find all subarrays in o(n) time 1D array using javascript - javascript

I want to collect all subarrays for further computation efficiently in javascript. I'm not sure this is possible, but it seems for a subarray sum kadane's formula is o(n) which is more efficient than other methods. But I'm not sure I how I can store the array at each step.
Similar to this quora question, for me the pseudo code was not enough. Thanks for the further breakdown.
another meta link
an example in action of this for [3, 3, 9, 9, 5]
[3], [9], [5], [9, 5], [9, 3], [9, 9], [3, 3],
[3, 9, 9], [3, 3, 9], [9, 9, 5], [3, 3, 9, 9],
[3, 9, 9, 5], [3, 3, 9, 9, 5]

I had done a work previously to calculate all combinations of amino acids total molecular weight. If you neglect the empty one you should have 2^n - 1 sub arrays. So there is no O(n) here. I've got two methods as binary and recursive.
function getSubArrays(arr){
var len = arr.length,
subs = Array(Math.pow(2,len)).fill();
return subs.map((_,i) => { var j = -1,
k = i,
res = [];
while (++j < len ) {
k & 1 && res.push(arr[j]);
k = k >> 1;
}
return res;
}).slice(1);
}
console.log(JSON.stringify(getSubArrays([1,2,3,4,5])));
function getSubArrays(arr){
if (arr.length === 1) return [arr];
else {
subarr = getSubArrays(arr.slice(1));
return subarr.concat(subarr.map(e => e.concat(arr[0])), [[arr[0]]]);
}
}
console.log(JSON.stringify(getSubArrays([1,2,3,4,5])));
I couldn't manage to get subarrays of an array with more than 23 items though.
Here are the performances. To be on the safe side i try with 22 items, first with recursive and then with binary route.
function getSubArrays(arr){
if (arr.length === 1) return [arr];
else {
subarr = getSubArrays(arr.slice(1));
return subarr.concat(subarr.map(e => e.concat(arr[0])), [[arr[0]]]);
}
}
var aminos = Array(22).fill().map((_,i) => i+1),
subarrs = [],
ps = 0,
pe = 0;
ps = performance.now();
subarrs = getSubArrays(aminos);
pe = performance.now();
console.log("recursive route took: " + (pe-ps) + "msec to produce " + subarrs.length + " sub arrays");
function getSubArrays(arr){
var len = arr.length,
subs = Array(Math.pow(2,len)).fill();
return subs.map((_,i) => { var j = -1,
k = i,
res = [];
while (++j < len ) {
k & 1 && res.push(arr[j]);
k = k >> 1;
}
return res;
}).slice(1);
}
var aminos = Array(22).fill().map((_,i) => i+1),
subarrs = [],
ps = 0,
pe = 0;
ps = performance.now();
subarrs = getSubArrays(aminos);
pe = performance.now();
console.log("binary route took: " + (pe-ps) + "msec to produce " + subarrs.length + " sub arrays");

This is fairly simple to do: https://jsfiddle.net/j1LuvxLq/
All you do is iterate possible lenghts and starting points and just print out the subsets. Complexity is O(n²) where n is the length of the original array. No way to improve it thought because that's the order of how many subsets there are.
var set = [3, 3, 9, 9, 5].join('')
var set_length = set.length
var subsets = []
for (var length_of_subset = 1; length_of_subset <= set_length; length_of_subset++) {
for (var start_of_subset = 0; start_of_subset <= set_length - length_of_subset; start_of_subset++) {
var current_subset = set.substring(start_of_subset, start_of_subset + length_of_subset)
if(subsets.indexOf(current_subset) == -1) {
subsets.push(current_subset.split(''))
}
}
}
// print the subsets out
for (s in subsets) {
$('body').append(subsets[s].join(', ') + '<br>')
}
Alternative, possibly nicer solution would be to use dynamic programming. Start with 3 and either remove last element or add next element. Check it out here: https://jsfiddle.net/p82fcs4m/
var set = [3, 3, 9, 9, 5].join('')
var subsets = []
take(set[0], set.substring(1))
function take(chosen, left) {
if(subsets.indexOf(chosen) != -1) {
return
}
subsets.push(chosen)
if (chosen.length > 1) {
take(chosen.substring(1), left)
}
if (left.length > 0) {
take(chosen.concat(left[0]), left.substring(1))
}
}
$('body').append(subsets.join('<br>'))

Try this:
function getSubArrays(arr) {
const subArrays = [];
for (let i = 0; i < arr.length; i++) {
for (let j = 0; j < arr.length - i; j++) {
const subArray = arr.slice(i, j + i + 1);
subArrays.push(subArray);
}
}
}
Still O(n²) tho

I believe using Array.slice is the most clean way to do it, isn't it?
function getSubArrays(arr) {
const subArrays = [];
for (var i = 0; i < arr.length; i++) {
for (var j = i; j < arr.length; j++) {
subArrays.push(arr.slice(i, j + 1));
}
}
return subArrays;
}

Related

Is there a best way to find the index value in a two dimensional array

Example:-
1234
2456
5678
I need to find the index of the value 7
Expected output:-
[3,3] // 3rd row and 3rd column.
Note:- Need to consider Time Complexity (ex:-Big(O) notation)
I am not looking for the below like answers
Sample 1
function findInArr(arr, elm) {
var occ = [];
for(var i = 0; i < arr.length; i++)
for(var j = 0; j < arr[i].length; j++)
if (arr[i][j] == elm)
occ.push(i+","+j);
return occ;
}
Test:
var numbers = [[1,2,3,4,5],[6,2,3,5,5],[9,8,3,4,9]];
var x = findInArr(numbers, 4);
console.log("found " + x.length + " occurences: " + x);
Here time complexity is 'n' rows * 'n' columns
Sample 2
var numbers=[[1,2,3,4,5],[6,2,3,5,5],[9,8,3,4,9]];
var searchItem=2;
numbers.forEach(function(parentItem,parentIndex){
parentItem.forEach(function(childItem,childIndex){
if(childItem===searchItem){
console.log(parentIndex);
console.log(childIndex);
}
})
});
this way ?
const numbers = [[1,2,3,4,5],[6,2,3,5,5],[9,8,3,4,9]];
function findInArr(arr, elm) {
let i = arr.findIndex(x=>x.includes(elm))
return (i<0) ? [null,null] : [i,arr[i].indexOf(elm)]
}
console.log( findInArr(numbers, 7))
console.log( findInArr(numbers, 8))
.as-console-wrapper { max-height: 100% !important; top: 0; }
you can use map, also index starts at 0, so your sample above would be [2,2] not [3,3], Below code, will find all indexes for the search item.
var numbers = [
[1, 2, 3, 4, 5],
[6, 2, 3, 5, 5],
[9, 8, 3, 4, 9]
];
var searchItem = 2;
console.log(numbers.map((v, i) => {
var res = v.filter(f => f === searchItem);
if (res.length) {
var ind=v.indexOf(searchItem);
return [i, ind]
}
}).filter(f => f !== undefined));

loop over array, but not every element

I would like to iterate over the array and getting an average from 5 next elements, not from the whole array. I try to do it by code bellow, but it doesn´t work. I appreciate any kind of help or suggestions.
function standardDeviation(array) {
let newArray = [];
for (let i = 0; i < array.length; i++) {
let tempArray = [];
const arrAvg = tempArray =>
tempArray.reduce((a, b) => a + b, 0) / tempArray.length;
newArray += tempArray;
for (let j = array[i]; (j = array[i + 5]); i++) {
tempArray += array[j];
}
}
console.log(newArray);
return newArray;
}
var arr = [1, 2, 3, 4, 5, 6, 7, 8, 9];
standardDeviation(arr);
You could slice a given array and take only five elements for getting an average.
function standardDeviation(array) {
const arrAvg = tempArray => tempArray.reduce((a, b) => a + b, 0) / tempArray.length;
return array.map((_, i, a) => arrAvg(a.slice(i, i + 5)));
}
var arr = [1, 2, 3, 4, 5, 6, 7, 8, 9];
console.log(standardDeviation(arr));
You could try using the slice() function of array element.
// simulated loop index
var curr_index_pos = 3;
// entire array
var array_full = [1,2,3,4,5,6,7,8,9,10];
// array with 5 next values from "curr_index_pos"
var array_5 = array_full.slice(curr_index_pos,curr_index_pos+5);
var sum = 0;
for( var i = 0; i < array_5.length; i++ ) {
sum += parseInt( array_5[i] );
}
var avg = sum/array_5.length;
console.log("array_5", array_5);
// [4,5,6,7,8]
console.log("sum", sum);
// 30
console.log("avg", avg);
// 6

How to find non repeated numbers in an Array using JavaScript?

Dear all I'm trying to find non repeated value in an array using javascript.I have written some code but it not working properly ..can you guys tell me where is the problem.thanks.
var arr = [-1, 2, 5, 6, 2, 9, -1, 6, 5, -1, 3];
var n = arr.length;
var result = '';
function nonrep() {
for (var i = 0; i < n; i++) {
var j;
for (j = 0; j < n; j++)
if (i != j && arr[i] == arr[j]) {
result = arr[i];
break;
}
if (j == n)
return arr[i];
}
return result;
}
console.log(nonrep())
There is possibly a more neat approach to this solution, but this works as expected by filtering the array and compare it's current value with the array items itself (expect current item's index value).
const sampleArray = [1,2,3,7,2,1,3];
const getNonDuplicatedValues = (arr) =>
arr.filter((item,index) => {
arr.splice(index,1)
const unique = !arr.includes(item)
arr.splice(index,0,item)
return unique
})
console.log("Non duplicated values: " , ...getNonDuplicatedValues(sampleArray))
Some changes:
Move all variable declarations inside of the function.
Use a function parameter for the handed over array, keep the function pure.
Declare all needed variables at top of the function in advance.
Take an array as result array unique.
Check i and j and if equal continue the (inner) loop.
Check the value at i and j and exit the (inner) loop, because a duplicate is found.
Take the check at the end of the inner loop and check the index j with the length of the array l, and if equal push the value to unique.
Use a single return statement with unique array at the end of the outer loop.
function getUnique(array) {
var l = array.length,
i, j,
unique = [];
for (i = 0; i < l; i++) {
for (j = 0; j < l; j++) {
if (i === j) {
continue;
}
if (array[i] === array[j]) {
break;
}
}
if (j === l) {
unique.push(array[i]);
}
}
return unique;
}
console.log(getUnique([-1, 2, 5, 6, 2, 9, -1, 6, 5, -1, 3]));
Another solution could be to check if indexOf and lastIndexOf returns the same value. Then you found a unique value.
var array = [-1, 2, 5, 6, 2, 9, -1, 6, 5, -1, 3],
unique = array.filter((v, i) => array.indexOf(v) === array.lastIndexOf(v));
console.log(unique);
You could first use reduce to get one object with count for each number element and then filter on Object.keys to return array of non-repeating numbers.
var arr=[-1,2,5,6,2,9,-1,6,5,-1,3];
var obj = arr.reduce((r, e) => (r[e] = (r[e] || 0) + 1, r), {});
var uniq = Object.keys(obj).filter(e => obj[e] == 1).map(Number)
console.log(uniq)
Solution with for loop.
var arr = [-1, 2, 5, 6, 2, 9, -1, 6, 5, -1, 3];
var uniq = [];
for (var i = 0; i < arr.length; i++) {
for (var j = 0; j < arr.length; j++) {
if (arr[i] == arr[j] && i != j) break;
else if (j == arr.length - 1) uniq.push(arr[i])
}
}
console.log(uniq)
Another simple approach
var arr = [1,1,2,3,3,4,4,5];
let duplicateArr = [];
var repeatorCheck = (item) => {
const currentItemCount = arr.filter(val => val=== item).length;
if(currentItemCount > 1) duplicateArr.push(item);
return currentItemCount;
}
var result = arr.filter((item,index) => {
var itemRepeaterCheck = !duplicateArr.includes(item) && repeatorCheck(item);
if(itemRepeaterCheck === 1){
return item;
}
});
console.log(result);
let arr = [1, 2, 1, 3, 3, 5];
function nonRepeatableNo(arr) {
let val = []
for (let i = 0; i < arr.length; i++) {
let count = 0;
for (let j = 0; j < arr.length; j++) {
if (arr[i] === arr[j]) {
count += 1
}
}
if (count === 1) {
val.push(arr[i])
}
}
console.log(val)
}
nonRepeatableNo(arr)
const arr = [-1, 2, 5, 6, 2, 9, -1, 6, 5, -1, 3];
const non_repeating = arr.filter(num => arr.indexOf(num) === arr.lastIndexOf(num))
console.log(non_repeating)
Filtering only unique elements according to OP request:
This uses for loops, as requested. It returns an array containing only elements appearing once in the original array.
var arr = [-1, 2, 5, 6, 2, 9, -1, 6, 5, -1, 3];
var n = arr.length;
var result = [];
function nonrep() {
for (var i = 0; i < n; i++) {
for (var j=0 ; j < n; j++)
if (i!=j && arr[i]==arr[j])
break;
if(j==n)
result.push(arr[i]);
}
return result;
}
console.log(nonrep())
var arr1 = [45, 4,16,25,45,4,16, 9,7, 16, 25];
var arr=arr1.sort();
console.log(arr);
var str=[];
arr.filter(function(value){
if( arr.indexOf(value) === arr.lastIndexOf(value))
{ str.push(value);
console.log("ntttttttttttttnnn" +str)
}// how this works ===============A
})
O/P
7,9
Please try the below code snippet.
var arr = [-1, 2, 5, 6, 2, 9, -1, 6, 5, -1, 3];
var uniqArr = [];
for (var i = 0; i < arr.length; i++) {
for (var j = 0; j < arr.length; j++) {
if (arr[i] == arr[j] && i != j) break;
else if (j == arr.length - 1){
uniqArr.push(arr[i])
}
}
}
console.log(uniqArr)
this ES6 code worked for me :
a.map(c=>a.filter(b=>c==b)).filter(e=>e.length<2).reduce((total, cur)=> total.concat(cur), [])
Here is a working method with loops.
var arr = [-1,2,5,6,2,9,-1,6,5,-1,3];
var len = arr.length;
const result = arr
.filter(value=>{
var count=0;
for(var i=0;i<len;i++)
{
if(arr[i]===value)
count++;
}
return count===1;
})
console.log(result);
const sampleArr = [-1, 2, 5, 6, 2, 9, -1, 6, 5, -1, 3];
function getUnique(arr){
const result=[]
const obj={}
for(let i=0;i<arr.length;i++){
if(!obj[arr[i]]){
obj[arr[i]]=true
result.push(arr[i])
}else{
const index= result.indexOf(arr[i])
if(index!==-1){
result.splice(result.indexOf(arr[i]),1)
}
}
}
return result
}
const uniqueArr= getUnique(sampleArr)
console.log(uniqueArr)
Here is the solution..
var x = [1,1,2,3,2,4]
var res = []
x.map(d => {
if(res.includes(d)) {
// remove value from array
res = res.filter((a) => a!=d)
} else {
// add value to array
res.push(d)
}
})
console.log(res) // [3,4]
//without using any filter also with minimum complexity
const array = [1 , 2, 3, 4, 2, 3, 1, 6, 8,1,1 ];
const unique = new Set();
const repetedTreses = new Set();
for(let i=0; i<array.length; i++) {
if(!unique.has(array[i]) && !repetedTreses.has(array[i])){
unique.add(array[i]);
}else{
repetedTreses.add(array[i]);
unique.delete(array[i]);
}
}
let uniqueElements=[...unique];
console.log(uniqueElements);
You can use filter and indexOf for that:
console.log(
[-1, 2, 5, 6, 2, 9, -1, 6, 5, -1, 3].filter((v, i, a) => a.indexOf(v, i + 1) === -1 )
);

function to repeat same numbers y times

I am trying to write a function that returns the number x. y number of times. The function should be able to take in multiple 2d arrays. There will be only 2 elements per input array.
So for example: function([4,3][2,2][12,5])
//output should be exactly: 444, 22, 1212121212
I have solved most of the problem but I am stuck when x is a 2 digit number.
would appreciate some help in solving this. I must return the solution as a string with a comma (' ,') that separates the different arrays.
Here is my code as of now:
function repeatnumber(data){
var result = " ";
for (var i = 0; i < data.length; i++){
if (Array.isArray(data[i])){
var x = data[i][0]
var y = data[i][1]
for(var j = 1; j <= y; j++){
result = result + x;
}
}
}
var number = result
var answer = number.match(/(\d)\1*/g)
return console.log(answer);
}
repeatnumber([[10, 2][11,1]])// DOESN'T WORK!! output must be: 1010,11
repeatnumber([[1, 2], [2, 3]]) //works
repeatnumber([[1, 4], [3, 6], [9, 2]]) //works
You could check if you got a nested array and map new values and join later all arrays with comma.
function repeatnumber(array) {
return (Array.isArray(array[0]) ? array : [array])
.map(([value, length]) => Array.from({ length }, _ => value).join(''))
.join(', ');
}
console.log(repeatnumber([42, 7]));
console.log(repeatnumber([[10, 2], [11,1]]));
console.log(repeatnumber([[1, 2], [2, 3]]));
console.log(repeatnumber([[1, 4], [3, 6], [9, 2]]));
If you are using es6 this could be the simplest.
let repeatnumber = (inputArr) => {
return inputArr.map((inp) => `${inp[0]}`.repeat(inp[1]));
}
Here's one of the solution.
function repeatNumber(data) {
var results = '';
data.forEach(function(arr) {
for(var i = 0; i< arr[1]; i++) results += arr[0].toString();
results += ',';
})
return results.substring(0, results.length - 1);
}
console.log(repeatNumber([[10, 2], [11,1]]));
console.log(repeatNumber([[1, 2], [2, 3]]));
console.log(repeatNumber([[1, 4], [3, 6], [9, 2]]));
As you've noticed, regex isn't the best option. You're making the question a lot more difficult than it needs to be. You can just make an array of each of those strings and then join them together. Also, more descriptive variable names will go a long way for you.
let data = [[10,2], [11,1]]
function repeatNumber(data) {
let resultArray = []
for (let i=0; i < data.length; i++) {
let num = data[i][0]
let times = data[i][1]
let newString = ""
for (let j=0; j < times; j++) {
newString += num
}
resultArray.push(newString)
}
return resultArray.join(", ")
}
console.log(repeatNumber(data))
If you're ok with a code golf solution, this will accomplish the same task:
data.map(([num, times]) => Array(times).fill(num).join("")).join(", ")
try this :
function repeatnumber(data) {
var result = "";
for (var i = 0; i < data.length; i++)
{
if (Array.isArray(data[i]))
{
var x = data[i][0]
var y = data[i][1]
for(var j = 1; j <= y; j++)
{
{
result = result + x;
}
}
}
if(i != data.length - 1)
result += " ";
}
var number = result
var split = number.split(" ");
//var answer = number.match(/(\d)\1*/g)
return console.log(split);
}
repeatnumber([[10, 2][11,1]]) should be repeatnumber([[10, 2],[11,1]]);
ES6 Solution
const repeat = (arr) => {
arr.map( v => {
const str = new Array(v[1] + 1).join(v[0]);
console.log(str); // save this in array or wherever u want.
})
};
repeat([[1, 4], [3, 6], [9, 2]])
You are missing a comma between two arrays, make it
repeatnumber([[10, 2],[11,1]])
Also, your regex won't work properly when you are adding two digit number, change your logic as
function repeatnumber(data)
{
var answer = [];
for (var i = 0; i < data.length; i++)
{
if (Array.isArray(data[i]))
{
var x = data[i][0]
var y = data[i][1]
result = "";
for(var j = 1; j <= y; j++)
{
result = result + x;
}
answer.push(result)
}
}
return answer;
}
Demo
function repeatnumber(data) {
var answer = [];
for (var i = 0; i < data.length; i++) {
if (Array.isArray(data[i])) {
var x = data[i][0]
var y = data[i][1]
result = "";
for (var j = 1; j <= y; j++) {
result = result + x;
}
answer.push(result)
}
}
return answer;
}
console.log(repeatnumber([
[10, 2],
[11, 1]
])) ;
console.log(repeatnumber([
[1, 2],
[2, 3]
])) ;
console.log(repeatnumber([
[1, 4],
[3, 6],
[9, 2]
])) ;
Edit
Or a more concise method using fill and reduce
function repeatnumber(data)
{
return data.map( s => Array( s[1] ).fill( "" ).reduce( (a,c) => a + s[ 0 ], "" ) )
}
Or using fill and join
function repeatnumber(data)
{
return data.map( s => Array( s[1] + 1 ).fill( "" ).join( s[ 0 ] ) )
}
Small code that works,
// iterate > create array and join with char > join parent array with comma
function repeatnumber(data){
if(!data || data.length < 1) return [];
return data.map(arr => Array(arr[1] + 1).join(arr[0])).join(',');
}
A better approach, thanks to Ariz
function repeatnumber(data) {
if (!data || data.length < 1) return [];
return data.map((arr) => `${arr[0]}`.repeat(arr[1])).join(',');
}
No creation of an extra array, so huge repetitions can be done with ease.
Single pass by nested template literals and .reduce() working hand to hand.
var a = [[10, 2], [11,1]],
b = [[1, 2], [2, 3]],
c = [[1, 4], [3, 6], [9, 2]],
f = a => a.reduce((r,[v,c],i) => `${i ? `${r},`:``}${`${v}`.repeat(c)}`,``);
console.log(f(a));
console.log(f(b));
console.log(f(c));

Sort an array containing numbers using a 'for' loop

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;
}
}
}

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