I am in need of help in this script. I am trying to write a program, where each array element is assign to a "point". I concated all the arrays and assigned it to a variable. The array runs through an if statement to tally up the score.
For this instance, I am trying to get a total of 6. However, when I run the program I am getting "2" rather than "6" in the console log. How should I write my if condition to get the result of 6?
var zero = [1,2,3,5,7]; // 0
var one = [0,4,6,9]; // 1
var two = 8; //2
function calculate(){
var NUMBERS = zero.concat(one,two);
var TOTAL = 0; // 6
for(var i = 0; i < NUMBERS.length; i++){
if(NUMBERS[i] === one[i]){
TOTAL += 1;
}else if(NUMBERS[i] == two){
TOTAL +=2;
}else {
TOTAL += 0;
}
}
console.log(TOTAL);
}
calculate();
This will only match if the value in NUMBERS[i] is in the same position of the array one.
if(NUMBERS[i] === one[i]){
Change it to this:
if (one.indexOf(NUMBERS[i]) !== -1){
to test for the presence of NUMBERS[i] in one.
Also, you have a four in both the zero and one array.
As #TJCrowder mentioned - the answer is seven.
You can omit:
TOTAL += 0
Adding zero doesn't change anything.
Related
I try to make an Array in javascript with number of elements and elements inputted by user through window.prompt().
This is what i got by now :
var n = window.prompt("Insert number of numbers", "")
var nr = new Array(n)
for (i = 0; i < n; i++) {
nr[i] = window.prompt("Number: ", "")
}
for (i = 0; i < n; i++) {
if ((i + 1) % 2 != 0) {
if (nr[i] % 2 == 0)
document.write("Even numbers on odd position are: ", nr[i], "<br/>")
}
}
document.write("The numbers are :")
for (i = 0; i < n; i++) {
document.write(nr[i] + "<br/>")
}
On web page shows nothing.
Defining elements.
The first thing in your code is that you should define an array as this:
var numbers = [];
In this array, you will handle every element you will receive from the prompt. So, with this said, you just need the total amount of numbers that you will use which can be retrieved by doing a prompt for a number:
var times = window.prompt("Insert number of numbers: ");
So, times would be our variable containing the amount of numbers we should ask the user, which will be saved on numbers.
Asking user for numbers X amount of times.
Now, what you could do is just a simple for loop which only job is to push the new number provided by the user:
for (let i = 0; i < times; i++) {
numbers.push(window.prompt(`Number ${i+1}: `))
}
This will give the user a prompt saying Number X:, which means the number that is being added.
Checking numbers for even and odds.
And for your functionality of giving a message when there is an even number in an odd index, you could do this:
numbers.forEach((n,i)=>{
if(n%2===0 && i%2!=0){
document.write(`Even number ${n} in odd position ${i} <br/>`);
}
});
Which will check every number you received from the user and check in one inline condition if even number is in odd position and will output the line only if this condition is true.
Printing the numbers.
And simply to output every number you can do:
numbers.forEach((n)=>{
document.write(n + "<br/>")
});
See how it works:
var times = window.prompt("Insert number of numbers"), numbers = [];
for (let i = 0; i < times; i++) {
numbers.push(window.prompt(`Number ${i+1}: `))
}
numbers.forEach((n,i)=>{
if(n%2===0 && i%2!=0){
document.write(`Even number ${n} in odd position ${i} <br/>`);
}
});
document.write("The numbers are: <br/>")
numbers.forEach((n)=>{
document.write(n + "<br/>")
});
To do this you must multiply 6*5*4*3*2*1. To verify your loop is working correctly, the value you are looking for as a result is: 720
var dvDDG = document.querySelector("#ddg");
for(var i = 0; i < 7; i++) {
//remainder..
if( (i*7) == 720 ) {
dvDDG.innerHTML += i + "<br />";
}
}
I'm not entirely certain what you're trying to do with the code you have, it will simply check all numbers zero through six inclusive, and output the value which, when multiplied by seven, is equal to 720.
Since the highest value you'll get is 6 x 7 = 42 (nowhere near 720), you'll see nothing.
The pseudo-code for what you're after would be along the lines of:
fact = 1
for i = 2 to N inclusive:
fact = fact * i
print fact
Turning that into Javascript (or any procedural language for that matter) should be fairly simple, such as with:
function fact(n) {
res = 1
for (var i = 2; i <= n; i++) {
res = res * i;
}
return res
}
alert(fact(6))
It's fairly simple:
var factorial = 1;
var num = 6;
for (var i = 1; i <= num; i++){
factorial *= i;
}
There you go, your answer is the variable factorial. Just copy it into any output function you want. Be careful though, factorial can get pretty huge very fast. Try not to experiment on numbers that much larger that 6.
I am new to Javascript, I am doing a coding challenge to learn more about the language. This is not school related or anything like that, totally for my own personal growth. Here is the challenge:
Return the sum of all odd Fibonacci numbers up to and including the
passed number if it is a Fibonacci number.
I have spent the past 2 evenings working on solving this challenge. When I run my code using underscore.js it works. When I use Ramda.js it says NaN. I would think both would return NaN. I'm very surprised that I can get the correct answer from one and not the other. Any insights would be greatly appreciated!
var R = require('ramda');
function sumFibs(num) {
var fib_Arr = [];
var new_Arr = [];
var total = 0;
// I use this to tell if the fib num is greater than 2
var the_Bit = "false";
// This is used to keep track of when to stop the loop
var fib_Num = 0;
// THIS WORKS FROM HERE
// This loop generates a list of fibonacci numbers then pushes them to the fib_Arr
for(var i = 0; total < num; i++){
if (i < 1){
fib_Arr.push(0);
}
else if (i === 1){
fib_Arr.push(i);
fib_Arr.push(1);
}
else if (i === 2){
fib_Arr.push(2);
the_Bit = "true";
}
else if (the_Bit === "true"){
temp_Arr = R.last(fib_Arr,2);
temp_Arr = temp_Arr[0] + temp_Arr[1];
fib_Arr.push(temp_Arr);
total = R.last(fib_Arr);
}
// Generating the fib Array works TO HERE!!!!
}
// console.log(fib_Arr); // Print out the generated fibonacci array
// if last Array element is greater than the original in
var last_Element = R.last(fib_Arr);
if (last_Element > num){
console.log("The last element of the array is bigger!");
fib_Arr.splice(-1,1); // This removes the last item from the array if it is larger than the original num input
}
// This loop removes all of the EVEN fibonacci numbers and leaves all of the ODD numbers
for (var j = 0; j < fib_Arr.length; j++){
if (fib_Arr[j] % 2 !== 0){
new_Arr.push((fib_Arr[j]));
}
}
// This checks if the original input num was a
if (num % 2 !== 0){
new_Arr.push(num);
}
else{
console.log("The original num was not a Fibonacci number!");
}
// if last Array element is the same as the original input num
var last = R.last(fib_Arr);
if (last === num){
console.log("Removing the last element of the array!");
new_Arr.splice(-1,1); // This removes the last item from the array if it is the same as the original num input
}
// Now to add all of the numbers up :-)
for (var k = 0; k < new_Arr.length; k++){
console.log("This is fib_Num: " + fib_Num);
// console.log(fib_N`);
fib_Num = fib_Num += new_Arr[k];
}
return fib_Num;
}
// TEST CASES:
// console.log(sumFibs(75025)); //.to.equal(135721);
console.log(sumFibs(75024)); //.to.equal(60696);
You have a problem on these lines :
temp_Arr = R.last(fib_Arr,2);
temp_Arr = temp_Arr[0] + temp_Arr[1];
Besides the fact that R.last does not take a second argument (that will not fail though), you are using temp_arr as an array, when it is a number. Therefore, temp_arr gets a NaN value.
You are probably looking for R.take (combined with R.reverse) or R.slice.
By changing :
temp_Arr = R.last(fib_Arr,2);
with :
temp_Arr = R.take(2, R.reverse(fib_Arr));
or with :
temp_Arr = R.slice(fib_Arr.length - 2, fib_Arr.length)(fib_Arr);
or with (bonus play with a reduce from the right) :
temp_Arr = R.reduceRight(function(arr, elem) {
return arr.length < 2 ? [elem].concat(arr) : arr;
}, [])(fib_Arr);
We get :
sumFibs(75024) === 60696
For the record, here's how you do this problem:
function fibSumTo(n) {
var f1 = 1, f2 = 1, sum = 1, t;
while (f2 <= n) {
if (f2 & 1) sum += f2;
t = f1 + f2;
f1 = f2;
f2 = t;
}
return sum;
}
There's really no need for any sort of library because there's really no need for any sort of data structure.
var _ = require('underscore');function sumUpFibs (number){
arr_of_fibs = [1,1];
current = 1; //cursor for previous location
while (true){
var num = arr_of_fibs[current] + arr_of_fibs[current - 1];
if (num <= number) {
arr_of_fibs.push(num);
current++;
} else {
break;
}
}
console.log(arr_of_fibs);
var total = 0;
_.each(arr_of_fibs, function(fib){
total += fib;
})
return total;}console.log(sumUpFibs(75025));
This may be a better implementation... Though I know you're just starting so I don't want to come off as mean : D.... Also, maybe check your test cases too.
I'm doing the Euler project problem 2 in which the objective is to sum the even numbers of the fibonacci sequence that have a value of less than 4 million. I've searched a bit and I've seen several solutions using a while loop but nothing simple using a for loop. I'm curious why I'm returning zero with the following code:
var array = [];
array[0] = 0;
array[1] = 1;
var total = 0;
for(var i=2;total<=4000000;i++) {
array[i] = array[i-1] + array[i-2];};
for(var x=0;x<array.length;x++){
if(array[x]%2 === 0){
total += array[x]};};
alert(total);
I'm guessing the problem is in my for loop using the total variable. I couldn't get it to work using array[i]<=4000000 either and I'm really curious behind the why here. Anyone know why this is? What can I change in the for loop condition (second statement) to get a correct total here?
First of all there is an infinite loop at first for. Your condition must be array[i-1] < 4000000. After that your second for loop will find the correct result.
Also for the problem, you don't need to store all fibonacci numbers then find sum of even numbers.
You can calculate sum when calculating fibonacci.
var first = 0;
var second = 1;
var sum = 0;
for(var current=first+second; current < 4000000; current = first+second){
if(current%2 === 0){
sum+=current;
}
first = second;
second = current;
}
I fixed it for you.
var i, data = [ 0, 1 ], total = 0;
for (i = 2; i <= 4000000; i++)
{
data[i] = data[i - 1] + data[i - 2];
if (data[i] % 2 === 0)
{
total += data[i];
}
}
alert(total);
I'm not sure what you termination condition should be like, you say have a value of less than 4 million, but this is ambiguous. Maybe it should be total <= 4000000 or data[i] <= 4000000. Your phrasing is not precise enough.
Sorry but for me your code going in a dead loop. first "for" use total as check but it's never incremented. If you want This is a solution for fibonacci sequence based on dinamic programming with memoization tecnic.
var f1 = 1;
var f2 = 1;
for(var i = 2; i < 40000; i++){
console.info(i, f1, f2);
var temp = f1 + f2;
f1 = f2;
f2 = temp;
}
alert(f2);
So here's my problem.. Might just be tired but, I want counter to ++ only if number has not occurred in array.
Meaning during the 4 iterations, counter should ++ only on iteration 1,3,4
var array = [], number, temp = [4,2,5,9], counter = 0;
for(var i = 0; i <= 3; i += 1) {
array.push(i);
number = temp[i];
}
document.write(counter);
But I'm drawing a blank here... Help?
(No this isn't homework)
if (array.indexOf(number) < 0)
counter++;
unfortunately JS doesn't have an "in_array", but it's pretty straight forward:
#MikeSamuel pointed out you can use indexOf (thanks Mike). So with that being said:
var array = [], number, temp = [4,2,5,9], counter = 0;
for(var i = 0; i <= 3; i += 1) {
array.push(i);
number = temp[i];
if (temp.indexOf(i)==-1) // much simpler, assuming you're checking if i is in temp.
counter++; // increase counter if it is not.
}
document.write(counter);
I'm not sure where you want the logic, so you'll have to figure that out or be more specific. Just know that you need to iterate through the array you're checking and check if the "needle" is in the "haystack" array.
EDIT Had the opposite, just added bool to check for existence.