Related
Here is a function for removing duplicates from an array.
function dedupe(arr) {
var seen = {};
arr.forEach((e,i)=>{
if (seen[e]) {
arr.splice(i, 1);
}
seen[e] = true;
});
return arr;
}
console.log(dedupe([1, 2, 1, 3, 4]));
I am interested in the time complexity of this function.
If we assume that Array is backed by a real array, does that the time complexity can be analysed as follows?
allocation of seen: O(1)
enumerate all elements: O(n)
removal of a duplicate: O(n) (because re-allocation required item by item?)
return O(1)
So is this an O(n^2) algorithm?
Edit:
Corrected for indexing issue.
function dedupe(arr) {
var seen = {};
for(let i = 0; i < arr.length; i++) {
const e = arr[i];
if (seen[e]) {
arr.splice(i, 1);
i--; // we have modified the array and need to continue from the current index
}
seen[e] = true;
}
return arr;
}
console.log(dedupe([1, 2, 1, 3, 1, 4, 4, 7, 6, 7, 7, 7, 1, 5]));
For those upset by the performance of the above, this is O(N) I think.
I wanted to de-dupe in-place. Use of Set maintains the order across host environments.
function dedupe(arr) {
var seen = new Set();
for(let i = 0; i < arr.length; i++) {
seen.add(arr[i]);
}
arr.length = 0; // empty the array
return arr.concat(...seen.keys());
}
console.log(dedupe([1, 2, 1, 3, 1, 4, 4, 7, 6, 7, 7, 7, 1, 5]));
One approach would be to use the Javascript Set. You could simply do this:
const removeDuplicates = array => (new Set(array)).values()
This will return an iterator, and not an array, however this can easily be fixed. Also, sets are not yet supported in most browsers. The complexity of this should be O(n).
Another approach more similar to yours (but probably identical to the Set, since I'm gonna guess it's implemented using the same underlying structure) would be like this:
const removeDuplicates = array =>
Object.keys(array.reduce((agg, x) => { agg[x] = true; return agg }, {}))
The time complexity of this should be O(m+n) where m will be the number of unique items, which will always be <= n, therefore O(n).
Also, the time complexity you worked out seems correct.
You could save seen by filtering by index:
var t1 = [1, 2, 1, 1, 3, 1, 1, 4];
function uniqueList(list) {
return list.filter(function (value, index, arr) {
return list.indexOf(value) == index;
});
}
console.log(t1);
console.log(uniqueList(t1));
My answer, builds a new array. Maybe is O(n).
function dedupe(arr) {
var result = [];
var seen = {};
for(let i = 0; i < arr.length; i++) {
const e = arr[i];
if (seen[e]) {
//skip
} else {
seen[e] = true;
result.push(e);
}
}
return result;
}
console.log(dedupe([1, 2, 1, 3, 1, 4, 4, 7, 6, 7, 7, 7, 1, 5]));
I have to find first duplicate value in array and then return its index in variable firstIndex. This has to be done with for loop which should stop after having found first duplicate. I know this is probably pretty simple but I got stuck. So far I have this but it doesn't seem to be working:
var numbers4 = [5, 2, 3, 4, 2, 6, 7, 1, 2, 3];
var firstIndex = "";
for (var a = 0; a < numbers4.length; a++) {
for (var b = a+1; b < numbers4.length; b++) {
if (numbers4[a] === numbers4[b])
firstIndex = numbers4.indexOf(numbers4[a]);
break;
}
}
console.log(firstIndex);
Console prints out 1 which is fine because 2 is first duplicate, but when I change numbers in array, the loop doesn't work. Can you advise what can be changed here?
Thanks in advance!
If I correctly understand your question, that's should help you...
Basically, you need for a double iteration.
const firstDupeIndex = list => list.findIndex(
(item, index) => list.lastIndexOf(item) !== index
);
console.log(
"First Dupe at index:",
firstDupeIndex([5, 2, 3, 4, 4, 6, 7, 1, 2, 3])
);
Thee above implementation comes with the drawback of being O(n2), due to nesting the lastIndexOf within the findIndex function.
A better solution would be to index your occurrences by building a dictionary, therefore keeping time complexity to just O(n) in the worst case. Probably a little bit less neat, but surely more efficient with big inputs.
const firstDupeIndex = (list) => {
const dict = {};
for (const [index, value] of list.entries()) {
if (dict.hasOwnProperty(value)) {
return dict[value];
}
dict[value] = index;
}
return -1;
};
console.log(
"First Dupe at index:",
firstDupeIndex(['a', 'b', 'c', 'd', 'e', 'b', 'z', 't', 'c'])
);
Change your code with the following
var numbers4 = [5, 2, 3, 4, 2, 6, 7, 1, 2, 3];
var firstIndex = "";
var isMatch=false;
for (var a = 0; a < numbers4.length; a++) {
for (var b = a+1; b < numbers4.length; b++) {
if (numbers4[a] === numbers4[b]){
firstIndex = numbers4.indexOf(numbers4[a]);
isMatch=true;
break;
}
}
if (isMatch) {break;}
}
console.log(firstIndex);
I would use an object remembering the values already found... Something like that should work ;)
var numbers4 = [5, 2, 3, 4, 4, 6, 7, 1, 2, 3];
function findFirstDuplicateIndex(arr){
var found = {};
for (var a = 0, aa = arr.length; a < aa ; a++) {
if (found[arr[a]])
return found[arr[a]];
found[numbers4[a]] = a
}
}
console.log(findFirstDuplicateIndex(numbers4));
It's quite fast because you just loop one time through the array. The rest of the time you just access an object property or you set the object property... Let me know if you have questions ;)
However maybe there something faster... It's just an idea ^^
PS: It also works with words, not just numbers
Your break; terminates b loop, because if() is not using parenthesis {}.
Additionally, firstIndex should be simply set to either a or b depending on whether you need to return the duplicate's first occurance or first repetition.
It should be:
if (numbers4[a] === numbers4[b])
{
firstIndex = a;
break;
}
Your problem is - you have two loops and only one break, you need to break out of both.
Why not simply return the index as soon as values match?
var numbers4 = [5, 2, 3, 4, 2, 6, 7, 1, 2, 3];
function getDuplicate(numbers4)
{
for (var a = 0; a < numbers4.length; a++) {
for (var b = a+1; b < numbers4.length; b++) {
if (numbers4[a] === numbers4[b]){
return a;
}
}
}
}
console.log(getDuplicate(numbers4 ));
However, you can optimize your code further by keeping a map
function getDuplicate( numbers )
{
var map = {};
for (var a = 0; a < numbers.length; a++)
{
if( map[ numbers[ a ] ] )
{
return a;
}
map[ numbers[ a ] ] = true;
}
return -1;
}
You can check if indexOf() is not equal to lastIndexOf() and return value and break loop
var numbers4 = [5, 2, 3, 4, 2, 6, 7, 1, 2, 3];
var firstIndex = "";
for (var i = 0; i < numbers4.length; i++) {
if (numbers4.indexOf(numbers4[i]) != numbers4.lastIndexOf(numbers4[i])) {
firstIndex = i;
break;
}
}
console.log(firstIndex)
You don't even need nested for loops.
var numbers4 = [5, 2, 3, 4, 2, 6, 7, 1, 2, 3];
var firstIndex = "";
for (var a = 1; a < numbers4.length; a++) { //start from second elem since first is never a duplicate
if (numbers4.lastIndexOf(numbers4[a])> a) {
firstIndex = a;
break;
}
}
console.log(firstIndex); //1
All you have to do during iterating is to check if current value exists somewhere further in array. That is done by checking last index of this value's occurrence using lastIndexOf().
var numbers = [5, 2, 3, 4, 2, 6, 7, 1, 2, 3];
var firstIndex;
var len = numbers.length;
for (var i = 0; i < len; i++) {
var tmpArray = numbers.slice(i + 1, len);
var index = tmpArray.indexOf(numbers[i]);
if (index !== -1) {
firstIndex = index + i + 1;
break;
}
}
console.log(firstIndex);
Update:
Actually your logic is right, but you missed braces for if condition and also if the condition satisfies then it means firstIndex is the same as a
This is your code with braces,
var numbers4 = [5, 2, 3, 4, 2, 6, 7, 1, 2, 3];
var firstIndex = "";
for (var a = 0; a < numbers4.length; a++) {
for (var b = a + 1; b < numbers4.length; b++) {
if (numbers4[a] === numbers4[b]) {
firstIndex = a
break;
}
}
}
console.log(firstIndex);
The question states the first dupe in the array has to be found along with it's index. So I return an object where the i property is the index and the e property is the first duplicate element itself. One way of performing this task would be
var numbers4 = [5, 2, 3, 4, 2, 6, 7, 1, 2, 3],
headDupe = (a,r={}) => (r.e = a.find((n,i) => (r.i = a.indexOf(n), r.i < i)),r);
console.log(headDupe(numbers4));
If I have the following array:
[0, 1, 3, 0, 4, 2]
And I'd like to sort it ascending order, barring zeros which I need on the end:
[1, 2, 3, 4, 0, 0]
Bear in mind I don't have access to underscore or linq.js for this solution.
My current solution works, but feels quite heavy, long, and not very elegant. Here's my code:
function sortNumbers(numbers) {
var zeroNumbers = [];
var notZeroNumbers = [];
for (var i = 0; i < numbers.length; i++) {
if (numbers[i] === 0) {
zeroNumbers.push(numbers[i]);
} else {
notZeroNumbers.push(numbers[i]);
}
}
var sortedNumbers = notZeroNumbers.sort(function (a, b) {
return parseFloat(a) - parseFloat(b);
});
for (var x = 0; x < zeroNumbers.length; x++) {
sortedNumbers.push(zeroNumbers[x]);
}
return sortedNumbers;
}
Can I improve on this solution?
This is not related to this question, but I was searching for "pure sort javascript" and this is the first answer.
Because sort mutates the original array, the best practice when sorting an array is to clone it first.
const sortedArray = [...array].sort(/* optional comparison function*/)
simply try
var output = [0, 1, 3, 0, 4, 2].sort(function(a, b) {
a = a || Number.MAX_SAFE_INTEGER; //if a == 0 then it will be a falsey value and a will be assigned Number.MAX_SAFE_INTEGER
b = b || Number.MAX_SAFE_INTEGER;
return a - b;
});
console.log(output)
var arr = [0, 1, 3, 0, 4, 2, 9, 8, 7, 0];
arr.sort(function (left, right) {
return left == right ? 0 : (left === 0 ? 1 : (left < right ? -1 : 1));
});
console.log(arr)
This will always put zeroes at the end regardless of the size of the number.
Another alternative solution using Array.sort, Array.splice and Array.push functions:
var arr = [0, 1, 3, 0, 4, 2];
arr.sort();
while(arr[0] === 0) { arr.splice(0,1); arr.push(0); }
console.log(arr); // [1, 2, 3, 4, 0, 0]
You can use sort for this, which takes a closure/callback.
var sortedArray = [0, 1, 3, 0, 4, 2].sort(function(currentValue, nextValue) {
if(currentValue === 0) {
return 1;
} else {
return currentValue - nextValue;
}
});
I am needing to find the correct way to have javascript loop through an array, find all numbers that are divisible by 3, and push those numbers into a new array.
Here is what I have so far..
var array = [],
threes = [];
function loveTheThrees(array) {
for (i = 0, len = array.length; i < len; i++) {
threes = array.push(i % 3);
}
return threes;
}
So if we pass through an array of [1, 2, 3, 4, 5, 6] through the function, it would push out the numbers 3 and 6 into the "threes" array. Hopefully this makes sense.
You can use Array#filter for this task.
filter() calls a provided callback function once for each element in an array, and constructs a new array of all the values for which callback returns a true value or a value that coerces to true. callback is invoked only for indexes of the array which have assigned values; it is not invoked for indexes which have been deleted or which have never been assigned values. Array elements which do not pass the callback test are simply skipped, and are not included in the new array.
function loveTheThrees(array) {
return array.filter(function (a) {
return !(a % 3);
});
}
document.write('<pre>' + JSON.stringify(loveTheThrees([1, 2, 3, 4, 5, 6, 7, 8, 9, 10]), 0, 4) + '</pre>');
console.log([1, 2, 3, 4, 5, 6, 7].filter(function(a){return a%3===0;}));
Array.filter() iterates over array and move current object to another array if callback returns true. In this case I have written a callback which returns true if it is divisible by three so only those items will be added to different array
var array = [],
three = [];
function loveTheThrees(array) {
for (i = 0, len = array.length; i < len; i++) {
if(array[i] % 3 == 0){
three.push(array[i]);
}
}
return three;
}
Using Filter like suggested by Nina is defiantly the better way to do this. However Im assuming you are a beginner and may not understand callbacks yet, In this case this function will work:
function loveTheThrees(collection){
var newArray = []
for (var i =0; i< collection.length;i++){
if (myArray[i] % 3 === 0){
newArray.push(collection[i])
}
}
return newArray;
}
loveTheThrees=(arr)=>arr.filter(el=>Boolean(parseFloat(el)) && isFinite(el) && !Boolean(el%3))
es6 version + skipping non numbers
loveTheThrees([null,undefined,'haha',100,3,6])
Result: [3,6]
Check if the number is divisible by 3 if so then add it to array. Try this
function loveTheThrees(array) {
for (i = 0, len = array.length; i < len; i++) {
if(array[i] % 3 == 0){
three.push(array[I]);
}
}
var originalArray = [1, 2, 3, 4, 5, 6, 7, 8, 9];
function loveTheThrees(array1) {
var threes = [];
for (var i = 0; i < array1.length; i++) {
if (array1[i] % 3 === 0) {
threes.push(array1[i]);
}
}
return threes;
}
loveTheThrees(originalArray);
In ES6:
const arr = [1, 33, 54, 30, 11, 203, 323, 100, 9];
// This single line function allow you to do it:
const isDivisibleBy3 = arr => arr.filter(val => val % 3 == 0);
console.log(isDivisibleBy3(arr));
// The console output is [ 33, 54, 30, 9 ]
I am trying to solve this challenge Seek and Destroy. I can't figure out what is wrong. Any help ?
Seek and Destroy
You will be provided with an initial array (the first argument in the destroyer function), followed by one or more arguments. Remove all elements from the initial array that are of the same value as these arguments.
This is the initial code below:
function destroyer(arr) {
// Remove all the values
return arr;
}
destroyer([1, 2, 3, 1, 2, 3], 2, 3);
This is my Code below:
function destroyer(arr) {
var letsDestroyThis = [];
var i =1 ; while (i<arguments.length) {
letsDestroyThis.push(arguments[i]);
i++;
}
for(var j=0 ; j< arguments[0].length; j++) {
for (var k= 0; k< letsDestroyThis.length; k++) {
if(arguments[0][j] === letsDestroyThis[k]){
arguments[0].splice(j, 1);
}
}
}
return arguments[0];
}
destroyer([2, 3, 2, 3], 2, 3);
Thanks in Advance!
You can create an array of all values that are supposed to be removed. Then use Array.filter to filter out these values.
Note: Array.splice will change original array.
function destroyer() {
var arr = arguments[0];
var params = [];
// Create array of all elements to be removed
for (var k = 1; k < arguments.length; k++)
params.push(arguments[k]);
// return all not matching values
return arr.filter(function(item) {
return params.indexOf(item) < 0;
});
}
console.log(destroyer([1, 2, 3, 1, 2, 3], 2, 3));
function destroyer(arr) {
/* Put all arguments in an array using spread operator and remove elements
starting from 1 using slice intead of splice so as not to mutate the initial array */
const args = [...arguments].slice(1);
/* Check whether arguments include elements from an array and return all that
do not include(false) */
return arr.filter(el => !args.includes(el));
}
console.log(destroyer([1, 2, 3, 1, 2, 3], 2, 3));
This worked for me:
function destroyer(arr) {
// Remove all the values
var args = Array.from(arguments);
var filter = [];
for (i = 0; i < args[0].length; i++) {
for (j = 1; j < args.length; j++) {
if (args[0][i] === args[j]) {
delete args[0][i];
}
}
}
return args[0].filter(function(x) {
return Boolean(x);
});
}
console.log(
destroyer([1, 2, 3, 1, 2, 3], 2, 3)
);
//two ways of resolving the Seek and Destroy challenge on the FreeCodeCamp
//I was trying to simplify this code, please post your solutions, simplifying the code
//as much has its possible
function destroyer1 (arr){
//get array from arguments
var args = Array.prototype.slice.call(arguments);
args.splice(0,1);
for (var i = 0; i < arr.length; i++){
for(var j = 0; j < args.length; j++){
if(arr[i]===args[j]){
delete arr[i];
}
}
}
return arr.filter(function(value){
return Boolean(value);
});
}
//--------------------------------------
function destroyer(arr) {
// Remove all the values
//Get values from arguments of the function to an array, index 0(arr[0] will be "arr",
//rest of indexes will be rest of arguments.
var args = Array.from(arguments);
for (var i = 0 ; i < args[0].length; i++){
for (var j = 1; j < args.length; j++){
if(args[0][i] === args[j]){
delete args[0][i];
}
}
}
return args[0].filter(function(value){
return Boolean(value);
});
}
console.log(destroyer([1, 2, 3, 1, 2, 3], 2, 3));
console.log(destroyer1([1,6,3,9,8,1,1], 3,1));
This is my Code:
function destroyer(arr) {
var argsBeRemove = [...arguments];
argsBeRemove.shift();
return arr.filter(val => {
return argsBeRemove.indexOf(val) == -1;
});
}
console.log(destroyer([1, 2, 3, 1, 2, 3], 2, 3));
Here is my version of Seek and Destroy. I assume that there are no zero elements in input (that assumption allows to pass the challenge). But that way I can make found elements equal zero and then just filter them out. It is pretty straight forward and no index mess when deleting elements in for loops.
function destroyer(arr) {
// Remove all the values
var args = Array.prototype.slice.call(arguments);
var temp = [];
temp = arguments[0].slice();
for (j = 1; j < args.length; j++) {
for (i = 0; i < arguments[0].length; i++) {
if (arguments[0][i] == arguments[j]) {
temp[i] = 0;
}
}
}
function isZero(value) {
return value !== 0;
}
var filtered = temp.filter(isZero);
return filtered;
}
console.log(destroyer([1, 2, 3, 1, 2, 3], 2, 3));
function destroyer(arr) {
var args = Array.prototype.slice.call(arguments, 1);
return arr.filter(destroyNum);
function destroyNum(element) {
return !args.includes(element);
}
}
console.log(destroyer([1, 2, 3, 1, 2, 3], 2, 3));
Give this a shot, it is way less convoluted:
function destroyer(arr) {
// arr1 is equal to the array inside arr
var arr1 = arr.slice(arguments);
// arr2 becomes an array with the arguments 2 & 3
var arr2 = Array.prototype.slice.call(arguments, 1);
// this function compares the two and returns an array with elements not equal to the arguments
return arr1.concat(arr2).filter(function(item) {
return !arr1.includes(item) || !arr2.includes(item)
})
}
destroyer([1, 2, 3, 1, 2, 3], 2, 3);
You will be provided with an initial array (the first argument in the destroyer function), followed by one or more arguments. Remove all elements from the initial array that are of the same value as these arguments.
The accepted solution returns a new array, instead of removing the elements from the existing array.
This can be achieved efficiently by iterating the array in reverse and removing any elements matching any of the filter arguments.
function destroy(original, ...matches) {
if('length' in original) {
let index = original.length
while(--index > -1) {
if(matches.includes(original[index])) {
original.splice(index, 1)
}
}
}
return original;
}
const original = [1, 2, 3, 1, 2, 3]
destroy(original, 2, 3)
console.log(original);
My answer is similar to previous one, but I didn't use indexOf. Instead of that I checked the values in cycle, but compiler gives me a warning to not to declare function in cycle.
function destroyer(arr) {
// Remove all the values
var temp = [];
for (var i = 1; i < arguments.length; i++) {
temp.push(arguments[i]);
arr = arguments[0].filter(function(value) {
return ( value !== temp[i - 1]) ;
});
}
return arr;
}
destroyer([1, 2, 3, 1, 2, 3], 2, 3);
We can get the arguments behind the array, which are the number required to be removed and store them to a list. Then, we can just use a filter to filter out the numbers that needed to be removed.
function destroyer(arr) {
let removeList=[...arguments].slice(1);
return arr.filter(e=>removeList.indexOf(e)===-1);
}
I know isn't the shortest way to do it, but i think is the more simple to understand in an intermediate level.
function destroyer(arr, ...elements) {
var i =0;
while(i<=arr.length){ //for each element into array
for(let j =0; j<elements.length; j++){ //foreach "elements"
if(arr[i]==elements[j]){ // Compare element arr==element
arr.splice(i,1); //If are equal delete from arr
i--; //Check again the same position
j=0;
break; //Stop for loop
}
}
i++;
}
return arr;
}
console.log(destroyer(["possum", "trollo", 12, "safari", "hotdog", 92, 65, "grandma", "bugati", "trojan", "yacht"], "yacht", "possum", "trollo", "safari", "hotdog", "grandma", "bugati", "trojan"));
console.log(destroyer([1, 2, 3, 5, 1, 2, 3], 2, 3));
function destroyer(arr) {
let newArray = Array.from(arguments).splice(1)
return arr.filter(item => !(newArray.includes(item)))
}