multiplying certain elements of an array javascript - javascript

var valid1=[4, 5, 3, 9, 6, 7, 7, 9, 0, 8, 0, 1, 6, 8, 0, 8]
function validateCred(array) {
let doubleArray =0
for(i=array.length-2; i >= 0; i-=2) {
doubleArray= array[i]*2
}
console.log(doubleArray);
}
validateCred(valid1) //prints 8 I want it to print entire loop
Im trying to code luhn Algorithm and this is the first step of doubling every second number from the back, i want my result to equal doubleArray however when i try print it only 8 prints

you have only 8 prints because you have step i-2. Also, see comments in code about reversing an array
const valid1 = [4, 5, 3, 9, 6, 7, 7, 9, 0, 8, 0, 1, 6, 8, 0, 8];
function validateCred(array) {
let doubleArray = [];
// we should move through all iterations that is why step i-=1
for(let i = array.length-1; i >= 0; i -= 1) {
// moving from the end of the array we will get reversed copy if using method push()
// that is why changed to unshift()
doubleArray.unshift((array.length - 1 - i) % 2 == 0 ? array[i] * 2 : array[i]);
// array.length - 1 - i - using because we need to move from right in case length not odd
}
console.log(doubleArray);
}
// ES6+ syntax version version
function esValidateCred1(array) {
const doubleArray = array.reverse().map((i, ind) => ind % 2 === 0 ? i*2 : i).reverse();
console.log(doubleArray);;
}
validateCred(valid1);
esValidateCred1(valid1);

You can refer the code below to get the desired output i.e. the an array with doubled value starting from back.
var valid1=[4, 5, 3, 9, 6, 7, 7, 9, 0, 8, 0, 1, 6, 8, 0, 8]
function validateCred(array) {
let doubleArray = [];
for(i=array.length-2; i >= 0; i-=2) {
doubleArray.push(array[i] * 2)
}
return(doubleArray);
}
console.log(validateCred(valid1))

the easiest way is to use .map() method for array
var valid1=[4, 5, 3, 9, 6, 7, 7, 9, 0, 8, 0, 1, 6, 8, 0, 8]
function validateCred(array) {
return array.map(item => item*2);
}
console.log( validateCred(valid1) ) //prints array doubled elements

Related

why it is returning only false? Luhn algorithm

What's wrong with my code? It should return true for this array. The invalid one should return false.
Please explain it to me because i'm just started with JS
//arrays :
const valid1 = [4, 5, 3, 9, 6, 7, 7, 9, 0, 8, 0, 1, 6, 8, 0, 8];
const invalid1 = [4, 5, 3, 2, 7, 7, 8, 7, 7, 1, 0, 9, 1, 7, 9, 5];
const validateCred = Array => {
let cardNum = 0
let reverseArray = Array.reverse()
for (let i = 0; i < reverseArray.length; i++){
let newVar = reverseArray[i]
if (i%2 !== 0){
newVar = reverseArray[i] * 2
if (newVar > 9){
newVar = newVar[i] - 9;
cardNum += newVar
} else {
cardNum += newVar
}
} else {
cardNum += reverseArray[i]
}
}
return(cardNum%10 === 0 ? true : false)
}
console.log(validateCred(valid1))
As you figured out and noted in the comments, this is not going to go well when newVar is a number:
newVar = newVar[i] - 9;
And as Pointy, um, pointed out, Array is a terrible name for a variable, shadowing an important constructor function. More than that, there is a strong convention in JS that InitialCapital variable names are reserved for constructor functions. I would suggest a name that describes what it's for, not its type. Perhaps "creditCard" would be useful, or, depending on your tolerance for short abbreviations, "cc".
But there's another, more subtle, problem with this code. It alters its input:
const valid1 = [4, 5, 3, 9, 6, 7, 7, 9, 0, 8, 0, 1, 6, 8, 0, 8];
console .log (validateCred (valid1)) //=> true
console .log (valid1) //=> [8, 0, 8, 6, 1, 0, 8, 0, 9, 7, 7, 6, 9, 3, 5, 4]
In a real application, this could cause you all sorts of problems, and maybe far away from this section, always frustrating.
It's easy enough to fix. Just clone the array before reversing it. There are many ways to do it (using myVariable.slice() or myVariable.concat(), for instance.) My preference these days is to spread it into a new array: [...myVariable].
In my answer to another Luhn's Algorithm question, I developed what I think of as an elegant version of this algorithm. If you're new to JS, this may use some features you're not familiar with, but I find it clear and useful. This is a slightly improved version:
const luhn = (ds) => ([...ds]
.filter(d => /^\d$/ .test (d))
.reverse ()
.map (Number)
.map ((d, i) => i % 2 == 1 ? (2 * d > 9 ? 2 * d - 9 : 2 * d) : d)
.reduce ((a, b) => a + b, 0)
) % 10 == 0
It's the same algorithm, just expressed a little more concisely. Between the spreading of the initial value and the filter call (removing non-digits), it allows us to pass various input formats:
const valid1 = [4, 5, 3, 9, 6, 7, 7, 9, 0, 8, 0, 1, 6, 8, 0, 8];
const valid2 = "4539677908016808"
const valid3 = "4539-6779-0801-6808"
const valid4 = "4539 6779 0801 6808"

Get the 5 items before and after the current index in an array

Say I have an array that looks like this
arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20]
These array represents page numbers in my scenario.
Say if I am on page 8, I'd like to create a seperate array that includes the following
The first 5 before page 8 and the first 5 after page 8.
i.e first 11 items array.
[3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]
Therefore having 11 items in an array including that page itself.
If the array is smaller than 5, then simply return the rest.
i.e
if the array looks like this [2,3,4,5,6,7,8]
and if the page is 4, since the before does not have 5 items exactly I'd like to get all of it.
You can use the slice method of the Array.
function paginatorNumbers(arr, currentIndex) {
return arr.length > 5 && currentIndex > 5
? arr.slice(currentIndex - 6, currentIndex + 5)
: arr.slice(0, currentIndex + 5)
}
Use the slice function. Given x=8 in your example, the following will create a new array with the indexes you want.
arr.slice(x-5, x+6)
The start index x-5 is inclusive (or it will include index 8-5 = 3). The end index x+6 is exclusive (it will not include index 8+6=14). So you get indices 3 - 13 like you wanted.
EDITED:
This should work now
const getRange = (array, index) => {
// You need constraints if index - 5 is < 0
const startIndex = Math.max(0, index - 5 - 1);
const endIndex = index+ 5;
return array.slice(startIndex, endIndex);
}
You can makeit very simple with Plus and Minus.
arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20]
function getRange(current) {
let start = (current -5)
let end = (current + 5)
let res = []
start = start < 0 ? 0 : start
for(start; start <= end; start++) {
res.push(start)
}
return res;
}
console.log(getRange(8)) // starts from 3
console.log(getRange(2)) // starts from 0
Using the slice method, you can get a specific subarray from the whole array. What you need to do is to slice 5 places before the page index, and then slice 5 places after the page index. Then, just concat both subarrays and get the result.
The Math.min and Math.max functions are to avoid range problems when slicing the array. So, when the subarray is smaller than 5, it just returns the rest of the array, and not an empty array.
const arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20];
let currentPage = parseInt(prompt("Current page"));
// page index
let i = arr.indexOf(currentPage);
// Using Math.max and Math.min to avoid range problems
const subarr = arr.slice(Math.max(i - 5, 0), i).concat(
arr.slice(i, Math.min(i + 6, arr.length)));
console.log(subarr);
This is a universal solution
count of items before and after can be set
.slice() is used for timming the array
checks that start index is not < 0 (otherwise it would be sliced from the end)
const arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20]
function getBeforeAfter(arr, index, count) {
let start = index - count -1
if(start < 0) start = 0
let end = index + count
return arr.slice(start, end)
}
console.log(getBeforeAfter(arr,8,5)) // [3,4,5,6,7,8,9,10,11,12,13]
You can use array#slice to generate number in a range relative to index.
const arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20],
getRangeItem = (arr, index, range) => {
const start = Math.max(index - range - 1, 0),
end = Math.min(index + range, arr.length),
result = arr.slice(start, end);
return result;
}
console.log(getRangeItem(arr, 8, 5));

How to sort an array starting from the middle?

Considering, I have an array like this [..., n-2, n-1, n, n+1, n+2, ...]. I would like to sort it in this way [n, n+1, n-1, n+2, n-2,...] with n equals to the middle of my array.
For example:
Input:
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
Output:
[5, 6, 4, 7, 3, 8, 2, 9, 1, 0]
let arrayNotSorted = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9];
let positionMiddleArray = Math.trunc(arrayNotSorted.length / 2);
let arraySorted = [arrayNotSorted[positionMiddleArray]];
for(let i=1; i <= positionMiddleArray; i++){
if(arrayNotSorted[positionMiddleArray + i] !== undefined){
arraySorted.push(arrayNotSorted[positionMiddleArray + i]);
}
if(arrayNotSorted[positionMiddleArray - i] !== undefined){
arraySorted.push(arrayNotSorted[positionMiddleArray - i]);
}
}
console.log('Not_Sorted', arrayNotSorted);
console.log('Sorted', arraySorted);
What I have done works properly, but I would like to know if there is a better way or a more efficient way to do so ?
You could take a pivot value 5 and sort by the absolute delta of the value and the pivot values an sort descending for same deltas.
var array = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9],
pivot = 5;
array.sort((a, b) => Math.abs(a - pivot) - Math.abs(b - pivot) || b - a);
console.log(...array); // 5 6 4 7 3 8 2 9 1 0
You can do that in following steps:
Create an empty array for result.
Start the loop. Initialize i to the half of the length.
Loop backwards means decrease i by 1 each loop.
push() the element at current index to the result array first and the other corresponding value to the array.
function sortFromMid(arr){
let res = [];
for(let i = Math.ceil(arr.length/2);i>=0;i--){
res.push(arr[i]);
res.push(arr[arr.length - i + 1])
}
return res.filter(x => x !== undefined);
}
console.log(sortFromMid([0, 1, 2, 3, 4, 5, 6, 7, 8, 9]))
[5, 6, 4, 7, 3, 8, 2, 9, 1, 0]

Shift the position of an element in an array

const numbers = [1, 2, 3, 4, 5, 6, 7, 8, 9];
const output = move(numbers, 3, -5);
console.log(output);
function move(array, index, offset) {
const output = [...array];
const element = output.splice(index, 1)[0];
output.splice(index + offset, 0, element)
return output;
}
The first line is an array of numbers.
At the second line, when calling the move function, we pass three arguments.
First, is the array itself called numbers.
Secondly, the index of the number we are trying to move (in the example, we have index 3 so we are passing the number 4).
Finally, we have the offset set to -5. The negative sign means we are moving the number to the left. The 5 means 5 positions.
But as you can see, we only have 3 positions to the left of the number 4 before reaching the beginning of the array. In this case, we have to go to the end of the array and count backwards. So, we are looking for a function which will turn the original array to [1, 2, 3, 5, 6, 7, 8, 4, 9].
As you can see, number 4 has shifted 3 positions to the left to reach the beginning of the array, then, 2 further positions from the end of the array.
A further example to clarify.
Let's say we write:
const output = move(numbers, 1, -4);
In this example, we want the number 2 from the array (index 1) to move 4 positions to the left. So, we should get [1, 3, 4, 5, 6, 7, 2, 8, 9].
You need to cover the edge cases when the updated index is less than 0 OR greater than the array length. You can try following
const numbers = [1, 2, 3, 4, 5, 6, 7, 8, 9];
function move(array, index, offset) {
const output = [...array];
const element = output.splice(index, 1)[0];
let updatedIndex = index + offset;
if(updatedIndex < 0) updatedIndex++;
else if (updatedIndex >= array.length) updatedIndex -= array.length;
output.splice(updatedIndex, 0, element);
return output;
}
console.log(move(numbers, 3, -5));
You could do this using while loop and iterating for the Math.abs() of the position you want to move to and then move in direction depending if parameter is positive or negative.
function move(arr, i, p) {
let left = p < 0,
counter = Math.abs(p),
newPos = i;
while (--counter > -1) {
newPos = (left ? (newPos - 1) : (newPos + 1));
if (newPos == -1) newPos = arr.length - 1;
if (newPos == arr.length) newPos = 0;
if (counter == 0) arr.splice(newPos, 0, arr.splice(i, 1)[0])
}
return arr;
}
console.log(move([1, 2, 3, 4, 5, 6, 7, 8, 9], 3, -5));
console.log(move([1, 2, 3, 4, 5, 6, 7, 8, 9], 5, 5));
console.log(move([1, 2, 3, 4, 5, 6, 7, 8, 9], 1, -25));

Get the N elements from array based on the position

I want a function that returns the sub array which takes a position & the no. of elements I want. I think there may be some algorithm to find the pivot point or something & from that I can get the sub array, but I totally forgot it.
Example: a = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
I want 6 elements
if position = 0, then I want [1, 2, 3, 4, 5, 6]
if position = 1, then [1, 2, 3, 4, 5, 6]
if position = 2, then [1, 2, 3, 4, 5, 6]
if position = 3, then [1, 2, 3, 4, 5, 6]
if position = 4, then [2, 3, 4, 5, 6, 7]
if position = 5, then [3, 4, 5, 6, 7, 8]
if position = 6, then [4, 5, 6, 7, 8, 9]
if position = 7, then [5, 6, 7, 8, 9, 10]
if position = 8, then [5, 6, 7, 8, 9, 10]
if position = 9, then [5, 6, 7, 8, 9, 10]
simply get the middle of N elements based on the position I pass.
I can write up my own loop which will contain multiple if-else conditions to get it done. But I feel there may be some easy way to do it.
I didnt include my incomplete code snippet because I strongly feel there must be some algorithm to do this.
What you want is : Array.prototype.slice(...)
It's neatly documented here : https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/slice
var n = 6;
var start = Math.max(0, Math.min(Math.floor(position-n/2), a.length-n));
return a.slice(start, start+n);
Simple way:
var a = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
function getSubArray(idx, _length, _array) {
return _array.slice(idx, idx + _length);
}
var subArray = getSubArray(3, 6, a);
You could use an offset for the postion and get the the start value first for slicing.
var a = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10],
n = 6,
i,
start;
for (i = 1; i < 12; i++) {
start = Math.max(Math.min(i - n / 2, a.length - n), 0);
console.log(i, ': ', a.slice(start, start + n).join());
}
Your only need is to check if you are not gonna check a pos that doesn't exist. Like :
var a = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
var n = 6; // Number of result you want
var x = 8; // Pos you want
// If you gonna exceed your length, we got only the n last element
if((x+(n/2)) > a.length) {
console.log(a.slice(a.length-n));
// Otherwise, if under 0, we got the n first
} else
if((x-(n/2)) < 0) { console.log(a.slice(0,n) );
// Default case
} else {
console.log(a.slice((x-(n/2)),(x+(n/2))));
}
This is not the smartest way, but he can give you some hint. I used the slice as other mentionned to avoid a lot of if, but you should do GENERIC test.
Something like this :
a = [1,2,3,4,5,6,7,8,9,10];
n = 6;
function split(position) {
var start = Math.min(Math.max(position - Math.floor(n/2), 0), a.length - n);
var stop = Math.min(start+n, a.length);
return a.slice(start, stop);
}
No need for the Math object at all. You may simply do as follows;
function getArr(a,n,d){
n = n - 4 < 0 ? 0
: a.length - d > n - 4 ? n - 3
: a.length - d;
return a.slice(n,n + d);
}
var arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10],
diff = 6;
for (var i = 0; i < 10; i ++) console.log(JSON.stringify(getArr(arr,i,diff)));
no need of if-else you can use arr[position] to arr[8]. have you got
function getArr(arr,position,requiredNumbers){
return arr.slice(position, position+requiredNumbers);
}

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