Develop Reference

Quickest way to check if 2 arrays contain same values in javascript

Is there a quicker or more efficient way to check that two arrays contain the same values in javascript?
Here is what I'm currently doing to check this. It works, but is lengthy.
var arraysAreDifferent = false;
for (var i = 0; i < array1.length; i++) {
if (!array2.includes(array1[i])) {
arraysAreDifferent = true;
}
}
for (var i = 0; i < array2.length; i++) {
if (!array1.includes(array2[i])) {
arraysAreDifferent = true;
}
}

To reduce computational complexity from O(n ^ 2) to O(n), use Sets instead - Set.has is O(1), but Array.includes is O(n).
Rather than a regular for loop's verbose manual iteration, use .every to check if every item in an array passes a test. Also check that both Set's sizes are the same - if that's done, then if one of the arrays is iterated over, there's no need to iterate over the other (other than for the construction of its Set):
const arr1Set = new Set(array1);
const arr2Set = new Set(array2);
const arraysAreDifferent = (
arr1Set.size === arr2Set.size &&
array1.every(item => arr2Set.has(item))
);

function same(arr1, arr2){
//----if you want to check by length as well
// if(arr1.length != arr2.length){
// return false;
// }
// creating an object with key => arr1 value and value => number of time that value repeat;
let frequencyCounter1 = {};
let frequencyCounter2 = {};
for(let val of arr1){
frequencyCounter1[val] = (frequencyCounter1[val] || 0) + 1;
}
for(let val of arr2){
frequencyCounter2[val] = (frequencyCounter2[val] || 0) + 1;
}
for(let key in frequencyCounter1){
//check if the key is present in arr2 or not
if(!(key in frequencyCounter2)) return false;
//check the number of times the value repetiton is same or not;
if(frequencyCounter2[key]!==frequencyCounter1[key]) return false;
}
return true;
}

Write a function that will return the two number array combination with value summed at 5. I can't get all the opportunities

A job posting wants me to write a an answer to a question which if I solve I am eligible for the next rownd.
Write a function that will return the array combination with value
summed at 5. Important: Use only one "for" loop. Example: var
rand_array = [1,3,5,2,4,6]; var target_sum = 5; Output = [1,4], [5],
[3,2], [2,3], [4,1];
I attempted to find a solution online and stumbled upon this:
https://www.geeksforgeeks.org/given-an-array-a-and-a-number-x-check-for-pair-in-a-with-sum-as-x/ as StackOverflow wants you to do your own research first.
However, when trying to convert it to JS, all that happened was that it returned just one case where it worked. I need it to return every case where it worked. I then make some other changes and it just stopped working now.
var ra = [1,3,5,2,4,6];
var target = 5
ra.sort();
lower = 0;
higher = ra.length -1;
var solutions = [];
var result;
while (lower < higher) {
if (ra[lower] + ra[higher] === target){
result = [ra[lower], ra[higher]];
solutions.push(result);
}
else if (ra[lower] + ra[higher] > target){
higher--;
}
else {
lower++;
}
}
return solutions;
}
console.log(solutions);
Can someone write an example for me?

Your code does not work at all at the moment because it doesn't always increment lower or higher (resulting in an infinite loop). It also has greater complexity than necessary (.sort has complexity O(n log n)), but the instructions indicate that low complexity is important. The array also isn't being sorted numerically. (To sort numerically, use .sort((a, b) => a - b))
If you want a solution with the least complexity possible, O(n), while iterating, create an object. On every iteration, check to see if the object has a key for which the current number would sum with to 5 (eg, when iterating on 1, look to see if a 4 property exists on the object). If one is found, add it to the solutions. Otherwise, set a new key on the object:
const ra = [1, 3, 5, 2, 4, 6];
const target = 5;
const solutions = [];
const obj = {};
for (const item of ra) {
const match = target - item;
if (obj[match]) {
solutions.push([item, match]);
delete obj[match];
} else {
obj[item] = true;
}
}
console.log(solutions);
If there may be repeated numbers, then store a count in the object instead of just true:
const ra = [1, 1, 1, 3, 5, 2, 4, 6, 4, 4];
const target = 5;
const solutions = [];
const obj = {};
for (const item of ra) {
const match = target - item;
if (obj[match]) {
solutions.push([item, match]);
obj[match]--;
} else {
obj[item] = (obj[item] || 0) + 1;
}
}
console.log(solutions);

I don't want to write the actual answer because its a job assignment, but I will say that a simple 2 loop function is the obvies solution, than try to think not checking the array from the top and the bottum, rether like the formation of a loop in a loop.
hint :
let i = 0;
let j = 0;
while (i < arr.langth) {
...
if (j < arr.langth) {
j++;
} else {
j = 0;
i++;
}
}

Your code as it stands just does not work at all. CertainPerformance has a solid answer, except that it doesn't do the task as required, i.e. treat the same numbers in a different order as different or get the values which are equal to target as solutions. Here is my solution to your problem:
const ra = [1,3,5,2,4,6];
const target = 5
function getSumsOfTarget(ra, target){
ra.sort();
lower = 0;
higher = ra.length -1;
const solutions = [];
let result;
while (lower < ra.length && higher >= 0) {
const sum = ra[lower] + ra[higher];
if (ra[lower] === target) {
result = [ra[lower]];
solutions.push(result);
break;
}
if (sum === target){
result = [ra[lower], ra[higher]];
solutions.push(result);
lower++;
}
else if (sum > target){
higher--;
}
else {
lower++;
}
}
return solutions;
}
console.log(getSumsOfTarget(ra, target));

Evaluating an Array for Uniformity

I am trying to evaluate an array for uniformity. I.e. var num =[1,1,1,1,1,] // true or var num =[1,1,2,1,1] = //false. It could also be for an array of strings. What I am trying to do is evaluate the whole array and return a single "true" or "false". My code so far evaluates each item in the array and returns the result for each item. I'm very new to coding and taking classes online and working through problems without cheating!
var num = [1,1,3,1,1];
var first = num[0];
num.forEach(function (num, i) {
if(first === num){
console.log(true);
} else {
console.log(false);
}
});

You're on the right track. Instead of logging a true or false for each answer, create a variable to store the overall answer, then print that when you're done. Something like this:
var num = [1,1,3,1,1];
var first = num[0];
var ans = true;
num.forEach(function (n) {
if(first !== n){
ans = false;
}
});
console.log(ans);

Using .some()
var num = [1,1,3,1,1];
var notUni = num.some(function(val, idx, arr){
return val != arr[0]; // (If arr[0] (1) is the base you're interested in)
});
console.log( notUni ); // true
P.S: some will exit as soon it finds a (true) mismatch.

Use Array.prototype.every() function
var num = [1, 1, 3, 1, 1];
var result = num.every(n => n == num[0]);
The every() method tests whether all elements in the array pass the test implemented by the provided function.
var num = [1, 1, 1, 1, 1];
var result = num.every(n => n == num[0]);
document.write(result);

Remove elements from array by type of element

I have an array and I want to remove all the string elements from it.
This is what I have so far. The result is not what I want since it returns only "bicycle"
Also, I am doing this in Test Complete so I need to have a main function that logs the result.
function ex06(){
var mailBox = "mailbox";
var twenty = 20;
var isItRaining = true;
var goat = "";
var stringsArray = ["bicycle", "pocket", 3, mailBox, twenty, isItRaining, goat];
var result = removeStrings();
Log.Message("stringsArray looks like this after the removal of all the string elements: " + result);
function removeStrings(){
var i;
var x
for(i = 0; i < stringsArray.length; i++){
if (typeof(stringsArray[i]) === 'string'){
x = stringsArray.splice(i, 1);
return x;
}
}
}
}

Version 1, with Array#filter
var a = [1, 2, "3", "4", true];
a = a.filter(function (e) {
return typeof e !== 'string';
});
document.write('<pre>' + JSON.stringify(a, 0, 4) + '</pre>');
Version 2, with Array#splice and running backwards.
var a = [1, 2, "3", "4", true],
i = a.length;
while (i--) {
if (typeof a[i] === 'string') {
a.splice(i, 1);
}
}
document.write('<pre>' + JSON.stringify(a, 0, 4) + '</pre>');

The Array.prototype.filter method is what you need:
var stringsArray = ["bicycle", "pocket", 3, mailBox, twenty, isItRaining, goat];
var result = stringsArray.filter(function(element) {
return typeof element !== 'string';
});

you need to reduce the counter variable and check the original array
try this simple example
var a = [1,2,"3", "4", true];
for( var counter = 0; counter < a.length; counter++)
{
if ( (typeof a[ counter ] ) == "string" )
{
a.splice(counter,1); counter--;
}
}
console.log(a); //output [1, 2, true]

try this code:
function ex06(){
var mailBox = "mailbox";
var twenty = 20;
var isItRaining = true;
var goat = "";
var stringsArray = ["bicycle", "pocket", 3, mailBox, twenty, isItRaining, goat];
var result = removeStrings();
Log.Message("stringsArray looks like this after the removal of all the string elements: " + result);
function removeStrings(){
var newarray = [];
var i;
var x
for(i = 0; i < stringsArray.length; i++){
if (typeof(stringsArray[i]) !== 'string'){
newarray.push(stringsArray[i]);
}
}
return newarray
}
}

JavaScript offers native methods to filter arrays, so that you can remove string elements more easily: Array.prototype.filter can make the process a lot easier (and prevents strange behaviours when using splice inside a loop).
function ex06(){
var mailBox = "mailbox";
var twenty = 20;
var isItRaining = true;
var goat = "";
var stringsArray = ["bicycle", "pocket", 3, mailBox, twenty, isItRaining, goat];
var result = removeStrings(stringsArray);
Log.Message("stringsArray looks like this after the removal of all the string elements: " + result);
function removeStrings(arrayWithString){
return arrayWithString.filter(function(item) {
return typeof item !== 'string'; // returns only items which are not strings
});
}
}
A small piece of advice: Pass in the array into your function instead of referencing it from the parent scope. This way you have a pure, reusable function (and no strange side effects you might not want).

I assume this is an exercise, and that's why you're not using Array#filter.
The problem is that you have your return x inside your for loop, so you return the first string you find.
You have at least three options:
Don't return anything, since removeStrings is modifying the original array. That one's easy: Just remove the return x; line.
Don't modify the original array; instead, create and return a new array with the strings left out. In that case, you'd start with x = [] before the loop, remove the splice call, and instead push any non-string onto x.
Modify the original array, and create and return a new array containing the strings you've removed. In that case, you'd remove return x from inside the loop, have x = [] before the loop, and push the entries you remove onto x. Then return x at the end.
In any of the places where you're modifying the original, note gurvinder372's point that when you remove an entry, you need to not increase the index counter, as you'll end up skipping the next entry.
I wouldn't do it the way he suggests, though; when I'm looping through an array modifying it, for isn't what I reach for, I reach for while:
i = 0;
while (i < stringsArray.length) {
if (typeof stringsArray[i] === 'string'){
stringsArray.splice(i, 1);
// We leave `i` alone here, because we need to process
// the new `stringsArray[i]` on the next pass
} else {
// Didn't remove this entry, move past it
++i;
}
}
Side note: typeof isn't a function, it's an operator, there's no need to put its operand in ():if (typeof stringsArray[i] === 'string'){

JavaScript: How to match out-of-order arrays

I'm trying to work out how to match arrays that share the same elements, but not necessarily in the same order.
For example, these two arrays share the same set of elements, even though they're in a different order.
Is there any way to determine whether two arrays contain the same elements?
var search1 = ["barry", "beth", "debbie"];
var search2 = ["beth", "barry", "debbie"];
if (search1 == search2) {
document.write("We've found a match!");
} else {
document.write("Nothing matches");
}
I've got a Codepen of this running at the moment over here: http://codepen.io/realph/pen/grblI

The problem with some of the other solutions is that they are of O(n²) complexity, if they're using a for loop inside of a for loop. That's slow! You don't need to sort either—also slow.
We can speed this up to O(2n) complexity1 by using a simple dictionary. This adds O(2n) storage, but that hardly matters.
JavaScript
var isEqual = function (arr1, arr2) {
if (arr1.length !== arr2.length) {
return false; // no point in wasting time if they are of different lengths
} else {
var holder = {}, i = 0, l = arr2.length;
// holder is our dictionary
arr1.forEach(function (d) {
holder[d] = true; // put each item in arr1 into the dictionary
})
for (; i < l; i++) { // run through the second array
if (!(arr2[i] in holder)) return false;
// if it's not in the dictionary, return false
}
return true; // otherwise, return true
}
}
Test Case
var arr1 = ["barry", "beth", "debbie"],
arr2 = ["beth", "barry", "debbie"];
console.log(isEqual(arr1,arr2));
// returns true
fiddle
Improvement
As Ahruss pointed out, the above function will return true for two arrays that are seemingly equal. For example, [1,1,2,3] and [1,2,2,3] would return true. To overcome this, simply use a counter in the dictionary. This works because !undefined and !0 both return true.
var isReallyEqual = function (arr1, arr2) {
if (arr1.length !== arr2.length) {
return false; // no point in wasting time if they are of different lengths
} else {
var holder = {}, i = 0, l = arr2.length;
// holder is our dictionary
arr1.forEach(function (d) {
holder[d] = (holder[d] || 0) + 1;
// checks whether holder[d] is in the dictionary: holder[d] || 0
// this basically forces a cast to 0 if holder[d] === undefined
// then increments the value
})
for (; i < l; i++) { // run through the second array
if (!holder[arr2[i]]) { // if it's not "in" the dictionary
return false; // return false
// this works because holder[arr2[i]] can be either
// undefined or 0 (or a number > 0)
// if it's not there at all, this will correctly return false
// if it's 0 and there should be another one
// (first array has the element twice, second array has it once)
// it will also return false
} else {
holder[arr2[i]] -= 1; // otherwise decrement the counter
}
}
return true;
// all good, so return true
}
}
Test Case
var arr1 = [1, 1, 2],
arr2 = [1, 2, 2];
isEqual(arr1, arr2); // returns true
isReallyEqual(arr1, arr2); // returns false;
1: It's really O(n+m) complexity, whereby n is the size of the first array and m of the second array. However, in theory, m === n, if the arrays are equal, or the difference is nominal as n -> ∞, so it can be said to be of O(2n) complexity. If you're feeling really pedantic, you can say it's of O(n), or linear, complexity.

you can use this function to compare two arrays
function getMatch(a, b) {
for ( var i = 0; i < a.length; i++ ) {
for ( var e = 0; e < b.length; e++ ) {
if ( a[i] === b[e] ){
return true;
}
}
}
}

Feed your arrays to the following function:
function isArrayEqual(firstArray, secondArray) {
if (firstArray === secondArray) return true;
if (firstArray == null || secondArray == null) return false;
if (firstArray.length != secondArray.length) return false;
// optional - sort the arrays
// firstArray.sort();
// secondArray.sort();
for (var i = 0; i < firstArray.length; ++i) {
if (firstArray[i] !== secondArray[i]) return false;
}
return true;
}
Now you may be thinking, can't I just say arrayOne.sort() and arrayTwo.sort() then compare if arrayOne == arrayTwo? The answer is no you can't in your case. While their contents may be the same, they're not the same object (comparison by reference).

You need to simply sort them, then compare them
function compareArrayItems(array1, array2){
array1 = array1.sort();
array2 = array2.sort();
return array1.equals(array2);
}
fiddle
You can use the equals function provided in How to compare arrays in JavaScript?

Sort them firstly. Secondly, if their length is different, then they're not a match.
After that, iterate one array and test a[i] with b[i], a being the first array, b the second.
var search1 = ["barry", "beth", "debbie"],
search2 = ["beth", "barry", "debbie"];
// If length are different, than we have no match.
if ((search1.length != search2.length) || (search1 == null || search2 == null))
document.write("Nothing matches");
var a = search1.sort(),
b = search2.sort(),
areEqual = true;
for (var i = 0; i < a.length; i++) {
// if any two values from the two arrays are different, than we have no match.
if (a[i] != b[i]) {
areEqual = false;
break; // no need to continue
}
}
document.write(areEqual ? "We've found a match!" : "Nothing matches");