 40, 60];
console.log(bubbleSort(numbers));
// Output: [ 10, 30, 40, 50, 60 ]
function bubbleSort(array) {
var done = false;
while (!done) {
//alert(1)
done = true;
for (var i = 1; i < array.length; i += 1) {
if (array[i - 1] > array[i]) {
//alert(2)
done = false;
var tmp = array[i - 1];
array[i - 1] = array[i];
array[i] = tmp;
}
}
}
return array;
}
Another way would be like this:
function bubbleSort(arr) {
let swapped;
do {
swapped = false;
for (var i = 0; i < arr.length; i++) {
if (arr[i] > arr[i + 1]) {
swapped = true;
var tmp = arr[i];
arr[i] = arr[i + 1];
arr[i + 1] = tmp;
}
}
} while (swapped);
return arr;
}
let myArray = [8, 1, 2, 5, 51, 13, 15, 33, 123, 100, 22];
console.log(bubbleSort(myArray));
Explanation:
In this function we are going to declare a swapped variable that is being set to false inside the DO WHILE loop, this is being done as a fail-safe not to end up with an infinite loop.
Inside the loop, we have another FOR loop which iterates through the given array and checks if the current value is greater than the next (which we don't want, we need ascending order).
When the IF the condition is true, we are going to swap the variables and assign true for the swapped variable, this is done because we want to keep on the DO WHILE loop untill everything is sorted.
package hasan;
public class hssd {
public static void main(String[] args) {
int t=9;
int g=20;
for (t=g;t>19;++t){
System.out.println(7);
int f=12;
int r=15;
for(r=f;r>5;++r)
System.out.println(r+1000000000+"*"+1000000000);
}
}
}