How would I get a standard for loop to output in pairs or other groups (like three's of four's) with the output shifting up one after the last digit of the group?
for(var i = 0: i < 8; i++){
console.log(i)
}
so instead of the output being; 0,1,2,3,4,5,6,7
In pairs it would be; 0,1,1,2,2,3,3,4
or if it went up in groups of four; 0,1,2,3,1,2,3,4
I did try doing something like this, but instead of going up in two's every time I need the loop to output the first 2 digits move up one then output the next two ect...
for(var i = 0: i < 8; i+= 2){
console.log(i)
}
Hope that makes sense
For each case you would need to come up with the right formula based on i:
so instead of the output being; 0,1,2,3,4,5,6,7 In pairs it would be; 0,1,1,2,2,3,3,4
for (let i = 1; i < 9; i++) {
console.log(i >> 1); // this bit shift is integer division by 2
}
or if it went up in groups of four; 0,1,2,3,1,2,3,4
for (let i = 0; i < 8; i++) {
// Perform division by 4 and add remainder to that integer quotient
console.log((i >> 2) + (i % 4));
}
You could work with a variable inside the loop to determine the index. This way you can specify how many times you want the loop to run:
for(let index = 0; index < 8; index++) {
const currentIndex = index - (index >> 1);
console.log(currentIndex);
}
It also makes it easy to implement it as immutable:
const array = new Array(8).fill(0).map((entry, index) => index - (index >> 1));
console.log(array);
I think a function like below where we specify the total n and the chunksize after which you want to increase a single step might work for us :-
function getByChunkSteps(n,chunkSize){
let step = -1;
let output = [];
for(let index = 0;index < n;index++){
if(index%chunkSize===0){
step+=1;
}
output.push((index%chunkSize)+step);
}
return output;
}
console.log(getByChunkSteps(10,2));
console.log(getByChunkSteps(8,4));
console.log(getByChunkSteps(9,3));
Related
I have the following code
let range = [1,2,3];
let multiples = [1,2,3,4,5,6,2,4,6,3,6];
I want to find the first number in the multiples array that occurs range.lenght times (3);
I want to start with multiples[0] check how many times it occurs in multiples, if it occurs 3 times I want to return multiples[0], if it is less than 3 times, I want to check how many times multiples[1] occurs in the multiples array. If multiples[1] occurs 3 times I want to return multiples[1], else I move on to check multiples[2], etc. until I find a number that occurs 3 times. In the code above I should return 6.
I've looked at
How to count the number of certain element in an array?
and
Idiomatically find the number of occurrences a given value has in an array
and
get closest number out of array
among other research but have not figured it out yet.
I tried to simplify the question as much as possible. But if more info is needed it relates to this challenge on freeCodeCamp. Where I am at with my code is
function smallestCommons(arr) {
let sortArr = arr.sort((a, b) => a - b);
console.log(sortArr);
let range = [];
for (let i = sortArr[0]; i <= sortArr[1]; i++) {
range.push(i);
}
console.log("range = " + range);
let maxNum = range.reduce( (a, b) => a * b);
console.log("maxNum = " + maxNum);
let multiples = [];
for (let i = 0; i < maxNum; i++) {
let j = 0;
do {
multiples.push(j + range[i]);
j += range[i];
} while (j < maxNum);
//j = 0;
}
for (let i = 0; i < multiples.length; i++) {
let numberToFind = multiples[i];
/*stuck here hence my question, maybe I shouldn't even start with a for loop*/
//tried reduce, forEach, filter, while loop, do while loop
}
console.log("multiples = " + multiples);
}
console.log(smallestCommons([1,3]));
The logs are
1,3
range = 1,2,3
maxNum = 6
multiples = 1,2,3,4,5,6,2,4,6,3,6,NaN,NaN,NaN
What you can do is, first split your string with , and then using below function loop for check.
function countLength(arr, checkNumber) {
var count = 0;
for (var i = 0; i < arr.length; i++) {
if (arr[i] === checkNumber) {
count++;
}
}
return count;
}
countLength(list, NUMBER YOU WANT TO CHECK);
And if you want to check first number occur for 3 time then you need to make change in function and introduce .map or .filter in action to count number.
Example
const multiples = [1,2,3,4,5,6,2,4,6,3,6];
let occurance_arr=[];
const aCount = [...new Set(multiples)].map(x => {
if(multiples.filter(y=> y==x).length == 3) {
occurance_arr.push(x);
}
});
console.log(occurance_arr);
Above code will give you 6 in console, if you have multiple value then 0th element is the answer you are looking for which is first three time occurrence of item.
You can loop through your list keeping an object that maps each number to the number of times you've seen it. You can check the counts object as you loop, so if you see a number and the count is one less than your target, you can return it. If you make it through the loop without returning you didn't find what you're looking for — return something sensible :
let range = [1,2,3]
let multiples = [1,2,3,4,5,6,2,4,6,3,6]
function findFirstMult(arr, len){
let counts = {} // to keep track of how many times you've seen something
for (let n of arr){ // loop throught the array
if (!counts[n]) counts[n] = 0 // if it's then first time you've seen n, defined that key
if (counts[n] == len - 1) return n // found it
counts[n] +=1 // otherwise increase the count
}
return undefined
}
console.log(findFirstMult(multiples, range.length))
This will require only one loop through the array in the worse case and will return early if if finds something.
Just started my uni course, struggling a little with javascript. I have been asked to display a square using any character, however, the solution must combine for loops and if statements.
This is what I have so far and I feel pretty close but I just can't get the second line to display. I know this can be done via two for loops, (one for iteration of the variable and another for spaces). But this is not how I have been asked to solve this problem.
Here is my code:
var size = 3;
let i;
for(i = 0; i < size; i++) {
print ("*");
if (size === i){
println ("");
}
}
For context, this is all taking place int he professors homemade learning environment.
You could use nested for loops and take a line break after each filled line.
function print(s) { document.getElementById('out').innerHTML += s; }
function println(s) { document.getElementById('out').innerHTML += s + '\n'; }
var size = 5,
i, j;
for (i = 0; i < size; i++) {
for (j = 0; j < size; j++) {
print("*");
}
println("");
}
<pre id="out"></pre>
Single loop with a check if i is unequal to zero and if the remainder is zero, then add a line break.
Using:
=== identity/strict equality operator checks the type and the value, for example if both are numbers and if the value is the same,
!== non-identity/strict inequality operator it is like above, but it checks the oposite of it,
% remainder operator, which returns a rest of a number which division returns an integer number.
&& logical AND operator, which check both sides and returns the last value if both a truthy (like any array, object, number not zero, a not empty string, true), or the first, if it is falsy (like undefined, null, 0, '' (empty string), false, the oposite of truthy).
function print(s) { document.getElementById('out').innerHTML += s; }
function println(s) { document.getElementById('out').innerHTML += s + '\n'; }
var size = 5,
i;
for (i = 0; i < size * size; i++) {
if (i !== 0 && i % size === 0) {
println("");
}
print("*");
}
<pre id="out"></pre>
Well the for loop is only iterating 3 times, printing the first line. If you want a square you'll have to print 9 stars total, right? So i'm assuming, is this is the approach you'd go for, you would need to iterate not until size, but until size * size.
I'm using console.log to 'print' the square:
var dimension = 10;
var edge = '*';
var inside = ' ';
var printLine;
for (var i = 1; i <= dimension; i++) {
if (i === 1 || i === dimension) {
printline = Array(dimension + 1).join(edge);
} else {
printline = edge + Array(dimension - 1).join(inside) + edge;
}
console.log(printline);
}
Note that in the following example, an array of length 11 gets you only 10 "a"s, since Array.join puts the argument between the array elements:
Array(11).join('a'); // create string with 10 as "aaaaaaaaaa"
You wanna make a square of * where the size is the number of * on its sides?
Let's split a task into 3 parts:
where you print top side like *****
where you print middle (left and right sides) like * *
where you print bottom (same as top)
Now let's code that, I kept the code as simple as possible, this can be done in fewer lines but I think this will be easier to understand for beginners:
var size = 5;
var i = 0;
// top
for (i = 0; i < size; i++)
console.log("*");
//middle
for (var j = 0; j < size - 2; j++){
console.log("\n"); // go to next row
// middle (2 on sides with size-2 in between)
console.log("*");
for (i = 0; i < size-2; i++)
console.log(" ");
console.log("*\n"); // goes to new row as well
}
// same as top
for (i = 0; i < size; i++)
console.log("*");
Full square is even simpler:
var size = 5;
var i = 0;
for (var i = 0; i < size; i++){ // iterates rows
for (var j = 0; j < size; j++) // iterates * in row
console.log("*");
console.log("\n") // moves to new row
}
In order to print a row, you print same sign X times. Well, to print X rows we can use just that 1 more time (only this time we are iterating over a different variable (j is * in a row, i is a number of rows).
After a row is made we go to go to next row with \n.
As for
it must contain if statement
Put this at the end:
if (youCanHandleTheTruth) console.log("It's a terrible practice to tell students their solution MUST CONTAIN CODEWORDS. If you need them to showcase something, write appropriate task that will require them to do so.");
I am going through one of the FreeCodeCamp challenges.
" Return the factorial of the provided integer.
If the integer is represented with the letter n, a factorial is the
product of all positive integers less than or equal to n.
Factorials are often represented with the shorthand notation n!
For example: 5! = 1 * 2 * 3 * 4 * 5 = 120 "
I already know that the easiest way is to use recursion but by the moment I've discovered this fact I was already trying to solve the problem by creating an array, pushing numbers in it and multiplying them. However I got stuck on this step. I have created an array with the number of digits depending on the function factorialize argument, but I can't get the product of those digits. What did I do wrong:
function factorialize(num) {
var array = [];
var product;
for(i = 0; i<=num;i++) {
array.push(i);
for (j=0; j < array.length; j++) {
product *= array[j];
}
return product;
}
}
factorialize(5);
I think the easiest way would be to create a range and reduce that:
var n = 5;
function factorize(max) {
return [...Array(max).keys()].reduce((a,b) => a * (b + 1), 1);
}
console.log(factorize(n));
It looks like you missed a close parenthesis
function factorialize(num) {
var array = [];
var product = 1;
for(i = 0; i<=num;i++) {
array.push(i);
} //right here!!! <----
for (j=0; j < array.length; j++) {
product *= array[j];
}
return product;
}
factorialize(5);
but as stated in the comments, you should change i = 0 to i = 1 not just because it would change the final result(which it does for all num ) but because it also doesn't follow the factorial algorithm.
1) You need initial value 'product' variable
2) You should change i = 0 to 1. You multiply by 0 in the loop
3) You don't need nested loop
function factorialize(num) {
var array = [];
var product = 1;
for(var i = 1; i <= num; i++) {
array.push(i);
}
for (j=0; j < array.length; j++) {
product *= array[j];
}
return product;
}
You only need one loop for that,
from 1 to the maximum number, then you multiply them up,
just a little clean up from your code
fact variable will contain the string version of the individual numbers making up the sum
if m is 5 you'll fact will be 1*2*3*4*5
function factorialize(num) {
var product = 1;
var fact = ""
for (i = 1; i <= num; i++) {
product *= i;
fact += i + "*"
}
fact = fact.substring(0, fact.length - 1)
console.log(fact)
return product;
}
console.log(factorialize(5));
I wrote a code for a "Heads or Tails" game below and:
var userInput = prompt("Enter maximum number output: ");
function coinFlip() {
return (Math.floor(Math.random() * 2) === 0) ? 'Heads' ; 'Tails';
}
for (var i = 0; i < 6; i++)
{
var result = [];
result["randomNum"] = (Math.floor(Math.random()*userInput);
result["coin"] = (coinFlip());
}
I'm trying to count the sum of total heads and sum of total tails each with the code:
var headsCount = 0;
var tailsCount = 0;
for (var j = 0; j < result["coin"].length; j++)
{
if (result["coin"] == 'Heads')
headsCount++;
else
tailsCount++;
}
The only problem is that it's counting each characters of 'Heads' and 'Tails' in the result["coin"] array as separate (such as 'H'-'e'-'a'-'d'-'s') and not into a full string (like "Heads"). Thus, instead of increment by 1 each time the loop above runs, it increments by +5.
I want it to increment by +1 only.
How do I make it so that the code reads the full string stored in result["coin"] and not character-by-character?
EDITED -- changed the <2 to *2
var result = []; is inside the for loop, so it is being overwritten with an empty array each time. So when you try to loop over the results, there's one one item in it; the last one. Pull the result array out of the loop so that you can add to it in each iteration.
It seems userInput should be the number of times to loop. Not sure why you're putting it in result["randomNum"]. result is an array, not an object, so it only has integer keys.
Instead of adding the result of the coin toss to result["coin"] I think you mean to add it to the array, so after tossing it six times it might look like this: ["Heads", "Heads", "Tails", "Heads", "Tails", "Tails"]. You can do this by calling result.push with the coin toss output.
To get one of two results randomly, compare the output of Math.random() against 0.5, which is half way between the limits. Numbers less than 0.5 can be considered heads, while numbers greater than or equal to 0.5 can be considered tails.
Putting it all together, this is what I think you were going for:
function coinFlip() {
return Math.random() < 0.5 ? 'Heads' : 'Tails';
}
var result = [];
var userInput = parseInt(prompt("Enter maximum number output: "), 10);
for (var i = 0; i < userInput; i++) {
result.push(coinFlip());
}
var headsCount = 0;
var tailsCount = 0;
for (var j = 0; j < result.length; j++) {
if (result[j] == 'Heads')
headsCount++;
else
tailsCount++;
}
console.log(headsCount, "heads and", tailsCount, "tails");
All that being said, there are definitely areas for improvement. You don't need to loop once to build the results, then loop a second time to read the results.
You can count the number of heads/tails as the coins are flipped. For example:
function isCoinFlipHeads() {
return Math.random() < 0.5;
}
var numFlips = parseInt(prompt("How many flips?"), 10);
var heads = 0;
var tails = 0;
for (var i = 0; i < numFlips; i++) {
isCoinFlipHeads() ? heads++ : tails++;
}
console.log(heads, "heads and", tails, "tails");
Here is the code in question:
var L1 = [];
var Q1 = [];
function populateListOne() {
var limit = prompt("please enter a number you would like to fill L1 to.");
for (i = 2; i <= limit; i++) {
L1[i] = i;
}
for (n = 2; n <= L1.length; n++) {
var count = 2;
if (n == count) {
var index = L1.indexOf(n);
L1.splice(index, 1);
Q1[n] = n;
count = count + 1;
}
for (j = 0; j <= L1.length; j++) {
if (L1[j] % 2 == 0) {
var secondIndex = L1.indexOf(j);
L1.splice(secondIndex, 1);
}
}
}
document.getElementById("demo").innerHTML = "iteration " + "1" + ": " + L1 + " Q1 = " + Q1;
}
I’m currently working on a homework assignment where I have to setup a queue. All is explained in my JSFiddle.
Problem description
Essentially, the part I’m stuck on is iterating through each instance of the array and then taking the value out if the modulus is identical to 0. However, as you can see when I run the program, it doesn’t work out that way. I know the problem is in the second for loop I just don’t see what I’m doing wrong.
The way I read it is, if j is less than the length of the array, increment. Then, if the value of the index of L1[j] modulus 2 is identical to 0, set the value of secondIndex to whatever the index of j is. Then splice it out. So, theoretically, only numbers divisible by two should be removed.
Input
A single number limit, which will be used to fill array L1.
L1 will be initialized with values 2, 3, ... limit.
Process
Get the starting element of array L1 and place it in array Q1.
Using that element, remove all values in array L1 that are divisible by that number.
Repeat until array L1 is empty.
You're going to have issues with looping over an array if you're changing the array within the loop. To help with this, I tend to iterate from back to front (also note: iterate from array.length - 1 as the length element does not exist, arrays are key'd from 0):
for(j = L1.length - 1; j >=0 ; j--)
For your first loop, you miss the elements L1[0] and L1[1], so I would change the first loop to:
L1 = [];
for(i = 2; i <= limit; i++)
{
L1.push(i);
}
In this section:
for(j = 0; j <= L1.length; j++){
if(L1[j] % 2 == 0)
{
var secondIndex = L1.indexOf(j);
L1.splice(secondIndex, 1);
}
}
you should splice with j instead of secondIndex.
Change L1.splice(secondIndex, 1); to L1.splice(j, 1);
Array indices and putting entries
You initial code used an array that was initialized to start at index 2. To avoid confusion, of what index to start at, start with index 0 and iterate until array.length instead of a predefined value limit to ensure that you go through each element.
The following still works but will be more of a headache because you need remember where to start and when you will end.
for (i = 2; i <= limit; i++) {
L1[i] = i; // 'i' will begin at two!
}
Here's a better way:
for (i = 2; i <= limit; i++) {
// 'i' starts at 2 and since L1 is an empty array,
// pushing elements into it will start index at 0!
L1.push(i);
}
Use pop and slice when getting values
When you need to take a peek at what value is at the start of your array, you can do so by using L1[0] if you followed my advice above regarding array keys.
However, when you are sure about needing to remove the starting element of the array, use Array.slice(idx, amt). idx specifies which index to start at, and amt specifies how many elements to remove beginning at that index (inclusive).
// Go to 1st element in L1. Remove (1 element at index 0) from L1.
var current = L1.splice(0, 1);
Use the appropriate loops
To make your life easier, use the appropriate loops when necessary. For loops are used when you know exactly how many times you will iterate. Use while loops when you are expecting an event.
In your case, 'repeat until L1 is empty' directly translates to:
do {
// divisibility checking
} while (L1.length > 0);
JSFiddle
Here's a complete JS fiddle with in-line comments that does exactly what you said.