square of two arrays numbers in javascript - javascript

Function that compares and array with its numerically sorted version in ascending order. Return the sum of the square differences between each term of the two arrays.
Expample:
For the array a= [2,3,1], the sorted version is b=[1,2,3],
.the sum of the square differences is (2-1)2 + (3-2)2 + (1-3)2 = 6
. or in terms of array elements (a[0]-b[0])2 + (a[1]-b[1])2 + (a[2]+b[2])2 = 6
const arr = [2, 3, 1];
function myFunction(arr) {
var newarr = [];
var a = arr;
var b = arr;
var sum = 0;
b.sort(function(a, b) {
return a - b
});
for (var i = 0; i < b.length; i++) {
var sub = a[i] - b[i];
sub = sub * 2;
sum += sub;
newarr.push(sum);
}
return newarr;
}
console.log(myFunction(arr));

Assigning arrays doesn't make copies. So a, b, and arr are all the same array, and you're just subtracting a number from itself, always getting zero. You need to make a copy of the array in the other variable.
You're also not squaring correctly, it should be sub * sub, not sub * 2.
You're supposed to return sum. There's no need for newArr.
const arr = [2, 3, 1];
function myFunction(arr) {
var sorted = [...arr]; // make copy
sorted.sort(function(a, b) {
return a - b
});
var sum = 0;
for (var i = 0; i < sorted.length; i++) {
var sub = arr[i] - sorted[i];
sub = sub * sub;
sum += sub;
}
return sum;
}
console.log(myFunction(arr));

Related

Binary addition algorithm using quotient and reminder

I'm trying to implement the binary addition algorithm using javascript ,
the problem is when i console.log the function it returns nothing
here is my code:
function max(a,b)
{
if(a>b)
return a;
return b;
}
function addb(A,B)
{
var result=[]
var l=max(A.length,B.length)
var carry=0
var sum=0;
for(var i=l.length-1;i>=0;i--)
{
sum=A[i]+B[i]+carry;
carry=Math.floor(sum/2);
result[i]=sum%2;
}
return result;
}
var a=[1,1,0];
var b=[1,1,1];
console.log(addb(a,b));
You're passing ints into the max function and then treating it like it's an array
e.g
var l=max(A.length,B.length)
....
for(var i=*l.length-1*;i>=0;i--)
just put the arrays in and it works.
var l=max(A,B)
You need the length l without length property and a final check of carry and unshift the array with this value.
function max(a, b) {
return a > b ? a: b;
}
function addb(A, B) {
var result = [],
l = max(A.length, B.length),
carry = 0,
sum = 0;
for (var i = l - 1; i >= 0; i--) {
sum = A[i] + B[i] + carry;
carry = Math.floor(sum / 2);
result[i] = sum % 2;
}
if (carry) result.unshift(carry); // the most left value!
return result;
}
var a = [1, 1, 0];
var b = [1, 1, 1];
console.log(addb(a, b));
l variable is a number already not an array hence just use that number inside the for loop as the starting case
function max(a, b) {
if (a > b)
return a;
return b;
}
function addb(A, B) {
var result = []
var l = max(A.length, B.length)
var carry = 0
var sum = 0;
for (var i = l- 1; i >= 0; i--) {
sum = A[i] + B[i] + carry;
carry = Math.floor(sum / 2);
result[i] = sum % 2;
}
return result;
}
var a = [1, 1, 0];
var b = [1, 1, 1];
console.log(addb(a, b))

remove NaN in my array display

I want to ask help how to remove NaN value in my array. Every time I hit submit button without a data it will display a NaN value.
Variables for number sort :
var array = [];
var arrayasc = [];
var arraydsc = [];
function myFunction(){
var newArray = parseInt(document.getElementById("number").value);
array.push(newArray);
arrayasc.push(newArray);
arraydsc.push(newArray);
if
//entered numbers dislay here
document.getElementById("demo").innerHTML = array;
console.log(array);
//entered numbers in ascending order
mySort_asc(arrayasc);
document.getElementById("asc").innerHTML = array;
console.log(mySort_asc(arrayasc));
//entered numbers in descending order
mySort_dsc(arraydsc);
document.getElementById("dsc").innerHTML = array;
console.log(mySort_dsc(arraydsc));
}
Function to sort numbers in ascending order
function mySort_asc(arr){
for(var i = 0; i < array.length; i++) {
var temp = array[i];
var b = i - 1;
while (b >= 0 && array[b] > temp) {
array[b + 1] = array[b];
b--;
}
array[b + 1] = temp;
}
return array;
}
Function to sort numbers in descending order
function mySort_dsc(arr){
for(var i = 0; i < array.length; i++) {
var temp = array[i];
var b = i - 1;
while (b >= 0 && array[b] < temp) {
array[b + 1] = array[b];
b--;
}
array[b + 1] = temp ;
}
return array;
}
Your syntax is invalid
if
//entered numbers dislay here
document.getElementById("demo").innerHTML = array;
but simply check to see if the new value is a number or not, and if it isn't, return immediately without changing anything in the HTML:
function myFunction(){
var newArray = parseInt(document.getElementById("number").value);
if (Number.isNaN(newArray)) return;
You also might consider naming your variables to correspond to what they represent. For example, the newArray is actually just a single number, not an array. Perhaps call it newValue, for ease of reading and debugging?

Given an array of integers, find the pair of adjacent elements that has the largest product and return that product

Given an array of integers, find the pair of adjacent elements that has the largest product and return that product.
and here is my code
function adjacentElementsProduct(inputArray) {
var arr = inputArray;
var x=0;
var y=0;
var p=0;
for(var i=0;i<arr.length;i++){
x=arr[i];
y=arr[i+1];
if(x*y>p){
p=x*y;
};
};
return p;
};
the problem is all the tests works fine but except the array with the negative product as it shown in the attached photo
can anyone help .. and thanks in advance
You could start with a really large negative value, instead of zero.
var p = -Infinity;
You are initializing the variable p to zero. That means any multiplication values smaller than that are not accepted. Rather set it to the smallest possible integer value:
var p = Number.MIN_SAFE_INTEGER;
function adjacentElementsProduct(inputArray) {
var arr = inputArray;
var x = 0;
var y = 0;
var p = Number.MIN_SAFE_INTEGER;
for (var i = 0; i < arr.length; i++) {
x = arr[i];
y = arr[i + 1];
if (x * y > p) {
p = x * y;
};
};
return p;
};
console.log(adjacentElementsProduct([-23, 4, -3, 8, -12]));
This is quite simple actually
function adjacentElementsProduct(inputArray) {
let max = -Infinity;
for (let i = 1; i < inputArray.length; i++) {
max = Math.max(inputArray[i] * inputArray[i - 1], max);
}
return max;
}
This is quite simple actually
const solution = (inputArray) => Math.max(...inputArray.slice(0, -1).map((n, index) => n * inputArray[index + 1]))
console.log(solution([3, 6, -2, -5, 7, 3]))
function solution(inputArray: number[]): number {
var max = -Infinity;
for(var i=0; i+1<inputArray.length; i++)
{
if(max<(inputArray[i]*inputArray[i+1])){
max=inputArray[i]*inputArray[i+1];
}
}
return max;
}
console.log(solution([2,3,6]))
I had the same problem at first, defining the first max as 0. Then i came up with this:
function solution(inputArray) {
let products = inputArray.map(function(x, index){
return inputArray[index+1] != undefined? x *inputArray[index+1] : -Infinity;
})
return Math.max(...products);
}
Problem:
Given an array of integers, find the pair of adjacent elements that has the largest product and return that product. #javascript #arraymethods
function solution(inputArray) {
let productsArr = []; // to hold the products of adjacent elements
let n = 0;
for (let i = 0; i < inputArray.length; i++) {
if (i < inputArray.length - 1)
{
productsArr[n] = inputArray[i] * inputArray[i + 1];
n++;
}
}
return productsArr.reduce((aggr, val) => Math.max(aggr, val)); // to find out the biggest product
}
Here's a very simple implementation without using any additional variables (actually less), and no special values. Just simple logic.
function adjacentElementsProduct(inputArray) {
var c =inputArray[0]*inputArray[1];
var p = c;
for(var i=1;i<inputArray.length;i++){
console.log(c);
var c=inputArray[i]*inputArray[i+1];
if(c > p){
p=c;
};
};
return p;
};
console.log("minimum product = " + adjacentElementsProduct([-23,4,-3,8,-12]));
What I did was, initialize a variable c (current product) with the product of first two elements of the array. And then I declared the variable p and initialize it to c. This way, all other products are compared to this product. Rest is simple.
Hope it helps. :)
you can try to initialize a integer as negative infinity value -math.inf and then use the python ternary operator var=true if condition else false to find the maximum value
code in python
def adjacentarray(a):
maximum=-math.inf
for i,in range(0,len(a)-1):
maximum=a[i]*a[i+1] if a[i]*a[i+1]>maximum else maximum
return maximum
code in javascript
function adjacentElementsProduct(a) {
var maximum=-Infinity;
for (var i=0;i<a.length-1;i++){
maximum= a[i]*a[i+1]>maximum?a[i]*a[i+1]:maximum;
}
return maximum;
}
function solution(inputArray) {
let first, second, sum = []
inputArray.map((arr,index)=>{
first = arr;
second = inputArray[index+1]
if(second == undefined){
return second
}
return sum.push(first * second)
})
let last = sum.sort().reduce((pre,next)=> {
return pre > next ? pre : next
})
return last;
}
//Kotlin
fun solution(inputArray: MutableList<Int>): Int {
var result: Int = Int.MIN_VALUE
for (i in 0..inputArray.size - 2) {
if (inputArray[i] * inputArray[i + 1] > result)
result = inputArray[i] * inputArray[i + 1]
}
return result
}
import 'dart:math';
int solution(List<int> inputArray) {
//assumption for highest number
int highestNumber = inputArray[0] * inputArray[1] ;
//we'll go through the array to campare the highestNumber
//with next index
for(var i = 1 ; i < inputArray.length ; i++){
highestNumber = max(highestNumber, inputArray[i] * inputArray[i - 1]);
}
return highestNumber;
}
In Javascript, you could use the reduce method from an array to avoid iterating in a for loop, just like this.
function solution(inputArray) {
let maxProd = []
inputArray.reduce((accumulator, currentValue) => {
maxProd.push(accumulator*currentValue)
return currentValue
},
);
return Math.max(...maxProd)
}
Once you have in the maxProd array the products, you use the spread operator to get the numbers and using Math.max() you get the largest
python solution
You can make a loop from 1 to end of your list and do the following arithmetic operations
def solution(inputArray):
list1 =[]
for i in range(1,len(inputArray)):
list1.append(inputArray[i]*inputArray[i-1])
return max(list1)
Here is a solution in PHP that is quite simple.
function solution($inputArray) {
$largest = null;
$pos = null;
for($i = 0; $i < count($inputArray) -1; $i++){
$pos = ($inputArray[$i] * $inputArray[$i+1]);
if($largest < $pos){
$largest = $pos;
}
}
return $largest ?? 0;
}
You can try to create a new array of length (arr.length-1) inside the function and append the products of adjacent numbers to this new array. Then find the largest number in the array and return it. This will solve the problem with negative product.
function adjacentElementsProduct(inputArray) {
var arr = inputArray;
var prodArr[];
var p;
for (var i = 0; i < arr.length-1; i++) {
prodArr[i] = arr[i]*arr[i+1];
};
for (j=prodArr.length; j--){
if (prodArr[j] > p) {
p = prodArr[j];
};
return p;
};
console.log(adjacentElementsProduct([-23, 4, -3, 8, -12]));
The var p which saves the max product should be initialized as small as possible instead of a 0. So that when the product is negative, it will still meet the if condition and save the value.
Here is a C# solution:
static void Main(string[] args)
{
int[] arr = { 1, -4, 3, -6, -7, 0 };
Console.WriteLine(FindMaxProduct(arr));
Console.ReadKey();
}
static int FindMaxProduct(int[] arr) {
int currentProduct = 0;
int maxProduct = int.MinValue;
int a=0, b = 0;
for (int i = 0, j = i + 1; i < arr.Length - 1 && j < arr.Length; i++, j++)
{
currentProduct = arr[i] * arr[j];
if (currentProduct>maxProduct) {
a = arr[i];
b = arr[j];
maxProduct = currentProduct;
}
}
Console.WriteLine("The max product is {0}, the two nums are {1} and {2}.",maxProduct,a,b);
return maxProduct;
}
function solution(inputArray) {
let f, s, arr = []
for(let i=0; i<inputArray.length; i++){
f = inputArray[i]
s = inputArray[i+1]
arr.push(f*s)
}
let max = arr.sort((a, b) => b - a)
return max[0]
}
console.log(solution([3, 6, -2, -5, 7, 3]))
This should help, wrote it in python. Concept: Pass an empty list, for every consecutive product keep storing it in the list. Then just return the max value.
def consecutive_product_max(a):
lst2 = []
for i in range(0, len(a)-1):
x = a[i] * a[i+1]
lst2.append(x)
return max(lst2)

Javascript array store each value in variable

How can I store each value of javascript array in variable such that I can use it to perform further actions?
Eg:
var plan = [100, 200];
var months = [3, 6];
for (var i = 0; i < plan.length; i++) {
p_det = plan[i];
}
for (var i = 0; i < months.length; i++) {
m_det = months[i];
}
console.log(p_det * m_det); //Gives me value of 200*6
Mathematical action I've to perform such that ((100 * 3) + (200 * 6))
Is storing each value in variable will help? How can I achieve this?
Storing each element in a variable won't help, you already can access those values using the array[index]. Assuming your arrays are the same length, you can calculate what you want in a loop:
for (var sum = 0, i = 0; i < plan.length; i++) {
sum += plan[i]*month[i];
}
console.log( sum );
you can achieve it with reduce assuming that both arrays have same length
var plan = [100, 200];
var months = [3, 6];
var sum = plan.reduce(function(sum, val, index) {
return sum + months[index] * val;
}, 0);
snippet.log(sum)
<script src="http://tjcrowder.github.io/simple-snippets-console/snippet.js"></script>
A simple while loop will do. It works for any length, because of the Math.min() method for the length.
function s(a, b) {
var i = Math.min(a.length, b.length),
r = 0;
while (i--) {
r += a[i] * b[i];
}
return r;
}
document.write(s([100, 200], [3, 6]));
In ES6:
sum(plan.map((e, i) => e * months[i])
where sum is
function sum(a) { return a.reduce((x, y) => x + y); }

JavaScript - Generating combinations from n arrays with m elements [duplicate]

This question already has answers here:
Cartesian product of multiple arrays in JavaScript
(35 answers)
Closed 1 year ago.
I'm having trouble coming up with code to generate combinations from n number of arrays with m number of elements in them, in JavaScript. I've seen similar questions about this for other languages, but the answers incorporate syntactic or library magic that I'm unsure how to translate.
Consider this data:
[[0,1], [0,1,2,3], [0,1,2]]
3 arrays, with a different number of elements in them. What I want to do is get all combinations by combining an item from each array.
For example:
0,0,0 // item 0 from array 0, item 0 from array 1, item 0 from array 2
0,0,1
0,0,2
0,1,0
0,1,1
0,1,2
0,2,0
0,2,1
0,2,2
And so on.
If the number of arrays were fixed, it would be easy to make a hard coded implementation. But the number of arrays may vary:
[[0,1], [0,1]]
[[0,1,3,4], [0,1], [0], [0,1]]
Any help would be much appreciated.
Here is a quite simple and short one using a recursive helper function:
function cartesian(...args) {
var r = [], max = args.length-1;
function helper(arr, i) {
for (var j=0, l=args[i].length; j<l; j++) {
var a = arr.slice(0); // clone arr
a.push(args[i][j]);
if (i==max)
r.push(a);
else
helper(a, i+1);
}
}
helper([], 0);
return r;
}
Usage:
cartesian([0,1], [0,1,2,3], [0,1,2]);
To make the function take an array of arrays, just change the signature to function cartesian(args) instead of using rest parameter syntax.
I suggest a simple recursive generator function:
// JS
function* cartesianIterator(head, ...tail) {
const remainder = tail.length ? cartesianIterator(...tail) : [[]];
for (let r of remainder) for (let h of head) yield [h, ...r];
}
// get values:
const cartesian = items => [...cartesianIterator(items)];
console.log(cartesian(input));
// TS
function* cartesianIterator<T>(items: T[][]): Generator<T[]> {
const remainder = items.length > 1 ? cartesianIterator(items.slice(1)) : [[]];
for (let r of remainder) for (let h of items.at(0)!) yield [h, ...r];
}
// get values:
const cartesian = <T>(items: T[][]) => [...cartesianIterator(items)];
console.log(cartesian(input));
You could take an iterative approach by building sub arrays.
var parts = [[0, 1], [0, 1, 2, 3], [0, 1, 2]],
result = parts.reduce((a, b) => a.reduce((r, v) => r.concat(b.map(w => [].concat(v, w))), []));
console.log(result.map(a => a.join(', ')));
.as-console-wrapper { max-height: 100% !important; top: 0; }
After doing a little research I discovered a previous related question:
Finding All Combinations of JavaScript array values
I've adapted some of the code from there so that it returns an array of arrays containing all of the permutations:
function(arraysToCombine) {
var divisors = [];
for (var i = arraysToCombine.length - 1; i >= 0; i--) {
divisors[i] = divisors[i + 1] ? divisors[i + 1] * arraysToCombine[i + 1].length : 1;
}
function getPermutation(n, arraysToCombine) {
var result = [],
curArray;
for (var i = 0; i < arraysToCombine.length; i++) {
curArray = arraysToCombine[i];
result.push(curArray[Math.floor(n / divisors[i]) % curArray.length]);
}
return result;
}
var numPerms = arraysToCombine[0].length;
for(var i = 1; i < arraysToCombine.length; i++) {
numPerms *= arraysToCombine[i].length;
}
var combinations = [];
for(var i = 0; i < numPerms; i++) {
combinations.push(getPermutation(i, arraysToCombine));
}
return combinations;
}
I've put a working copy at http://jsfiddle.net/7EakX/ that takes the array you gave earlier ([[0,1], [0,1,2,3], [0,1,2]]) and outputs the result to the browser console.
const charSet = [["A", "B"],["C", "D", "E"],["F", "G", "H", "I"]];
console.log(charSet.reduce((a,b)=>a.flatMap(x=>b.map(y=>x+y)),['']))
Just for fun, here's a more functional variant of the solution in my first answer:
function cartesian() {
var r = [], args = Array.from(arguments);
args.reduceRight(function(cont, factor, i) {
return function(arr) {
for (var j=0, l=factor.length; j<l; j++) {
var a = arr.slice(); // clone arr
a[i] = factor[j];
cont(a);
}
};
}, Array.prototype.push.bind(r))(new Array(args.length));
return r;
}
Alternative, for full speed we can dynamically compile our own loops:
function cartesian() {
return (cartesian.cache[arguments.length] || cartesian.compile(arguments.length)).apply(null, arguments);
}
cartesian.cache = [];
cartesian.compile = function compile(n) {
var args = [],
indent = "",
up = "",
down = "";
for (var i=0; i<n; i++) {
var arr = "$"+String.fromCharCode(97+i),
ind = String.fromCharCode(105+i);
args.push(arr);
up += indent+"for (var "+ind+"=0, l"+arr+"="+arr+".length; "+ind+"<l"+arr+"; "+ind+"++) {\n";
down = indent+"}\n"+down;
indent += " ";
up += indent+"arr["+i+"] = "+arr+"["+ind+"];\n";
}
var body = "var res=[],\n arr=[];\n"+up+indent+"res.push(arr.slice());\n"+down+"return res;";
return cartesian.cache[n] = new Function(args, body);
}
var f = function(arr){
if(typeof arr !== 'object'){
return false;
}
arr = arr.filter(function(elem){ return (elem !== null); }); // remove empty elements - make sure length is correct
var len = arr.length;
var nextPerm = function(){ // increase the counter(s)
var i = 0;
while(i < len)
{
arr[i].counter++;
if(arr[i].counter >= arr[i].length){
arr[i].counter = 0;
i++;
}else{
return false;
}
}
return true;
};
var getPerm = function(){ // get the current permutation
var perm_arr = [];
for(var i = 0; i < len; i++)
{
perm_arr.push(arr[i][arr[i].counter]);
}
return perm_arr;
};
var new_arr = [];
for(var i = 0; i < len; i++) // set up a counter property inside the arrays
{
arr[i].counter = 0;
}
while(true)
{
new_arr.push(getPerm()); // add current permutation to the new array
if(nextPerm() === true){ // get next permutation, if returns true, we got them all
break;
}
}
return new_arr;
};
Here's another way of doing it. I treat the indices of all of the arrays like a number whose digits are all different bases (like time and dates), using the length of the array as the radix.
So, using your first set of data, the first digit is base 2, the second is base 4, and the third is base 3. The counter starts 000, then goes 001, 002, then 010. The digits correspond to indices in the arrays, and since order is preserved, this is no problem.
I have a fiddle with it working here: http://jsfiddle.net/Rykus0/DS9Ea/1/
and here is the code:
// Arbitrary base x number class
var BaseX = function(initRadix){
this.radix = initRadix ? initRadix : 1;
this.value = 0;
this.increment = function(){
return( (this.value = (this.value + 1) % this.radix) === 0);
}
}
function combinations(input){
var output = [], // Array containing the resulting combinations
counters = [], // Array of counters corresponding to our input arrays
remainder = false, // Did adding one cause the previous digit to rollover?
temp; // Holds one combination to be pushed into the output array
// Initialize the counters
for( var i = input.length-1; i >= 0; i-- ){
counters.unshift(new BaseX(input[i].length));
}
// Get all possible combinations
// Loop through until the first counter rolls over
while( !remainder ){
temp = []; // Reset the temporary value collection array
remainder = true; // Always increment the last array counter
// Process each of the arrays
for( i = input.length-1; i >= 0; i-- ){
temp.unshift(input[i][counters[i].value]); // Add this array's value to the result
// If the counter to the right rolled over, increment this one.
if( remainder ){
remainder = counters[i].increment();
}
}
output.push(temp); // Collect the results.
}
return output;
}
// Input is an array of arrays
console.log(combinations([[0,1], [0,1,2,3], [0,1,2]]));
You can use a recursive function to get all combinations
const charSet = [["A", "B"],["C", "D", "E"],["F", "G", "H", "I"]];
let loopOver = (arr, str = '', final = []) => {
if (arr.length > 1) {
arr[0].forEach(v => loopOver(arr.slice(1), str + v, final))
} else {
arr[0].forEach(v => final.push(str + v))
}
return final
}
console.log(loopOver(charSet))
This code can still be shorten using ternary but i prefer the first version for readability 😊
const charSet = [["A", "B"],["C", "D", "E"],["F", "G", "H", "I"]];
let loopOver = (arr, str = '') => arr[0].map(v => arr.length > 1 ? loopOver(arr.slice(1), str + v) : str + v).flat()
console.log(loopOver(charSet))
Another implementation with ES6 recursive style
Array.prototype.cartesian = function(a,...as){
return a ? this.reduce((p,c) => (p.push(...a.cartesian(...as).map(e => as.length ? [c,...e] : [c,e])),p),[])
: this;
};
console.log(JSON.stringify([0,1].cartesian([0,1,2,3], [[0],[1],[2]])));

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