I got an array
var myArray = [5,8,1,4,2,9,3,7,6];
I want the output to be [ 9, 1, 8, 2, 7, 3, 6, 4, 5 ]. I tried the following code:
function firstAndLast(array) {
var arr= [];
array = myArray.sort().reverse();
for(var i = 0; i < array.length; i++){
var firstItem = myArray[i];
var lastItem = myArray[myArray.length - 1];
if(lastItem > firstItem){
arr.push(array[i]);
}}
var display = firstAndLast(myArray);
console.log(display);
Can anyone suggest what am I missing to achieve the targeted result?
What I want to acheive is to arrange the array in even odd indexes where odd indexes contain larger values in descending order and even indexes contain values in ascending order
Your code actually fits your description, except this part:
if(lastItem > firstItem){
arr.push(array[i]);
}
Why don't you just push both items to the array:
if(lastItem > firstItem){
arr.push(firstItem, lastItem);
}
And the lastItem should be dependent on i:
var lastItem = array[array.length - i - 1];
Them you only have to
return arr;
At the end and it should work :)
function firstAndLast(array) {
const result = [];
array = array.sort((a, b) => a - b).reverse();
for(var i = 0; i < array.length; i++){
var firstItem = array[i];
var lastItem = array[array.length - i - 1];
if(lastItem < firstItem){
result.push(firstItem, lastItem);
}
}
return result;
}
var myArray = [5,8,1,4,2,9,3,7,6];
console.log(firstAndLast(myArray));
Now this only omits the value in the middle, which you can easily add like this in the loop:
if(firstItem === lastItem) {
result.push(firstItem);
}
Apparently you want to shuffle the array?
If that is the case the simplest way of doing it is just using this
function shuffle(array) {
var m = array.length, t, i;
// While there remain elements to shuffle…
while (m) {
// Pick a remaining element…
i = Math.floor(Math.random() * m--);
// And swap it with the current element.
t = array[m];
array[m] = array[i];
array[i] = t;
}
return array;
}
It is Fisher-Yates shuffle.
More on it on the link : https://bost.ocks.org/mike/shuffle/
If that is not the case post a reply to the comment so we can find some new sorting logic!
function firstAndLast(array) { //You're declaring array here but you're using it in line 3 using the same array without, you're pasing myArray in line 12, my sugestion is to declare array inside de function
var arr = [];
array = myArray.sort().reverse();
for(var i = 0; i < array.length; i++) {
var firstItem = myArray[i];
var lastItem = myArray[myArray.length - 1]; //You're always using the last item of your array, if i'm not wrong (or confused) you want decrement the position, right? you have to use myArray.length-i or a new variable to decrement the position
if(lastItem > firstItem)
arr.push(array[i]); //Again, you're using array when you want to use myArray (it is what you're using for the position in line 5 and 6)
}
var display = firstAndLast(myArray);
console.log(display);
I want to reverse an array without using reverse() function like this:
function reverse(array){
var output = [];
for (var i = 0; i<= array.length; i++){
output.push(array.pop());
}
return output;
}
console.log(reverse([1,2,3,4,5,6,7]));
However, the it shows [7, 6, 5, 4] Can someone tell me, why my reverse function is wrong? Thanks in advance!
array.pop() removes the popped element from the array, reducing its size by one. Once you're at i === 4, your break condition no longer evaluates to true and the loop ends.
One possible solution:
function reverse(array) {
var output = [];
while (array.length) {
output.push(array.pop());
}
return output;
}
console.log(reverse([1, 2, 3, 4, 5, 6, 7]));
You can make use of Array.prototype.reduceright and reverse it
check the following snippet
var arr = ([1, 2, 3, 4, 5, 6, 7]).reduceRight(function(previous, current) {
previous.push(current);
return previous;
}, []);
console.log(arr);
In ES6 this could be written as
reverse = (array) => array.map(array.pop, [... array]);
No need to pop anything... Just iterate through the existing array in reverse order to make your new one.
function reverse(array){
var output = [];
for (var i = array.length - 1; i> -1; i--){
output.push(array[i]);
}
return output;
}
console.log(reverse([1,2,3,4,5,6,7]));
Edit after answer got accepted.
A link in a comment on your opening post made me test my way VS the accepted answer's way. I was pleased to see that my way, at least in my case, turned out to be faster every single time. By a small margin but, faster non the less.
Here's the copy/paste of what I used to test it (tested from Firefox developer scratch pad):
function reverseMyWay(array){
var output = [];
for (var i = array.length - 1; i> -1; i--){
output.push(array[i]);
}
return output;
}
function reverseTheirWay(array) {
var output = [];
while (array.length) {
output.push(array.pop());
}
return output;
}
function JustDoIt(){
console.log("their way starts")
var startOf = new Date().getTime();
for(var p = 0; p < 10000; p++)
{
console.log(reverseTheirWay([7,6,5,4,3,2,1]))
}
var endOf = new Date().getTime();
console.log("ran for " + (endOf - startOf) + " ms");
console.log("their way ends")
}
function JustDoIMyWay(){
console.log("my way starts")
var startOf = new Date().getTime();
for(var p = 0; p < 10000; p++)
{
console.log(reverseMyWay([7,6,5,4,3,2,1]))
}
var endOf = new Date().getTime();
console.log("ran for " + (endOf - startOf) + " ms");
console.log("my way ends")
}
JustDoIt();
JustDoIMyWay();
Solution to reverse an array without using built-in function and extra space.
let arr = [1, 2, 3, 4, 5, 6, 7];
let n = arr.length-1;
for(let i=0; i<=n/2; i++) {
let temp = arr[i];
arr[i] = arr[n-i];
arr[n-i] = temp;
}
console.log(arr);
Do it in a reverse way, Because when you do .pop() every time the array's length got affected.
function reverse(array){
var output = [];
for (var i = array.length; i > 0; i--){
output.push(array.pop());
}
return output;
}
console.log(reverse([1,2,3,4,5,6,7]));
Or you could cache the length of the array in a variable before popping out from the array,
function reverse(array){
var output = [];
for (var i = 0, len= array.length; i< len; i++){
output.push(array.pop());
}
return output;
}
console.log(reverse([1,2,3,4,5,6,7]));
You are modifying the existing array with your reverse function, which is affecting array.length.
Don't pop off the array, just access the item in the array and unshift the item on the new array so that the first element of the existing array becomes the last element of the new array:
function reverse(array){
var output = [],
i;
for (i = 0; i < array.length; i++){
output.unshift(array[i]);
}
return output;
}
console.log(reverse([1,2,3,4,5,6,7]));
If you'd like to modify the array in-place similar to how Array.prototype.reverse does (it's generally inadvisable to cause side-effects), you can splice the array, and unshift the item back on at the beginning:
function reverse(array) {
var i,
tmp;
for (i = 1; i < array.length; i++) {
tmp = array.splice(i, 1)[0];
array.unshift(tmp);
}
return array;
}
var a = [1, 2, 3, 4, 5];
console.log('reverse result', reverse(a));
console.log('a', a);
This piece allows to reverse the array in place, without pop, splice, or push.
var arr = [1, 2, 3, 4, 5];
function reverseArrayInPlace(arr2) {
var half = Math.floor(arr2.length / 2);
for (var i = 0; i < half; i++) {
var temp = arr2[arr2.length - 1 - i];
arr2[arr2.length - 1 - i] = arr2[i];
arr2[i] = temp;
}
return arr2;
}
As you pop items off the first array, it's length changes and your loop count is shortened. You need to cache the original length of the original array so that the loop will run the correct amount of times.
function reverse(array){
var output = [];
var len = array.length;
for (var i = 0; i< len; i++){
output.push(array.pop());
}
return output;
}
console.log(reverse([1,2,3,4,5,6,7]));
You're modifying the original array and changing it's size. instead of a for loop you could use a while
function reverse(array){
var output = [];
while(array.length){
//this removes the last element making the length smaller
output.push(array.pop());
}
return output;
}
console.log(reverse([1,2,3,4,5,6,7]));
function rvrc(arr) {
for (let i = 0; i < arr.length / 2; i++) {
const buffer = arr[i];
arr[i] = arr[arr.length - 1 - i];
arr[arr.length - 1 - i] = buffer;
}
};
const reverse = (array)=>{
var output = [];
for(let i=array.length; i>0; i--){
output.push(array.pop());
}
console.log(output);
}
reverse([1, 2, 3, 4, 5, 6, 7, 8]);
This happens because every time you do array.pop(), whilst it does return the last index in the array, it also removes it from the array. The loop recalculates the length of the array at each iteration. Because the array gets 1 index shorter at each iteration, you get a much shorter array returned from the function.
This piece of code will work without using a second array. It is using the built in method splice.
function reverse(array){
for (let i = 0; i < array.length; i++) {
array.splice(i, 0, array.splice(array.length - 1)[0]);
}
return array;
}
Here, let's define the function
function rev(arr) {
const na = [];
for (let i=0; i<arr.length; i++) {
na.push(arr[arr.length-i])
}
return na;
}
Let's say your array is defined as 'abca' and contains ['a','b','c','d','e','foo','bar']
We would do:
var reva = rev(abca)
This would make 'reva' return ['bar','foo','e','d','c','b','a'].
I hope I helped!
You can use .map as it is perfect for this situation and is only 1 line:
const reverse = a =>{ i=a.length; return a.map(_=>a[i-=1]) }
This will take the array, and for each index, change it to the length of the array - index, or the opposite side of the array.
with reverse for loop
let array = ["ahmet", "mehmet", "aslı"]
length = array.length
newArray = [];
for (let i = length-1; i >-1; i--) {
newArray.push(array[i])
}
console.log(newArray)
And this one:
function reverseArray(arr) {
let top = arr.length - 1;
let bottom = 0;
let swap = 0;
while (top - bottom >= 1) {
swap = arr[bottom];
arr[bottom] = arr[top];
arr[top] = swap;
bottom++;
top--;
}
}
function reverse(arr) {
for (let i = 0; i < arr.length - 1; i++) {
arr.splice(i, 0, arr.pop())
}
return arr;
}
console.log(reverse([1, 2, 3, 4, 5]))
//without another array
reverse=a=>a.map((x,y)=>a[a.length-1-y])
reverse=a=>a.map((x,y)=>a[a.length-1-y])
console.log(reverse(["Works","It","One","Line"]))
One of shortest:
let reverse = arr = arr.map(arr.pop, [...arr])
This is an old question, but someone may find this helpful.
There are two main ways to do it:
First, out of place, you basically push the last element to a new array, and use the new array:
function arrReverse(arr) {
let newArr = [];
for(let i = 0; i<arr.length; i++){
newArr.push(arr.length -1 -i);
}
return newArr;
}
arrReverse([0,1,2,3,4,5,6,7,8,9]);
Then there's in place. This is a bit tricky, but the way I think of it is like having four objects in front of you. You need to hold the first in your hand, then move the last item to the first place, and then place the item in your hand in the last place.
Afterwards, you increase the leftmost side by one and decrease the rightmost side by one:
function reverseArr(arr) {
let lh;
for(let i = 0; i<arr.length/2; i++){
lh = arr[i];
arr[i] = arr[arr.length -i -1];
arr[arr.length -i -1] = lh;
}
return arr;
}
reverseArr([0,1,2,3,4,5,6,7,8,9]);
Like so. I even named my variable lh for "left hand" to help the idea along.
Understanding arrays is massively important, and figuring out how they work will not only save you from unnecessarily long and tedious ways of solving this, but will also help you grasp certain data concepts way better!
I found a way of reversing the array this way:
function reverse(arr){
for (let i = arr.length-1; i >= 0; i--){
arr.splice(i, 0, arr.shift());
}
return arr;
}
Without Using any Pre-define function
const reverseArray = (array) => {
for (let i = 0; i < Math.floor(array.length / 2); i++) {
[array[i], array[array.length - i - 1]] = [
array[array.length - i - 1],
array[i]
];
}
return array;
};
let array = [1,2,3,4,5,6];
const reverse = (array) => {
let reversed = [];
for(let i = array.length - 1; i >= 0; i--){
reversed[array.length - i] = array[i];
}
return reversed;
}
console.log(reverse(array))
you can use the two pointers approach
example
function reverseArrayTwoPointers(arr = [1, 2, 3, 4, 5]) {
let p1 = 0;
let p2 = arr.length - 1;
while (p2 > p1) {
const temp = arr[p1];
arr[p1] = arr[p2];
arr[p2] = temp;
p1++;
p2--;
}
return arr;
}
to return [5,4,3,2,1]
example on vscode
let checkValue = ["h","a","p","p","y"]
let reverseValue = [];
checkValue.map((data, i) => {
x = checkValue.length - (i + 1);
reverseValue[x] = data;
})
function reverse(str1) {
let newstr = [];
let count = 0;
for (let i = str1.length - 1; i >= 0; i--) {
newstr[count] = str1[i];
count++;
}
return newstr;
}
reverse(['x','y','z']);
Array=[2,3,4,5]
for(var i=0;i<Array.length/2;i++){
var temp =Array[i];
Array[i]=Array[Array.length-i-1]
Array[Array.length-i-1]=temp
}
console.log(Array) //[5,4,3,2]
Wondering why i needed to add 4 to the array length in order for it to print out the entire array in reverse?
before i added 4 it was just using the .length property and it was only printing out 6543.
thanks in advance!
function reverseArray(array) {
var newArray =[];
for(var i = 0; i <= array.length+4; i++) {
newArray += array.pop(i);
}
return newArray;
}
var numbers = [1,2,3,4,5,6];
console.log(reverseArray(numbers));
array.pop removes (and returns) the last element. This affects the length of the array. The length is checked on every iteration, so since the array is getting shorter every time, the loop is ended early.
You can create a loop and pop items until it is empty, but another thing to take into account, is that it is the original array you are altering. I think a function like reverseArray shouldn't alter the array numbers that was passed to it if it returns another one. So a better solution would be a simple loop that iterates over all items without modifying the array.
function reverseArray(array)
{
var newArray =[];
for (var i = array.length-1; i >= 0; i--) {
newArray.push(array[i]);
}
return newArray;
}
var numbers = [1,2,3,4,5,6];
console.log(reverseArray(numbers));
console.log(numbers); // Should be unaltered.
If you don't mind modifying the array, you can use the reverse() method of the array:
var numbers = [1,2,3,4,5,6];
numbers.reverse();
console.log(numbers);
In Javascript, pop always removes the last element of the array. This shortens length, meaning that i and array.length were converging.
You can do a few things to avoid this behavior:
Store the original length when you start the loop: for (var i = 0 , l = array.length; i < l; i++)
Copy over values without modifying the original array
When you pop the items from the array, the item is removed from the array. As you increase the counter and decrease the length, they will meet halfway, so you get only half of the items.
Use push to put the items in the result. If you use += it will produce a string instead of an array.
If you use pop, then you can just loop while there are any items left in the array:
function reverseArray(array) {
var newArray = [];
while (array.length > 0) {
newArray.push(array.pop());
}
return newArray;
}
You can leave the original array unchanged by looping through it backwards and add items to the new array:
function reverseArray(array) {
var newArray = [];
for (var i = array.length - 1; i >= 0; i--) {
newArray.push(array[i]);
}
return newArray;
}
use following method for same output
function reverseArray(array)
{
var newArray =[];
var j = array.length-1;
for(var i = 0; i < array.length; i++)
{
newArray[j]= array[i]; j--;
}
return newArray;
}
var numbers = [1,2,3,4,5,6];
console.log(reverseArray(numbers));
I have a comma separated string, out of which I need to create a new string which contains a random order of the items in the original string, while making sure there are no recurrences.
For example:
Running 1,2,3,1,3 will give 2,3,1 and another time 3,1,2, and so on.
I have a code which picks a random item in the original string, and then iterates over the new string to see if it does not exist already. If it does not exist - the item is inserted.
However, I have a feeling this can be improved (in C# I would have used a hashtable, instead of iterating every time on the new array). One improvement can be removing the item we inserted from the original array, in order to prevent cases where the random number will give us the same result, for example.
I'd be happy if you could suggest improvements to the code below.
originalArray = originalList.split(',');
for (var j = 0; j < originalArray.length; j++) {
var iPlaceInOriginalArray = Math.round(Math.random() * (originalArray.length - 1));
var bAlreadyExists = false;
for (var i = 0; i < newArray.length; i++) {
if (newArray[i].toString() == originalArray[iPlaceInOriginalArray].toString()) {
bAlreadyExists = true;
break;
}
}
if (!bAlreadyExists)
newArray.push(originalArray[iPlaceInOriginalArray]);
}
Thanks!
You can still use a 'hash' in javascript to remove duplicates. Only in JS they're called objects:
function removeDuplicates(arr) {
var hash = {};
for (var i=0,l=arr.length;i<l;i++) {
hash[arr[i]] = 1;
}
// now extract hash keys... ahem...
// I mean object members:
arr = [];
for (var n in hash) {
arr.push(n);
}
return arr;
}
Oh, and the select random from an array thing. If it's ok to destroy the original array (which in your case it is) then use splice:
function randInt (n) {return Math.floor(Math.random()*n)}
function shuffle (arr) {
var out = [];
while (arr.length) {
out.push(
arr.splice(
randInt(arr.length),1 ));
}
return out;
}
// So:
newArray = shuffle(
removeDuplicates(
string.split(',') ));
// If you sort the first array, it is quicker to skip duplicates, and you can splice each unique item into its random position as you build the new array.
var s= 'Function,String,Object,String,Array,Date,Error,String,String,'+
'Math,Number,RegExp,Group,Collection,Timelog,Color,String';
var A1= s.split(',').sort(), A2= [], tem;
while(A1.length){
tem= A1.shift();
while(A1[0]== tem) tem= A1.shift();
if(tem) A2.splice(Math.floor(Math.random()*A2.length), 0, tem);
}
alert(A2.join(', '))
With your solution, you are not guaranteed not to pick same number several times, thus leaving some others of them never being picked. If the number of elements is not big (up to 100), deleting items from the source array will give the best result.
Edit
originalArray = originalList.split(',');
for (var j = 0; j < originalArray.length; j++) {
var iPlaceInOriginalArray = Math.round(Math.random() * (originalArray.length - 1 - j));
var bAlreadyExists = false;
for (var i = 0; i < newArray.length; i++) {
if (newArray[i].toString() == originalArray[iPlaceInOriginalArray].toString()) {
bAlreadyExists = true;
break;
}
}
var tmp = originalArray[originalArray.length - 1 - j];
originalArray[originalArray.Length - 1 - j] = originalArray[iPlaceInOriginalArray];
originalArray[iPlaceInOriginalArray] = tmp;
if (!bAlreadyExists)
newArray.push(originalArray[iPlaceInOriginalArray]);
}