Right now I have for loop that is returning an array with the numbers 1 - 90, but am having trouble replacing the numbers 1 and 90 with the strings stored in my variables far and close (respectively).
const far = "01/01/20"
const close = "03/01/20"
const pushNinetyDays = () => {
let arr = [];
for (i = 1; i <= 90 ; i++) {
arr.push(i)
}
return arr
}
// => [1, 2, 3, 4,...89, 90]
I tried something along these lines for awhile, but wasn't getting the correct result.
const pushNinetyDays = () => {
let arr = [];
for (i = 1; i <= 90 ; i++) {
if (arr[i] === 1) {
arr.push(far)
} else if (arr[i] === 90) {
arr.push(close)
} else {
arr.push(i)
}
}
return arr
}
// still returning this array => [ 1, 2, 3, 4,...89,90]
// when i'm looking for this array => ["01,01,20", 2, 3, 4...89, "03/01/20"]
Can anyone help me out?
const pushNinetyDays = () => {
let arr = [];
for (i = 1; i <= 90 ; i++) {
if (i == 1) {
arr.push(far)
} else if (i == 90) {
arr.push(close)
} else {
arr.push(i)
}
}
return arr
}
All you had to do is check if i ==1 or 90 instead of what was in the array which was nothing. A better way would have been using the array and just setting arr[0] and arr[89].
So the question reads:
Given an array a that contains only numbers in the range from 1 to a.length, find the first duplicate number for which the second occurrence has the minimal index. In other words, if there are more than 1 duplicated numbers, return the number for which the second occurrence has a smaller index than the second occurrence of the other number does. If there are no such elements, return -1.
Write a solution with O(n) time complexity and O(1) additional space complexity.
I have a solution, but apparently it's not fast enough and stalls when there are over a thousand items in the array.
This is what I have:
function firstDuplicate(arr) {
let dictionary = {};
for(let i = 0, ii = arr.length; i < ii; i++) {
for(let z = i+1, zz = arr.length; z < zz; z++) {
if(arr[i] === arr[z]) {
if(dictionary.hasOwnProperty(arr[i])) {
if(dictionary[arr[i]] !== 0 && dictionary[arr[i]] > z) {
dictionary[i] = z;
}
} else {
dictionary[arr[i]] = z;
}
}
}
}
let answer = [];
for(key in dictionary) {
// [array number, position];
answer.push([key, dictionary[key]]);
};
if(answer.length > 0) {
return Number(answer.sort((a, b) => {
return a[1]-b[1];
})[0][0]);
}
return -1;
}
I think converting the object into an array and then sorting the array after the answers are complete slows down the whole function. Using built in JS methods like forEach, map and sort (like I did above), slows the code/function down even more. There is obviously a better and more accurate way to do this, so I'm asking for some JS masterminds to help me out on this.
you can keep adding numbers to a dictionary as keys with values as their index, and as soon as you find a repeating key in the dictionary return its value. This will be O(n) time complexity and O(n) space complexity.
function firstDuplicate(arr) {
var dictionary = {};
for(var i = 0; i < arr.length; i++) {
if(dictionary[arr[i]] !== undefined)
return arr[i];
else
dictionary[arr[i]] = i;
}
return -1;
}
console.log(firstDuplicate([2, 3, 3, 1, 5, 2]));
Since the numbers are between 1 to arr.length you can iterate on the array. For each arr[i] use arr[i] as index and make the element present and arr[arr[i]] negative, then the first arr[arr[i]] negative return arr[i]. This give O(1) space complexity and O(n) time complexity you can do this:
function firstDuplicate(arr) {
for(var i = 0; i < arr.length; i++) {
if(arr[Math.abs(arr[i])] < 0)
return Math.abs(arr[i]);
else
arr[Math.abs(arr[i])] = 0 - arr[Math.abs(arr[i])];
}
return -1;
}
console.log(firstDuplicate([2, 3, 3, 1, 5, 2]));
function firstDuplicate(arr) {
for(var i = 0; i < arr.length; i++) {
var num = Math.abs(arr[i]);
if(arr[num] < 0)
return num;
arr[num] = - arr[num];
}
return -1;
}
console.log(firstDuplicate([2,2,3,1,2]));
function firstDuplicate(arr) {
var numMap = {};
for (var i = 0; i < arr.length; i++) {
if (numMap[arr[i]]) {
return arr[i];
}
numMap[arr[i]] = true;
}
return -1;
}
Answer mentioned by #dij is great, but will fail for [2,2] or [2,3,3],
a little change
for input [2,2], i=0 we get a[ Math.abs[a[0] ] = a[2] = undefined
so we remove 1 from a[ Math.abs[a[0] -1 ] this will work now
function firstDuplicate(arr) {
for(var i = 0; i < arr.length; i++) {
if(arr[Math.abs(arr[i])-1] < 0)
return Math.abs(arr[i]);
else
arr[Math.abs(arr[i])-1] = 0 - arr[Math.abs(arr[i])-1];
}
return -1;
}
Please try the code below:
function getCountOccurence(A) {
let result = [];
A.forEach(elem => {
if (result.length > 0) {
let isOccure = false;
result.some((element) => {
if (element.element == elem) {
element.count++;
isOccure = true;
}
});
if (!isOccure) {
result.push({element: elem, count: 1});
}
} else {
result.push({element: elem, count: 1});
}
});
return result;
}
function getFirstRepeatingElement(A) {
let array = getCountOccurence(A);
array.some((element)=> {
if (element.count > 1) {
result = element.element;
return true;
}
});
return result;
}
I know it's a stupid question, but I only learning programming 3 months now.
How would you solve this problem, if you can't use higher order functions and built-in methods, like filter or indexOf?
Create a function that takes a list of numbers and returns a new list where all the duplicate values are removed
I got this so far, but I think It's a dead end...
const array = [1, 2, 3, 3, 1];
const removeDuplicate = () => {
let shortArray = [];
let index = 0;
for (let i = 0; i < array.length; i++) {
for (let j = 0; j < array.length; j++) {
if (i != j) {
if (array[i] == array[j]) {
shortArray[index] += array[i]
console.log(array[i]);
}
}
}
}
return shortArray;
}
console.log(removeDuplicate());
return this:
1
3
3
1
[ NaN ]
thanks!
Use an object as a helper. If a value appears in the helper, it's not unique and can be ignored. If it's not in the helper it's unique, push it into the result array, and add it to the helper object.
const array = [1, 2, 3, 3, 1];
const removeDuplicate = (arr) => {
const helperMap = {};
const result = [];
for (let i = 0; i < arr.length; i++) {
const item = arr[i];
if (!helperMap[item]) {
result[result.length] = item;
helperMap[item] = true;
}
}
return result;
};
console.log(removeDuplicate(array));
function unique(arr) {
var obj = {};
for (var i = 0; i < arr.length; i++) {
var value = arr[i];
obj[value] = true; // set as key
}
return Object.keys(obj); //return all keys
}
Use below function:
function RemoveDuplicate(array){
let shortArray = [];
let index = 0;
for (let i = 0; i < array.length; i++) {
let exist=false;
for(let j=0;j<shortArray.length;j++){
if(array[i]==shortArray[j]){
exist=true;
break;
}
}
if(!exist){
shortArray[shortArray.length]=array[i];
}
}
return shortArray;
}
This question already has answers here:
In Javascript, how do I check if an array has duplicate values?
(9 answers)
Closed 10 months ago.
I wanted to write a javascript function which checks if array contains duplicate values or not.
I have written the following code but its giving answer as "true" always.
Can anybody please tell me what am I missing.
function checkIfArrayIsUnique(myArray)
{
for (var i = 0; i < myArray.length; i++)
{
for (var j = 0; j < myArray.length; j++)
{
if (i != j)
{
if (myArray[i] == myArray[j])
{
return true; // means there are duplicate values
}
}
}
}
return false; // means there are no duplicate values.
}
An easy solution, if you've got ES6, uses Set:
function checkIfArrayIsUnique(myArray) {
return myArray.length === new Set(myArray).size;
}
let uniqueArray = [1, 2, 3, 4, 5];
console.log(`${uniqueArray} is unique : ${checkIfArrayIsUnique(uniqueArray)}`);
let nonUniqueArray = [1, 1, 2, 3, 4, 5];
console.log(`${nonUniqueArray} is unique : ${checkIfArrayIsUnique(nonUniqueArray)}`);
let arr = [11,22,11,22];
let hasDuplicate = arr.some((val, i) => arr.indexOf(val) !== i);
// hasDuplicate = true
True -> array has duplicates
False -> uniqe array
This should work with only one loop:
function checkIfArrayIsUnique(arr) {
var map = {}, i, size;
for (i = 0, size = arr.length; i < size; i++){
if (map[arr[i]]){
return false;
}
map[arr[i]] = true;
}
return true;
}
You got the return values the wrong way round:
As soon as you find two values that are equal, you can conclude that the array is not unique and return false.
At the very end, after you've checked all the pairs, you can return true.
If you do this a lot, and the arrays are large, you might want to investigate the possibility of sorting the array and then only comparing adjacent elements. This will have better asymptotic complexity than your current method.
Assuming you're targeting browsers that aren't IE8,
this would work as well:
function checkIfArrayIsUnique(myArray)
{
for (var i = 0; i < myArray.length; i++)
{
if (myArray.indexOf(myArray[i]) !== myArray.lastIndexOf(myArray[i])) {
return false;
}
}
return true; // this means not unique
}
Here's an O(n) solution:
function hasDupes(arr) {
/* temporary object */
var uniqOb = {};
/* create object attribute with name=value in array, this will not keep dupes*/
for (var i in arr)
uniqOb[arr[i]] = "";
/* if object's attributes match array, then no dupes! */
if (arr.length == Object.keys(uniqOb).length)
alert('NO dupes');
else
alert('HAS dupes');
}
var arr = ["1/1/2016", "1/1/2016", "2/1/2016"];
hasDupes(arr);
https://jsfiddle.net/7kkgy1j3/
Another solution:
Array.prototype.checkIfArrayIsUnique = function() {
this.sort();
for ( var i = 1; i < this.length; i++ ){
if(this[i-1] == this[i])
return false;
}
return true;
}
function hasNoDuplicates(arr) {
return arr.every(num => arr.indexOf(num) === arr.lastIndexOf(num));
}
hasNoDuplicates accepts an array and returns true if there are no duplicate values. If there are any duplicates, the function returns false.
Without a for loop, only using Map().
You can also return the duplicates.
(function(a){
let map = new Map();
a.forEach(e => {
if(map.has(e)) {
let count = map.get(e);
console.log(count)
map.set(e, count + 1);
} else {
map.set(e, 1);
}
});
let hasDup = false;
let dups = [];
map.forEach((value, key) => {
if(value > 1) {
hasDup = true;
dups.push(key);
}
});
console.log(dups);
return hasDup;
})([2,4,6,2,1,4]);
Late answer but can be helpful
function areThereDuplicates(args) {
let count = {};
for(let i = 0; i < args.length; i++){
count[args[i]] = 1 + (count[args[i]] || 0);
}
let found = Object.keys(count).filter(function(key) {
return count[key] > 1;
});
return found.length ? true : false;
}
areThereDuplicates([1,2,5]);
The code given in the question can be better written as follows
function checkIfArrayIsUnique(myArray)
{
for (var i = 0; i < myArray.length; i++)
{
for (var j = i+1; j < myArray.length; j++)
{
if (myArray[i] == myArray[j])
{
return true; // means there are duplicate values
}
}
}
return false; // means there are no duplicate values.
}
Returns the duplicate item in array and creates a new array with no duplicates:
var a = ["hello", "hi", "hi", "juice", "juice", "test"];
var b = ["ding", "dong", "hi", "juice", "juice", "test"];
var c = a.concat(b);
var dupClearArr = [];
function dupArray(arr) {
for (i = 0; i < arr.length; i++) {
if (arr.indexOf(arr[i]) != i && arr.indexOf(arr[i]) != -1) {
console.log('duplicate item ' + arr[i]);
} else {
dupClearArr.push(arr[i])
}
}
console.log('actual array \n' + arr + ' \nno duplicate items array \n' + dupClearArr)
}
dupArray(c);
const containsMatches = (a1, a2) => a1.some((v) => a2.includes(v));
If your array nests other arrays/objects, using the Set approach may not be what you want since comparing two objects compares their references. If you want to check that their contained values are equal, something else is needed. Here are a couple different approaches.
Approach 1: Map using JSON.stringify for keys
If you want to consider objects with the same contained values as equal, here's one simple way to do it using a Map object. It uses JSON.stringify to make a unique id for each element in the array.
I believe the runtime of this would be O(n * m) on arrays, assuming JSON.stringify serializes in linear time. n is the length of the outer array, m is size of the arrays. If the objects get very large, however, this may slow down since the keys will be very long. Not a very space-efficient implementation, but it is simple and works for many data types.
function checkArrayDupeFree(myArray, idFunc) {
const dupeMap = new Map();
for (const el of myArray) {
const id = idFunc(el);
if (dupeMap.has(id))
return false;
dupeMap.set(id, el);
}
return true;
}
const notUnique = [ [1, 2], [1, 3], [1, 2] ];
console.log(`${JSON.stringify(notUnique)} has no duplicates? ${checkArrayDupeFree(notUnique, JSON.stringify)}`);
const unique = [ [2, 1], [1, 3], [1, 2] ];
console.log(`${JSON.stringify(unique)} has no duplicates? ${checkArrayDupeFree(unique, JSON.stringify)}`);
Of course, you could also write your own id-generator function, though I'm not sure you can do much better than JSON.stringify.
Approach 2: Custom HashMap, Hashcode, and Equality implementations
If you have a lot of big arrays, it may be better performance-wise to implement your own hash/equality functions and use a Map as a HashMap.
In the following implementation, we hash the array. If there is a collision, map a key to an array of collided values, and check to see if any of the array values match according to the equality function.
The downside of this approach is that you may have to consider a wide range of types for which to make hashcode/equality functions, depending on what's in the array.
function checkArrayDupeFreeWHashes(myArray, hashFunc, eqFunc) {
const hashMap = new Map();
for (const el of myArray) {
const hash = hashFunc(el);
const hit = hashMap.get(hash);
if (hit == null)
hashMap.set(hash, [el]);
else if (hit.some(v => eqFunc(v, el)))
return false;
else
hit.push(el);
}
return true;
}
Here's a demo of the custom HashMap in action. I implemented a hashing function and an equality function for arrays of arrays.
function checkArrayDupeFreeWHashes(myArray, hashFunc, eqFunc) {
const hashMap = new Map();
for (const el of myArray) {
const hash = hashFunc(el);
const hit = hashMap.get(hash);
if (hit == null)
hashMap.set(hash, [el]);
else if (hit.some(v => eqFunc(v, el)))
return false;
else
hit.push(el);
}
return true;
}
function arrayHasher(arr) {
let hash = 19;
for (let i = 0; i < arr.length; i++) {
const el = arr[i];
const toHash = Array.isArray(el)
? arrayHasher(el)
: el * 23;
hash = hash * 31 + toHash;
}
return hash;
}
function arrayEq(a, b) {
if (a.length != b.length)
return false;
for (let i = 0; i < a.length; i++) {
if ((Array.isArray(a) || Array.isArray(b)) && !arrayEq(a[i], b[i]))
return false;
else if (a[i] !== b[i])
return false;
}
return true;
}
const notUnique = [ [1, 2], [1, 3], [1, 2] ];
const unique = [ [2, 1], [1, 3], [1, 2] ];
console.log(`${JSON.stringify(notUnique)} has no duplicates? ${checkArrayDupeFreeWHashes(notUnique, arrayHasher, arrayEq)}`);
console.log(`${JSON.stringify(unique)} has no duplicates? ${checkArrayDupeFreeWHashes(unique, arrayHasher, arrayEq)}`);
function checkIfArrayIsUnique(myArray)
{
isUnique=true
for (var i = 0; i < myArray.length; i++)
{
for (var j = 0; j < myArray.length; j++)
{
if (i != j)
{
if (myArray[i] == myArray[j])
{
isUnique=false
}
}
}
}
return isUnique;
}
This assume that the array is unique at the start.
If find two equals values, then change to false
i think this is the simple way
$(document).ready(function() {
var arr = [1,2,3,9,6,5,6];
console.log( "result =>"+ if_duplicate_value (arr));
});
function if_duplicate_value (arr){
for(i=0;i<arr.length-1;i++){
for(j=i+1;j<arr.length;j++){
if(arr[i]==arr[j]){
return true;
}
}
}
return false;
}
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
var c=[2,2,3,3,5,5,4,4,8,8];
for(var i=0; i<b.length; i++){
for(var j=i+1; j<b.length; j++){
if(c[i]==c[j]){
console.log(c[j]);
}
}
}
I'm looking for something similar to Groovy's every() method, which tests every element of a list if it meets some criteria. If they all meet the criteria, the function returns true. Otherwise, false. I've tried something like this:
var arr = [1, 0, 1, 0, 1, 1];
var allOnes = $.grep(arr, function(ind) {
return this == 1;
}).length == arr.length;
..but its not very clean. I haven't had any luck while searching through the API. Is using grep() the only way to do it?
if it is a plain js array, you have $.grep()
.filter() is for use with jQuery or DOM Elements
Here is a plugin I made that might make it easier:
(function($) {
$.fn.allOnes = function() {
var allVal = true;
this.each(function(ind, item) {
if (item != 1) {
allVal = false;
return allVal;
}
});
return allVal;
};
})(jQuery);
var arr = [1, 1, 0, 1, 1, 1];
console.log($(arr).allOnes());
Fiddle: http://jsfiddle.net/maniator/NctND/
The following plugin is an expansion of the above and lets you search for a specific number: http://jsfiddle.net/maniator/bFNnn/
(function($) {
$.fn.allValue = function(pred) {
var allOnes = true;
this.each(function(ind, item) {
if (item != pred) {
allOnes = false;
return allOnes;
}
});
return allOnes;
};
})(jQuery);
var arr = [1, 1, 1, 1, 1, 1];
console.log($(arr).allValue(1));
here is example of function you can use.
var arr = [1, 0, 1, 0, 1, 1];
var allOnes = arr.check(1);
//this function compares all elements in array and if all meet the criteria it returns true
Array.prototype.chack = function(cond)
{
var ln = 0;
for(i=0; i<this.length; i++)
{
if(bond === this[i])
{
ln++
}
}
if(ln == this.length)
return true;
else
return false;
}
How about just turning your working code into a method on array, to ease its reuse:
Array.prototype.every = function(predicate){
return $.grep(this,predicate).length == this.length;
}
usage:
alert([1,0,1,0].every(function(i) { return i == 1; }));
Working example: http://jsfiddle.net/59J5A/
Edit: changed to grep
You could always implement an allOnes method:
function allOnes(array) {
var result = [];
for(var i = 0; i < array.length; i += 1) {
if (array[i] === 1) { result.push(true); }
}
return array.length == result.length;
}
or you could be a bit more abstract and test for true/false:
function all(array) {
var result = [];
for(var i = 0; i < array.length; i += 1) {
if (array[i]) { result.push(true); }
}
return array.length == result.length;
}
var arr = [1, 0, 1, 0, 1, 1];
var allOnes = all(arr);
or even better, maybe have a changeable predicate:
function all(array, predicate) {
var result = [],
predicate = predicate || function(x) { if (x) { return true; } };
for(var i = 0; i < array.length; i += 1) {
if (predicate(array[i])) { result.push(true); }
}
return array.length == result.length;
}
var allOnes = all(arr, function(x) { return x === 1; })