How can I search for a value in an array and delete just one of those values if 2 or more is found? - javascript

I have an array
var aos = ["a","a","a","b","b","c","d","d"];
I want to know if I can remove just 1 item if it finds 2 or more of the same value in the array. So for instance if it finds
"a", "a"
it will remove one of those "a"
This is my current code:
var intDennis = 1;
for (var i = 0; i < aos.length; i++) {
while (aos[i] == aos[intDennis]) {
aos.splice(i, 1);
intDennis++;
console.log(aos[intDennis], aos[i]);
}
intDennis = 1;
}
NOTE: My array is sorted.


Edited after better understanding of OP use-case.
Updated solution and fiddle test to incorporate suggestion from pst in comments.
(Not for nothing, but this method does not require the original array be sorted.)
Try this...
var elements = [];
var temp = {};
for (i=0; i<aos.length; i++) {
temp[aos[i]] = (temp[aos[i]] || 0) + 1;
}
for (var x in temp) {
elements.push(x);
for (i=0; i<temp[x]-2; i++) {
elements.push(x);
}
}
Fiddle Test


Because you said you have a sorted array, you only need to remove the second time a element is found. You only need one for.
The splice() function returns the removed element so, just use it to not remove more elements of that kind.
This solution is more clean and efficient.
var aos = ["a","a","a","b","b","c","d","d"];
var lastRemoved = "";
for (var i = 1; i < aos.length; i++) {
if (aos[(i-1)] == aos[i] && lastRemoved != aos[i]) {
lastRemoved = aos.splice(i, 1);
}
}
Code tested and working. Result: ["a", "a", "b", "c", "d"]


I don't believe there's any better way to do this on an unsorted array than an approach with O(n^2) behaviour. Given ES5 array-builtins (supported in all modern browsers, though not in IE prior to IE9), the following works:
aos.filter(function(value, index, obj) { return obj.indexOf(value) === index; })


UPDATED ANSWER TO REMOVE ONLY 1 DUPLICATE:
Assuming that each object will resolve to a unique String, here's a potential solution. The first time the object is detected, it sets a counter for that object to one. If it finds that object again, it splices that element out and increments the associated counter. If it finds that element more times, it will leave it alone.
var elements = {};
for (var i = 0; i < aos.length; i++) {
if(elements[aos[i]]){
if(elements[aos[i]] == 1){
aos.splice(i,1);//splice the element out of the array
i--;//Decrement the counter to account for the reduced array
elements[aos[i]]++;//Increment the count for the object
}
} else {
elements[aos[i]] = 1;//Initialize the count for this object to 1;
}
}
Here's the test fiddle for this.


I would not mutate the input -- that is, don't use splice. This will simplify the problem a good deal. Using a new array object here may actually be more efficient. This approach utilizes the fact that the input array is sorted.
Consider: (jsfiddle demo)
var input = ["a","a","a","b","b","c","d","d"]
var result = []
for (var i = 0; i < input.length; i++) {
var elm = input[i]
if (input[i+1] === elm) {
// skip first element (we know next is dup.)
var j = i + 1
for (; input[j] === elm && j < input.length; j++) {
result.push(input[j])
}
i = j - 1
} else {
result.push(elm)
}
}
alert(result) // a,a,b,c,d
Happy coding.
Replace === with a custom equality, as desired. Note that it is the first item is omitted from the output, which may not always be "correct".


REVISED EXAMPLE
function removeDuplicate(arr) {
var i = 1;
while(i < arr.length) {
if(arr[i] == arr[i - 1]) {
arr.splice(i, 1);
}
while(arr[i] == arr[i - 1] && i < arr.length) {
i += 1;
}
i += 1;
}
return arr;
}
alert(removeDuplicate(["a","a","a","b","b","c","d","d"]));

Related

If condition vs loop one item

I was just curious, is it worth to have if condition before looping some array, that in 90% will be array of 1 item?
Code example:
const a = [3];
const aLength = a.length;
if(aLength > 1) {
for(let i = 0; i < aLength; i++) {
func(i);
}
} else {
func();
}
function func(position = 0) {
console.log('hi' + position);
}

I agree with Federico's comment, a single for loop is the most readable in this case.
Also, even though you reuse it, there is not much point to extracting a.length into aLength
const a = [3];
for(let i = 0; i < a.length; i++) {
func(i);
}
function func(position) {
console.log('hi' + position);
}
Warning: very personal perspective down there, you could achieve the same level of clarity with comments too.
Well, unless the single element case has a very specific meaning in your domain. In which case, I would separate them with two functions with very specific names as follows:
const a = [3];
if(a.length > 1) {
handleMultiple(a);
} else {
handleSingleAndWhyItIsASpecialCase(a)
}
handleMultiple(array) {
for(let i = 0; i < array.length; i++) {
func(i);
}
}
handleSingleAndWhyItIsASpecialCase(array) {
func();
}
function func(position = 0) {
console.log('hi' + position);
}
As Hamid said below, you can easily turn it into a oneliner:
[45,63,77].forEach((element, index) => console.log(index));
Consider using forEach instead of map to make your intent clear though.

Write clean code and make everyone happy.
you can eliminate if and loop:
const a=[5,6,3]
a.forEach((value,index)=>console.log('hi'+index));

Codewars - Delete occurrences of an element if it occurs more than n times

I just completed this challenge on Codewars.
I have completed the question with my own method, but can someone explain to me the Best Practice answer. Can somehow explain how this code functions?
function deleteNth(arr,x) {
var cache = {};
return arr.filter(function(n) {
cache[n] = (cache[n]||0) + 1;
return cache[n] <= x;
});
}
I did this:
function deleteNth(arr,n){
var count = 0;
//loop backwards so it removes duplicates from the right
for(let i= arr.length; i > 0; i--){
for(let j=0; j < arr.length; j++){
if (arr[i] == arr[j]){
count += 1
}
}
if(count > n){
arr.splice(i,1);
i = arr.length;
}
count = 0;
}
return arr;
}

This is how the first code works. It is a quite nice solution and quite easy to read if you understand the filter function.
function deleteNth(arr, x) {
// Create an empty object to store how many times each object exists
var cache = {};
// Call the filter function, the delegate is called once for each item
// in the array and you return true or false depending on if it should
// be kept or not
return arr.filter(function(n) {
// Use the item as key and store the number of times the item has appeared.
// (cache[n]||0) fetches the current value of cache[n] or zero if it doesn't
// exist. Then add one to it and store it.
cache[n] =(cache[n]||0) + 1;
// If the number of times it has appeared in the array is less or equal to
// the limit then return true so the filter function keeps it.
return cache[n] <= x;
});
}

Understanding get second lowest and second highest value in array

Good day fellow Stack-ers,
I must ask your pardon if this question has been asked before or if it seems elementary (I am only a Javascript novice).
I have been doing w3c js challenges lately: Write a JavaScript function which will take an array of numbers stored and find the second lowest and second greatest numbers.
Here is my answer:
var array = [3,8,5,6,5,7,1,9];
var outputArray = [];
function arrayTrim() {
var sortedArray = array.sort();
outputArray.push(sortedArray[1],array[array.length-2]);
return outputArray;
}
arrayTrim();
and here is the answer that they have provided:
function Second_Greatest_Lowest(arr_num) {
arr_num.sort(function(x,y) {
return x-y;
});
var uniqa = [arr_num[0]];
var result = [];
for(var j=1; j < arr_num.length; j++) {
if(arr_num[j-1] !== arr_num[j]) {
uniqa.push(arr_num[j]);
}
}
result.push(uniqa[1],uniqa[uniqa.length-2]);
return result.join(',');
}
alert(Second_Greatest_Lowest([1,2,3,4,5]));
I know that the for loop runs through until the length of the input, but I don't understand the if statement nested within the for loop. It seems like a long way around to the solution.
Thank you!

Your answer does not perform correct for input such as f.e. [3,8,5,6,5,7,1,1,9]. Your proposed solution returns 1 as the second lowest number here – whereas it should actually be 3.
The solution suggested by the site takes that into account – that is what the if inside the loop is for, it checks if the current number is the same as the previous one. If that’s the case, it gets ignored. That way, every number will occur once, and that in turn allows to blindly pick the second element out of that sorted array and actually have it be the second lowest number.
It seems like a long way around to the solution
You took a short cut, that does not handle all edge cases correctly ;-)

The loop in question:
for(var j=1; j < arr_num.length; j++) {
if(arr_num[j-1] !== arr_num[j]) {
uniqa.push(arr_num[j]);
}
}
Provides some clue as to what it's doing by using a (reasonably) descriptive variable name: uniqa - or "unique array". The if statement is checking that the current element is not the same as the previous element - having sorted the array initially this works to give you a unique array - by only filling a new array if the element is indeed unique.
Thereafter the logic is the same as yours.

import java.util.Arrays;
public class BubbleWithMax_N_Min
{
public static void main(String[] agrs)
{
int temp;
int[] array = new int[5];
array[0] = 3;
array[1] = 99;
array[2] = 55;
array[3] = 2;
array[4] = 1;
System.out.println("the given array is:" + Arrays.toString(array));
for (int i = 0; i < array.length; i++)
{
System.out.println(array[i] + "");
}
for (int i = 0; i < array.length; i++)
{
for (int j = 1; j < array.length - i; j++)
{
if (array[j - 1] > array[j])
{
temp = array[j - 1];
array[j - 1] = array[j];
array[j] = temp;
}
}
}
System.out.println(" 2nd Min and 2nd Highest:");
for (int i = 0; i < 1; i++)
{
System.out.println(array[i+1]);
}
for (int i = 0; i < 1; i++)
{
int a= array.length-2;
System.out.println(array[a]);
}
}
}

Why is my duplicate removal script not capturing unique values

I'm trying to removal duplicates from a sorted list.
I wrote this script.
for(var i = 0; i < duplicateAuthors.length - 1; i++){
if(duplicateAuthors[i] == duplicateAuthors[i + 1]) { continue; }
else{
uniqAuthors.push(duplicateAuthors[i]);
}
}
It works except it does not capture unique values in the list. What am I doing wrong?

var uniqAuthors = [];
for(var i = 0; i < duplicateAuthors.length; i++){
if(uniqAuthors.indexOf(duplicateAuthors[i]) == -1) {
uniqAuthors.push(duplicateAuthors[i]);
}
}
The above code checks whether an object exist in an array or not, if not add it to array. Hence, at the end we will have an array of unique values.

Since object keys have to be unique I realized that you could also do the following to isolate unique values.
var uniqs = {};
for(var i = 0; i < duplicateAuthors.length ; i++){
uniqs[duplicateAuthors[i]] = null;
}
var uniqAuthors = Object.keys(uniqs);

This will take advantage of the fact that it's already sorted.
var previous;
var current;
for(var i = 0; i < duplicateAuthors.length; i++){
current = duplicateAuthors[i];
if (current && current != previous) {
uniqAuthors.push(current);
previous = current;
}
}

Searching a 2D Javascript Array For Value Index

I am trying to write a jQuery that will find the index of a specific value within a 7x7 2D array.
So if the value I am looking for is 0 then I need the function to search the 2D array and once it finds 0 it stores the index of the two indexes.
This is what I have so far, but it returns "0 0" (the initial values set to the variable.
Here is a jsFiddle and the function I have so far:
http://jsfiddle.net/31pj8ydz/1/
$(document).ready( function() {
var items = [[1,2,3,4,5,6,7],
[1,2,3,4,5,6,7],
[1,2,3,0,5,6,7],
[1,2,3,4,5,6,7],
[1,2,3,4,5,6,7],
[1,2,3,4,5,6,7],
[1,2,3,4,5,6,7]];
var row = 0;
var line = 0;
for (i = 0; i < 7; ++i) {
for (j = 0; i < 7; ++i) {
if (items[i, j] == '0,') {
row = i;
line = j;
}
}
}
$('.text').text(row + ' ' + line);
});
HTML:
<p class="text"></p>

Your if statement is comparing
if (items[i, j] == '0,')
Accessing is wrong, you should use [i][j].
And your array has values:
[1,2,3,4,5,6,7]
....
Your value '0,' is a string, which will not match numeric values inside the array, meaning that your row and line won't change.

First, you are accessing your array wrong. To access a 2D array, you use the format items[i][j].
Second, your array doesn't contain the value '0'. It doesn't contain any strings. So the row and line variables are never changed.
You should change your if statement to look like this:
if(items[i][j] == 0) {
Notice it is searching for the number 0, not the string '0'.

You access your array with the wrong way. Please just try this one:
items[i][j]
When we have a multidimensional array we access the an element of the array, using array[firstDimensionIndex][secondDimensionIndex]...[nthDimensionIndex].
That being said, you should change the condition in your if statement:
if( items[i][j] === 0 )
Please notice that I have removed the , you had after 0. It isn't needed. Also I have removed the ''. We don't need them also.

There are following problems in the code
1) items[i,j] should be items[i][j].
2) You are comparing it with '0,' it should be 0 or '0', if you are not concerned about type.
3) In your inner for loop you should be incrementing j and testing j as exit condition.
Change your for loop like bellow and it will work
for (i = 0; i < 7; i++) {
for (j = 0; j < 7; j++) {
if (items[i][j] == '0') {
row = i;
line = j;
}
}
}
DEMO
Note:-
1) Better to use === at the place of ==, it checks for type also. As you see with 0=='0' gives true.
2) Better to say i < items.length and j<items[i].length instead of hard-coding it as 7.

var foo;
items.forEach(function(arr, i) {
arr.forEach(function(val, j) {
if (!val) { //0 coerces to false
foo = [i, j];
}
}
}
Here foo will be the last instance of 0 in the 2D array.

You are doing loop wrong
On place of
for (i = 0; i < 7; ++i) {
for (j = 0; i < 7; ++i) {
if (items[i, j] == '0,') {
row = i;
line = j;
}
}
}
use this
for (i = 0; i < 7; i++) {
for (j = 0; j < 7; j++) {
if (items[i][j] == 0) {
row = i;
line = j;
}
}
}
Here is the demo

looks like you are still learning how to program. But here is an algorithm I've made. Analyze it and compare to your code ;)
var itens = [[1,2,3,4,5,6,7],
[1,2,3,4,5,6,7],
[1,2,3,0,5,6,7],
[1,2,3,4,5,6,7],
[1,2,3,4,5,6,7],
[1,2,3,4,5,6,7],
[1,2,3,4,5,6,7]];
var row = null;
var collumn = null;
for (var i = 0; i < itens.length; i++) {
for (var j = 0; j < itens[i].length; j++) {
if (itens[i][j] == 0) {
row = i;
collumn = j;
}
}
}
console.log(row, collumn);

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