Find the correct passcode in the array and we'll do the rest. We can't disclose more information on this one, sorry.
Each entry in the first array represents a passcode
- Find the passcode that has no odd digits.
- For each passcode, show us the amount of even digits.
- If it has no odd digits, show us that you've found it and increase the number of terminals by one.
var passcodes = [
[1, 4, 4, 1],
[1, 2, 3, 1],
[2, 6, 0, 8],
[5, 5, 5, 5],
[4, 3, 4, 3],
];
so, i've tried almost everything i could think of. modulo, function, for loop and i can't seem to get it. i'm a beginner and this is an important exercise i have to do. but what do i do? it asks for the amount of even digits in each passcode, so i have to get the array within the array and then code something that i don't know to find even values. i'm stuck
Your question is not really suitable for StackOverflow, you should at least try to write something and see how far you get.
Anyhow, you seem to want to iterate over the elements in passcodes to find the array with no odd numbers.
The first task is to how to determine if a number is even. That is as simple as looking for the remainder from modulus 2. If the remainder is zero, then the number is even, otherwise it's odd.
So a simple test is:
var isEven;
if (x % 2 == 0) {
isEven = true;
} else {
isEven = false;
}
Since 0 type converts to false, and the not (!) operator reverses the truthiness of values and converts the result to boolean, the above can be written:
var isEven = !(x % 2);
There are many ways to iterate over an array, if your task was just to find the element with no odd numbers, I'd use Array.prototype.every, which returns as soon as the test returns false, or Array.prototype.some, which returns as soon as the test returns true.
However, in this case you want to count the number of even numbers in each element and find the first with all even numbers. One way is to iterate over the array and write out the number of even numbers in the element, and also note if its all even numbers. You haven't said what the output is expected to be, so I've just made a guess.
var passcodes = [
[1, 4, 4, 1],
[1, 2, 3, 1],
[2, 6, 0, 8],
[5, 5, 5, 5],
[4, 3, 4, 3], // this last comma affects the array length in some browsers, remove it
];
// Set flag for first element with all even numbers
var foundFirst = false;
// Iterate over each element in passcodes
passcodes.forEach(function(code) {
// Count of even numbers in current array
var evenCount = 0;
// Test each element of code array and increment count if even
code.forEach(function(num) {
if (!(num % 2)) ++evenCount;
});
// If all elements are even and haven't found first yet, write out elements
if (code.length == evenCount && !foundFirst) {
console.log('Passcode (first all even): ' + code.join());
// Set flag to remember have found first all even array
foundFirst = true;
}
// Write count of even numbers in this array
console.log('Even number count: ' + evenCount + ' of ' + code.length + ' numbers.');
});
I have no idea what you meant ...but it does all what i could understand from your question. Hope it will help :)
var passcodes = [
[1, 4, 4, 1],
[1, 2, 3, 1],
[2, 6, 0, 8],
[5, 5, 5, 5],
[4, 3, 4, 3],
];
var aPassCode;
while(aPassCode = passcodes.shift()){
for(var i=0,evenCount=0,totalCount=aPassCode.length;i<totalCount;i++){
if(aPassCode[i] % 2 == 0)evenCount++;
}
if(evenCount == totalCount){
console.log('all digits even here: ' + aPassCode.join(','));
}else{
console.log(aPassCode.join(',') + ' contains ' + evenCount + ' even digits.');
}
}
Related
I'm trying to understand the following solution for finding the largest adjacent product in any given array.
Example:
For inputArray = [3, 6, -2, -5, 7, 3], the output should be
adjacentElementsProduct(inputArray) = 21.
7 and 3 produce the largest product.
Possible solution in JS:
function adjacentElementsProduct(arr) {
return Math.max(...arr.slice(1).map((x,i)=>[x*arr[i]]))
}
I am having a hard time understanding two things:
What do the three dots exactly do and how does this get passed into the function? Is there any way to write this in a more understandable way? I know that is the "spread syntax" feature in ES6, but still don't understand completely.
Why do we insert "1" as argument to slice? My first though was to input "0", because we want to start at the start, then loop through everything, and see which adjacent product is the largest.
I'd appreciate any advice, links and explanations.
Thanks.
Cheers!
1. What do the three dots exactly do and how does this get passed into the function? Is there any way to write this in a more understandable way? I know that is some kind of "spread" feature in ES6, but still don't understand completely.
The Math#max needs a list of numbers as parameters, and map produces an array. The spread syntax is used to convert an array to be expanded to a list of parameters.
const arr = [1, 2, 3];
console.log('max on array', Math.max(arr));
console.log('max on list of parameters', Math.max(...arr));
In this case you can use Function#apply to convert the array to a list of parameters. I find it less readable, however.
const arr = [1, 2, 3];
console.log(Math.max.apply(Math, arr));
2. Why do we insert "1" as argument to slice? My first though was to input "0", because we want to start at the start, then loop through everything, and see which adjacent product is the largest.
Lets break down the iteration order of the 2 arrays.
[3, 6, -2, -5, 7, 3] // inputArray
[6, -2, -5, 7, 3] // inputArray.slice(1)
Now on each iteration of inputArray.slice(1):
x: 6, i = 0, arr[0] = 3
x: -2, i = 1, arr[1] = 6
x: -5, i = 2, arr[2] = -2
Since the inputArray.slice(1) array starts from the 2nd element of the inputArray, the index (i) points to the 1st element of the inputArray. And the result is an array of products of 2 adjacent numbers.
var biggestProduct = inputArray[0] * inputArray[1];
for (i=0; i<inputArray.length-1 ; ++i)
{
console.log(biggestProduct)
if ((inputArray[i] * inputArray[i+1] ) > biggestProduct)
{
biggestProduct = inputArray[i] * inputArray[i+1]
}
}
return biggestProduct;
Note: I've declared a variable that consists of 2 input arrays with index number then starts a for loop that indicates input array with his index number, so by that he will go throw all the index number of the array (one of them raised by one so that they won't be at the same value). and at the end of the code, you have the if statement.
You may simply do as follows;
function getNeigboringMaxProduct([x,...xs], r = -Infinity){
var p = x * xs[0];
return xs.length ? getNeigboringMaxProduct(xs, p > r ? p : r)
: r;
}
var arr = [3, 6, -2, -5, 7, 3],
res = getNeigboringMaxProduct(arr);
console.log(res);
I'm trying to write a function that continually adds together the first and last elements of an array using forEach with array.shift() + array.pop().
The problem is that the for-loop doesn't complete the innermost numbers, and so the array is always left with 2 values inside of it.
Code:
function choreAssignment(chores) {
chores.sort(function(a, b) {return a - b});
var assignment = [];
chores.forEach(function() {
assignment.push((chores.pop() + chores.shift()));
});
return assignment.sort(function(a, b) {return a - b});
}
The above code works as expected, but it leaves the innermost two values inside the chores array.
For example if I run:
Code:
var arr = [1, 4, 7, 2, 5, 9, 4, 3];
choreAssignment(arr);
I get:
[8, 9, 10]
Ie, it adds 9 & 1, 7 & 2, 5 & 3, but it leaves [4, 4] inside the array.
I'm not sure why this is. Thank you.
Try changing the forEach to:
while (chores.length) {
assignment.push((chores.pop() + chores.shift()));
}
Note this assumes there are always an even number of elements in array
I don't know how to make my code more efficient. It completes the examples, but it will time out when it tackles the big array.
I ask you for some advice for performance on loops, or if you can tell me where my error is. If my approach is completely wrong don't give me the answer please.
What i have to do:
Given a list of integers and a single sum value, return the first two values (parse from the left please) in order of appearance that add up to form the sum.
Examples:
sum_pairs([11, 3, 7, 5], 10) => [3, 7]
sum_pairs([4, 3, 2, 3, 4], 6)=>[2, 4],[3, 3],[2, 4]
(indices:0,2|1,3|2,4)=>
entire pair is earlier, and therefore is the correct answer===[2, 4]
sum_pairs([0, 0, -2, 3], 2)
there are no pairs of values that can be added to produce 2.
== None/nil/undefined (Based on the language)
sum_pairs([10, 5, 2, 3, 7, 5],10)=>[5, 5][3, 7]
--------------------1------------5
--------------------------3--4
indicies:(1,5|3,4)
entire pair is earlier, and therefore is the correct answer
== [3, 7]
My attemt:
var sum_pairs=function(ints, s){
for(t=1; t <=ints.length-1;t++){
for(i=0; i <=ints.length-t-1;i++){
if(ints[i]+ints[i+t]===s){
return [ints[i],ints[i+t]];
}
}
}
}
If we get something like
array=[5,5,5,5,3,2];
return Math.max.Apply(Math,array);
How do I get it to return the numbers from first to last if such a case occurs.
To answer the question in the title:
what does max() function do in javascript if array has several equally
large numbers
The answer is, nothing. Math.max() doesn't act on arrays.
You can pass an array by spreading the items as arguments to max():
Math.max(...[1,2,3]) // 3
Or as you've seen, with apply():
Math.max.apply(Math, [1,2,3]) // 3
If the question is more:
What does Math.max() do when more than one of the same maximum number is given?
The answer is, it returns that number:
const a = [5, 5, 5, 5, 3, 2]
const max = Math.max(...a)
console.log(max) // 5
This question is confusing:
How do I get it to return the numbers from first to last if such a case occurs.
You want it to return a sorted array? From [5, 5, 5, 5, 3, 2] to [2, 3, 5, 5, 5, 5]?
a.sort() // [2, 3, 5, 5, 5, 5]
You want dupes removed? From [5, 5, 5, 5, 3, 2] to [2, 3, 5]?
Array.from(new Set(a)) // [2, 3, 5]
Could you clarify your question?
The best way to do this is the following:
var a = [5,5,5,5,3,2];
var largest = Math.max.apply(null,a)
var filtered = a.filter(function(item) {
item === largest
});
Where filtered will have contain all the largest elements.
In #Clarkie's example, he's calling Math.max more frequently than needed.
In both Dan and Clarkie's example they're capitalizing Apply which is incorrect, the correct function to call is Math.max.apply and Math need not be passed in as the first argument.
See the following for a working example:
https://jsfiddle.net/fx5ut2mm/
Modifying #Clarkie's very nice idea. We can boil it down to...
var a = [5,5,5,5,3,2],
m = Math.max(...a),
f = a.filter(e => e == m);
document.write("<pre>" + JSON.stringify(f) + "</pre>");
I am trying to find the min value of an array, and am trying to do it by sorting the array, and then reversing the array, and then calling the very first index of the array.
Unfortunately with what I have been trying, I keep getting 9. (don't know why) Can anybody take a quick look at what I have been doing and bail me out here? (i'm using js)
var minny = [4, 3, 5, 2, 6, 3, 4, 5, 2, 3, 4, 6, 7, 8, 9, 9, 1, 11, 25];
var smallest = function (minny){
minny = minny.sort('');
var sorted = minny + " ";
sorted = minny.reverse('').join('');
return sorted[0];
}
console.log(smallest(minny))
By default the sort method sorts elements alphabetically(11 comes before 9) and therefore you need to add a compare function as a param.
var smallest = function (minny) {
minny = minny.sort(function(a, b) { return a - b; });
return minny[0];
}
console.log(smallest(minny))
JSFIDDLE.
Based on your code, you could just do
return minny.sort()[0];
So, your full code example becomes
var minny = [4, 3, 5, 2, 6, 3, 4, 5, 2, 3, 4, 6, 7, 8, 9, 9, 1, 11, 25];
var smallest = function (minny){
return minny.sort()[0];
}
console.log(smallest(minny))
You're calling minny.sort('') which is using the default natural sort, so 11 and 25 end up near the beginning because of the 1 and 2.
What you have to do is call sort with a function that compares numbers, such as:
minny.sort(function(a,b) { return b-a; });
This will sort minny the way you want it.
There is no need to even call reverse and join afterwards, just return the first item. "return sorted[0]" is fine but will fail if there are no items, so you might just want to call "return sorted.shift()" instead. This will return the first item too, but won't fail if the array is empty.
PS. your call to minny.reverse also has an empty string as a parameter. That's not needed, reverse takes no parameters.
sort() sorts alphabetically by string representation, so in your case it would result in 1, 11, 2, 2, 25, .... You have to provide a comparison function for correct integer sorting, although in your specific case it doesn't really make a difference.
var smallest = function (minny){
minny = minny.sort(function(a, b){return a-b});
return minny[0];
}
See jsfiddle
Using sort is fairly short code to write, and it will return the correct number if you use minny.sort(function(a,b){return a-b})[0].
If you have a large unordered array you are running the comparison many times and you are sorting the array, which is not usually what you want to do to an array.
It may be better to just iterate the members and compare each just once to the lowest fond so far.
var minny= [4, 3, 5, 2, 6, 3, 4, 5, 2, 3, 4, 6, 7, 8, 9, 9, 1, 11, 25];
var smallest= function(minny){
var min= Infinity;
minny.forEach(function(next){
if(next<min) min= next;
});
return min;
}
Or use Math.min, if this is code golf:
Math.min.apply(Array,minny);