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I have three arrays and I need to create a set of rules based on these three arrays, but I'm struggling with the logic of how to write a function that will give me every possible combination of every entry in each array. So, I have, for example:
var array 1 = [1, 2];
var array 2 = [3, 4, 5];
var array 4 = [6, 7, 8, 9, 10];
And I'd wan't get back a string, object etc of all possible combinations (which I wont attempt to work out here). So for example:
var result = ["1-3-6", "2-3-6", "1,4,6"];
And so on, so far I've tried sitting down and composing a For Loop but I'm just really not sure where to start. I also looked at maps, but could not find any examples that went this deep, so I wasn't sure if a map would get the job done either.
The actual data I want to load in, the first array has 2 entries, the second have 7 and the last one had 6, so for the workings out I've done there should be 84 entries. That was based on (Array 3 * Array 2) * Array 1.
Hope that all makes sense I know it's a bit confusing. Also worth mentioning that I'm using Angular JS so an angular solution or vanilla JS solution is preferred but not essential.
What you are looking is the Cartesian product of arrays. You can use a function like this (extracted from here):
function cartesian() {
var r = [], arg = arguments, max = arg.length-1;
function helper(arr, i) {
for (var j=0, l=arg[i].length; j<l; j++) {
var a = arr.slice(0); // clone arr
a.push(arg[i][j]);
if (i==max)
r.push(a);
else
helper(a, i+1);
}
}
helper([], 0);
return r;
}
There are lot of examples, like:
JavaScript - Generating combinations from n arrays with m elements
With recursive:
Finding All Combinations of JavaScript array values
Cartesian product of multiple arrays in JavaScript
And with multiple (N) arrays:
Combine 2 arrays into 1 in all possible ways in JavaScript
Hope it helps!
Nested for loops will do
function arrComb(arr1, arr2, arr3) {
var l1 = arr1.length,
l2 = arr2.length,
l3 = arr3.length,
i, j, k, res = [];
for (i = 0; i < l1; i++) {
for (j = 0; j < l2; j++) {
for (k = 0; k < l3; k++) {
res.push(arr1[i] + '-' + arr2[j] + '-' + arr3[k]);
}
}
}
console.log(res)
}
arrComb([1, 2], [3, 4, 5], [6, 7, 8, 9, 10]);
A bit more elegant:
var array_1 = [1, 2];
var array_2 = [3, 4, 5];
var array_4 = [6, 7, 8, 9, 10];
var result = [];
for (var a1 of array_1)
{
for (var a2 of array_2)
{
for (var a3 of array_4)
{
result.push("\""+a1+"-"+a2+"-"+a3+"\"")
}
}
}
alert("["+result+"]")
Per Codefighters:
Note: Write a solution with O(n) time complexity and O(1) additional
space complexity, since this is what you would be asked to do during a
real interview.
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.
Example
For a = [2, 3, 3, 1, 5, 2], the output should be firstDuplicate(a) =
3.
There are 2 duplicates: numbers 2 and 3. The second occurrence of 3
has a smaller index than than second occurrence of 2 does, so the
answer is 3.
For a = [2, 4, 3, 5, 1], the output should be firstDuplicate(a) = -1.
So here is what I came up with. It works but fails on the final test because it ran over 4000ms. I'm at a loss as to what else I can do. Any Ideas to improve speed?
function firstDuplicate(a) {
var test = [],
lowest = undefined;
for (var i=0; i<a.length; i++) {
if (test.indexOf(a[i]) > -1) {
lowest = lowest || i;
if (i < lowest) {
lowest = i;
}
}
else {
test.push(a[i]);
}
}
return lowest ? a[lowest] : -1;
}
Here was my second attempt but still failing on the last test...
function firstDuplicate(a) {
var low = undefined,
last = -1;
for (var i=0; i<a.length; i++) {
last = a.lastIndexOf(a[i])
if (last > i && (low === undefined || last < low)) {
low = last;
}
}
return low !== undefined ? a[low] : -1;
}
The requirements give a clue of how to solve this. The set of numbers contained in the array must match the following critera:
only numbers in the range from 1 to a.length
In other words, only positive numbers that are less than or equal to the length of the array. If the array contains ten numbers, none of them will be greater than 10.
With that insight, we have a means of keeping track of numbers that we have already seen. We can treat the numbers themselves as indexes into the array, modify the element at that index (in this case by making it negative) and if we run into the same number and the element at that index is less than zero, then we know we have seen it.
console.clear()
const test1 = [2, 3, 3, 1, 5, 2]
const test2 = [2, 4, 3, 5, 1]
function firstDuplicate(a) {
for (let i of a) {
let posi = Math.abs(i) - 1
if (a[posi] < 0) return posi + 1
a[posi] = a[posi] * -1
}
return -1
}
console.log(firstDuplicate(test1))
console.log(firstDuplicate(test2))
console.log(firstDuplicate([2,2]))
console.log(firstDuplicate([2,3,3]))
console.log(firstDuplicate([3,3,3]))
Original Incorrect Answer
Keep track of what numbers have already been seen and return the first one that has been seen before.
console.clear()
const test1 = [2, 3, 3, 1, 5, 2]
const test2 = [2, 4, 3, 5, 1]
function firstDuplicate(a){
const seen = {}
for (let v of a){
if (seen[v]) return v
seen[v] = v
}
return -1
}
console.log(firstDuplicate(test1))
console.log(firstDuplicate(test2))
As pointed out in the comments, however, this answer takes O(n) additional space, not O(1) additional space.
We will take advantage of the fact that the array a contains only numbers in the range from 1 to a.length, to remember that a value has been seen by reversing the sign of whatever is in that position in the array.
function lowestDuplicate(arr) {
for (let i = 0; i < arr.length; i++) {
const val = Math.abs(arr[i]);
if (arr[val - 1] < 0) return val;
arr[val - 1] = -arr[val - 1];
}
return -1;
}
console.log(lowestDuplicate([1, 2, 3, 4, 3, 2, 1]));
console.log(lowestDuplicate([1, 2, 3, 4, 5]));
console.log(lowestDuplicate([5, 4, 3, 2, 2]));
console.log(lowestDuplicate([2, 2]));
console.log(lowestDuplicate([2, 3, 3]));
console.log(lowestDuplicate([3, 3, 3]));
console.log(lowestDuplicate([2, 3, 3, 1, 5, 2]));
Python 3 version that passes the tests.
def firstDuplicate(a):
oldies={}
notfound=True
for i in range(len(a)):
try:
if oldies[a[i]]==a[i]:
notfound=False
return a[i]
except:
oldies[a[i]]=a[i]
if notfound:
return -1
You are iterating n times in both examples.
What if the array length was 200,000,000 and the first duplicate was found at index 3? The loop is still running 200,000,000 times unnecessarily.
So the idea is to break out of the loop once you find the first duplicate.
you can use break or just return.
A simple solution in JS. As the question title focus, you just need to find the first duplicate of the number you already traversed. So i guess this will do:
function solution(a) {
let duplicateArray = []
for(let i=0;i<a.length;i++){
if(duplicateArray.includes(a[i])){
return a[i];
}
duplicateArray.push(a[i])
}return -1;}
Hope this helps.
I am having trouble with a problem and how to address it. Mind you, I am relatively new to JavaScript and the problem feels like I am over complicating it.
The problem:
Given a sequence of integers as an array, determine whether it is possible to obtain a strictly increasing sequence by removing no more than one element from the array.
Example
For sequence = [1, 3, 2, 1], the output should be
almostIncreasingSequence(sequence) = false;
There is no one element in this array that can be removed in order to get a strictly increasing sequence.
For sequence = [1, 3, 2], the output should be
almostIncreasingSequence(sequence) = true.
You can remove 3 from the array to get the strictly increasing sequence [1, 2]. Alternately, you can remove 2 to get the strictly increasing sequence [1, 3].
Thanks for all your comments! I wanted to update this to be a better question, and also inform you I have found a solution if someone would like to check it and see if there is a cleaner way to put it. :)
function almostIncreasingSequence(sequence) {
if(sequence.length == 2) return true;
var error = 0;
for(var i = 0; i < sequence.length - 1; i++){
if(sequence[i] >= sequence[i+1]){
var noStepBack = sequence[i-1] && sequence[i-1] >= sequence[i+1];
var noStepFoward = sequence[i+2] && sequence[i] >= sequence[i+2];
if(i > 0 && noStepBack && noStepFoward) {
error+=2;
}else{
error++;
}
}
if(error > 1){
return false;
}
}
return true;
}
Think about what your code:
sequence[i+1] - sequence[i] !== 1, variable++;
would do for the following array:[1,2,3,8,8].
From the problem description, it is not clear weather the program must remove one character. But if that is the case, the code below should do it.
function canGetStrictlyIncreasingSeq(numbers) {
var counter = 0;
var lastGreatestNumber = numbers[0];
for (var i = 1; i < numbers.length; i++) {
if (lastGreatestNumber >= numbers[i]) {
counter++;
lastGreatestNumber = numbers[i];
} else {
lastGreatestNumber = numbers[i];
}
}
if (counter <= 1)
return true;
return false;
}
var nums1 = [1, 2, 3, 4, 5]; //true
var nums2 = [1, 2, 2, 3, 4]; //true
var nums3 = [1, 3, 8, 1, 9]; //true
var nums4 = [3, 2, 5, 6, 9]; //true
var nums5 = [3, 2, 1, 0, 5]; //false
var nums6 = [1, 2, 2, 2, 3]; //false
var nums7 = [1, 1, 1, 1, 1]; //false
var nums8 = [1, 2]; //true
var nums9 = [1, 2, 2]; //true
var nums10 = [1, 1, 2, 3, 4, 5, 5]; //false
var nums11 = [10, 1, 2, 3, 4, 5]; //true
var nums12 = [1, 2, 3, 4, 99, 5, 6]; //true
console.log(canGetStrictlyIncreasingSeq(nums1));
console.log(canGetStrictlyIncreasingSeq(nums2));
console.log(canGetStrictlyIncreasingSeq(nums3));
console.log(canGetStrictlyIncreasingSeq(nums4));
console.log(canGetStrictlyIncreasingSeq(nums5));
console.log(canGetStrictlyIncreasingSeq(nums6));
console.log(canGetStrictlyIncreasingSeq(nums7));
console.log(canGetStrictlyIncreasingSeq(nums8));
console.log(canGetStrictlyIncreasingSeq(nums9));
console.log(canGetStrictlyIncreasingSeq(nums10));
console.log(canGetStrictlyIncreasingSeq(nums11));
console.log(canGetStrictlyIncreasingSeq(nums12));
Lets take a step back and think about the problem: "Given a sequence of Integers as an array" - we're dealing with arrays of data...but you already knew that.
"determine whether it is possible to obtain a strictly increasing sequence" ok, we need to make something that checks for valid sequence.
"by removing no more than one element from the array." so we can try plucking each element one-by-one and if at least one resulting array is sequential, its possible.
Now instead of one large problem we have two smaller ones
First, we're dealing with arrays, so avail yourself to JavaScript's built-in Array functions to make things easier. In the below, we use 'every()', 'forEach()', 'splice()', 'push()', and 'some()' You can read into how they work here https://www.w3schools.com/jsref/jsref_obj_array.asp It's not long and well worth your time.
Lets deal with the first problem: Determining if an array is sequential. The below function does this
function checkSequence(inputArray){
return inputArray.every(function(value, index, arr){
if (index == 0 && value < arr[index + 1]) {return true}
else if (index < arr.length && value < arr[index + 1] && value > arr[index - 1]) {return true}
else if (index = arr.length - 1 && value > arr[index - 1]) {return true}
else {return false}
});
}
It takes an input array, and uses an Array built-in function called every(), which runs a test on each element in an array
and returns 'true' if all the elements test true. Our test expects the first element to always be lower less than the second
for any given element to be greater than the previous, and less than the next, and for the last element to be greater than the next-to-last
if any element does not satisfy this test, the whole thing returns false
Now we have a means of seeing of an array is sequential, which will make the next part much easier
Now we make another function to pluck out individual elements and see if anythign works
function isPossible(input){
var results = []; //we will store our future results here
input.forEach(function(value, index, arr){
copy = Array.from(arr); //we work on a copy of 'arr' to avoid messing it up (splice mangles what you give it, and we need the un-maimed version for later iterations)
copy.splice(index, 1); //remove an element from copy (a copy of 'arr')
results.push(checkSequence(copy)); //see if it is still in sequence
});
return results.some(function(value){return value});
}
We first make an array to store the results of each attempt into the array 'results' we will use it later.
Then, we take a supplied array 'input' and use "forEach()", which performs a function with each element in an array.
for each element, we make a new array with that element removed from it, and we run the "checkSequence()"
function we made before on it, and finally store the result in the results array.
When the forEach is done, we take the results array and use 'some()' on it, which works just like 'every()'
only it returns true if at least one value is true
Now, you simply call isPossible(your_array) and it will satisfy the problem
Taking into account Patrick Barr's suggestion and assuming that es6 and arrow functions are fine, this solution using Array.prototype.filter could work. The filter itself will return array of elements that should be removed to satisfy conditions of the problem:
UPDATED
function isSequential(array) {
return array && array.length > 0 ? array.filter((x,i) => x >= array[i + 1] || array[i + 1] <= array[i - 1]).length < 2 : false;
}
console.log(isSequential([1]));
console.log(isSequential([1,2,4,5,6]));
console.log(isSequential([1,2,2,3,4,5,6]));
console.log(isSequential([1,4,3,2,5,6]));
console.log(isSequential([1,2,3,4,5,6]));
console.log(isSequential([1,2,0,3,4]));
console.log(isSequential([1,1]));
console.log(isSequential([1,2,0,1,2]));
console.log(isSequential([1,2,3,1,2,3]));
console.log(isSequential([]));
console.log(isSequential([1,0,0,1]));
console.log(isSequential([1,2,6,3,4])); //should be true, but return false
I was trying to solve some of the programming challenges at "free code camp" website, The problem was to find the symmetric difference between multiple arrays and returning an array of the symmetric difference of the provided arrays.
for example the following arrays:
[1, 2, 5], [2, 3, 5], [3, 4, 5]
should return [ 1, 4, 5 ]
so that's what I came up with:
function sym() {
var final = [];
var current_array = [];
for (var i = 0; i < arguments.length; i++) {
current_array = arguments[i];
//ensures duplicates inside each array are removed first
current_array = current_array.filter(function (element, index) {
return current_array.indexOf(element) == index;
});
for (var j = 0, end = current_array.length; j < end; j++) {
if(final.indexOf(current_array[j]) < 0)
final.push(current_array[j]);
else
// for some reason "splice" isn't working properly..
// final.splice(final.indexOf(current_array[j], 1));
delete final[final.indexOf(current_array[j])];
}
}
var final_2 = [];
// Removing the empty slots caused by the "delete" keyword usage
for (var m = 0; m < final.length; m++) {
if(typeof final[m] !== 'undefined')
final_2.push(final[m]);
}
return final_2;
}
in the previous logic I created an array called final that is supposed to hold all of the elements that only exist once in all of the passed arrays, firstly I loop over the arguments parameter which represents here the arrays and for each array I loop over its elements and check if that element exists in the final array or not. If it exists I remove it from the final array, else I push it to the array.
The problem here is if I use the splice method as given in the code above, it behaves very strangely, for example for the following arrays
[1, 2, 3], [5, 2, 1, 4], the result should be: [3, 5, 4]
when I use this line
final.splice(final.indexOf(current_array[j], 1));
instead of
delete final[final.indexOf(current_array[j])];
and return the final array it returns this [ 4 ]
here is the array values at each iteration
round (0, 0): 1
round (0, 1): 1,2
round (0, 2): 1,2,3
round (1, 0): 1,2,3,5
round (1, 1): 1
round (1, 2):
round (1, 3): 4
once it gets to an element that exists in the array it removes all of the elements starting from this element until the end of the array.
I don't know exactly if I'm missing something, I tried to search for any similar problems but most of what I came up with was a problem of removing elements from an array that the person was looping over and hence missing with its indices .. In my case the array I'm trying to modify got nothing to do with the arrays I'm iterating through.
Also I believe splice modifies the array in place and returns the removed elements.. please correct me if I'm not getting it well.
You've misplaced a ), here's the correction:
final.splice( final.indexOf(current_array[j]), 1 );
An additional note: the algorithm adds 5 for the first array, removes it for the second, and adds it again for the third (since it isn't present in final anymore), resulting in [1,4,5].
With an odd number of arguments, the value is preserved, with an even number, it is removed.
A simpler way to get all unique values from all arrays (if that is the intent), is to count the occurrences and filter on a single occurrence:
function sym2() {
var count = {};
for ( var i in arguments ) {
console.log("Processing ", i );
for ( var k = 0; k < arguments[i].length; k ++)
count[ arguments[i][k] ] = (count[ arguments[i][k] ]||0) + 1;
}
var final = [];
for ( var i in count )
if ( count[i] == 1 )
final.push( i );
return final;
}
sym2([1, 2, 5], [2, 3, 5], [3, 4, 5]);
Note that this will return [1,4] rather than [1,4,5].
Here is the way I am using to return duplicate elements.. But I am facing most dangerous performance issues like browser close etc when my array have large number of items with long texts..
var arr = [9, 9, 111, 2, 3, 4, 4, 5, 7];
var sorted_arr = arr.sort();
var results = [];
for (var i = 0; i < arr.length - 1; i++) {
if (sorted_arr[i + 1] == sorted_arr[i]) {
results.push(sorted_arr[i]);
}
}
alert(results);
Please suggest me a best way of doing this
i don't get exactly what you want, but if you need to return duplicates you could use a cache object. this works with number or string or whatever.
var arr = [9, 9, 111, 2, 3, 4, 4, 5, 7];
var cache = {};
var results = [];
for (var i = 0, len = arr.length; i < len; i++) {
if(cache[arr[i]] === true){
results.push(arr[i]);
}else{
cache[arr[i]] = true;
}
}
console.log(results);//returns an array with 9 and 4
Of course you can do other things like deleting multiple items etc. etc.
EDIT - i've written a blog entry on how to remove duplicates from an array
If you have array filter, you also have indexOf and lastIndexOf,
and you can return the duplicates without doing the sort.
var results, arr= [9, 9, 111, 2, 3, 4, 4, 5, 4, 7];
if(arr.filter){
results= arr.filter(function(itm, i){
return arr.lastIndexOf(itm)== i && arr.indexOf(itm)!= i;
});
}
else// use your loop method
alert(results)
/* returned value: (Array)
9,4
*/
Assuming Nicola's solution doesn't work for you (since it uses about as much memory as the original solution: two elements stored per element in the input, worst-case), you can use the slower process of repeatedly searching your input.
This requires the Array.indexOf method from ECMAScript 5. A lot of browsers have it. For alternatives, see How do I check if an array includes an object in JavaScript?.
var arr = [9, 9, 111, 2, 3, 4, 4, 5, 7];
var results = [];
for (var i = 0, len = arr.length - 1; i < len; i++) {
if((results.indexOf(arr[i]) == -1) && (arr.indexOf(arr[i], i + 1) != -1)) {
results.push(arr[i]);
}
}
console.log(results);
This uses no more memory than the input arr plus the output results, but it's an O(N^2) algorithm and doesn't have to modify arr.
Your method relies on a sort, which may or may not be one reason you run out of space/time.
The canonical way to remove duplicates is to keep a hash map of the keys (an object in JS). The object keys you get back won't necessarily be in the order you want; you don't specify if you want the results ordered as well, but they are now.
You could null out the original array, since you no longer require it; when it gets collected is up to the JS engine though.
You could remove duplicates "in place" by keeping a "current index" into the sorted array, and increment it only when you move a non-duplicated element "down" from the counter index, then truncate the array that you return.
Combining the last two techniques should mean that in general you'll only have a single array with a valid reference.
Edit Example. Setting length explicitly, as .slice() creates a new array.
var have = {};
var arr = [9, 9, 111, 2, 3, 4, 4, 5, 7];
arr = arr.sort();
for (var rIdx = 0, i = 0; i < arr.length; i++) {
if (have[arr[i]]) {
arr[rIdx++] = arr[i];
} else {
have[arr[i]] = true;
}
}
arr.length = rIdx;
console.log(arr);