Develop Reference

Fill matrix with snake pattern using javascript

I want to fill in a 2D array of numbers with snake pattern, I tried this algorithm but the result it's not the expected:
function fill2d(c,l) {
var arr = new Array[l][c];
let counter = 0;
for (let col = 0; col < arr.l; col++) {
if (col % 2 == 0) {
for (let row = 0; row < arr.l; row++) {
arr[row][col] = counter++;
}
} else {
for (let row = arr.length - 1; row >= 0; row--) {
arr[row][col] = counter++;
}
}
}
return arr;
}
fill2d(4,4)
some thing like that :
1 2 3 4
8 7 6 5
9 10 11 12
16 15 14 13

Okay so basically all you need is to traverse row-by-row but alternating between left-to-right and right-to-left. Here's my solution:
function fill2d(c, l) {
let arr = new Array(l);
for (var i = 0; i < l; i++) {
arr[i] = new Array(c);
}
let count = 1;
for (let i = 0; i < l; i++) {
console.log("i: ", i);
let right_to_left = i % 2;
for (let j = (right_to_left * (l - 1));
(right_to_left ? (j >= 0) : (j < c)); j += (right_to_left ? -1 : 1)) {
console.log("j: ", j);
arr[i][j] = count++;
}
}
return arr;
}
console.log(fill2d(4, 4));
Also, you weren't allocating a 2d array properly, you need to allocate a 1d array first and then allocate an array for each of the elements of that array.
Note: I added logs for each i and j so that you can see how the array is traversed.

Why does function reassign 2Darray argument as sideeffect from calling the latter

When I run the third line alone and log test2darr it returns a 2D array filled with 6's in a 3x3 matrix
But when I run the fourth line and log test2darr again, it returns:
[4, 5, 4]
​
[5, 6, 5]​
[4, 5, 4]
(as well as for secondtest)
Though it should return the same array of 6's for test2darr and on assign the 2d array to secondtest
const n = 3;
const filler = new Array(n * n);
const test2darr = fill2DarrFromArr(filler.fill(6));
const secondtest = pileReduce(test2darr);
Here is my code for fill2DarrFromArr and pileReduce:
function pileReduce(_cells) {
_cells = fillEmpty(_cells);
for (let j = 0; j < _cells.length; j++) { //The Algorithm itself is not important
for (let i = 0; i < _cells.length; i++) { // But there might be some assignment problem that I missed
if (_cells[j][i] >= 4) {
_cells[j][i] = _cells[j][i] - 4;
if (j !== _cells.length - 1) _cells[j + 1][i]++;
if (j !== 0) _cells[j - 1][i]++;
if (i !== _cells.length - 1) _cells[j][i + 1]++;
if (i !== 0) _cells[j][i - 1]++;
}
}
}
return _cells;
}
function fill2DarrFromArr(_arr) {
let sideLength = Math.sqrt(_arr.length);
let out = create2DArr(sideLength, sideLength);
for (let j = 0; j < sideLength; j++) {
for (let i = 0; i < sideLength; i++) {
out[j][i] = _arr[j * sideLength + i];
}
}
return out;
}
function create2DArr(_n, _m) {
let _arr = new Array(_n);
for (let j = 0; j < _m; j++) {
_arr[j] = new Array(_m);
}
return _arr;
}
function fillEmpty(_arr) {
for (let j = 0; j < _arr.length; j++) {
for (let i = 0; i < _arr.length; i++) {
if (!_arr[j][i]) _arr[j][i] = 0;
}
}
return _arr;
}

Passing an array into a function doesn't create a copy of that array. Your functions are modifying the contents of the passed arrays, therefore they have side effects.

How many times needed to Sort a number in an Array Javascript

I want to count how many times needed for an array to be sorted
var array = [4,2,3,1]
var yourCounter = 0;
for (var i = 0; i < array.length; i++) {
for (var j = 1; j < array.length-j; j++)
if (array[j - 1] > array[j]) {
yourCounter++;
} }
it will return 4 , it should be 5
but if I input array [1,2,3] will correctly return 0 , and if I input array [3,2,1] it will correctly return 3

You could take the given code and swap the values while counting.
for (int i = 0; i < n; i++) {
for (int j = 0; j < n - 1; j++) {
// Swap adjacent elements if they are in decreasing order
if (a[j] > a[j + 1]) {
swap(a[j], a[j + 1]);
}
}
}
var array = [4, 2, 3, 1],
counter = 0,
i, j, n = array.length;
for (i = 0; i < n; i++) {
for (j = 0; j < n - 1; j++) {
if (array[j] > array[j + 1]) {
[array[j + 1], array[j]] = [array[j], array[j + 1]];
++counter;
}
}
}
console.log(counter);
console.log(array);

I found the solution
var a = [4,2,3,1]
function sortArray(a){
let swapCount = 0;
let swapOccurred = true;
let index = 0;
while (swapOccurred == true && index < a.length){
swapOccurred == false;
if (a[index] > a[index+1]){
let holder = a[index]
a[index] = a[index+1];
a[index+1] = holder;
swapOccurred == true;
swapCount ++;
index = -1;
}
index ++
}
function countSwaps(a) {
let swapCount = 0;
[a, swapCount] = sortArray(a)
console.log(swapCount)
}
return [a, swapCount]
}

Creating a function to combine a nested array without recursion

I have the following array as an example;
let arr = [['red','blue','pink],['dog','cat','bird'],['loud', 'quiet']]
I need to write a generalized function that prints all combinations of one word from the first vector, one word from the second vector, etc. I looked up some codes on here but they are all recursion or working only with the specific array. How can I write this code without recursion?
let allComb = function(arr) {
if (arr.length == 1) {
return arr[0];
} else {
let result = [];
let arrComb = allComb(arr.slice(1));
for (let i = 0; i < arrComb.length; i++) {
for (let j = 0; j < arr[0].length; j++) {
result.push(arr[0][j] + ' ' + arrComb[i]);
}
}
return result;
}
}
allComb(arr)

This version uses a single increment per cycle technique with no recursion.
let arr = [
['red', 'blue', 'pink'],
['dog', 'cat', 'bird'],
['loud', 'quiet']
];
function allComb(arr) {
var total = 1;
var current = [];
var result = [];
for (var j = 0; j < arr.length; j++) {
total *= arr[j].length;
current[j] = 0;
}
for (var i = 0; i < total; i++) {
var inc = 1;
result[i] = "";
for (var j = 0; j < arr.length; j++) {
result[i] += arr[j][current[j]] + ' ';
if ((current[j] += inc) == arr[j].length)
current[j] = 0;
else
inc = 0;
}
}
return (result);
}
console.log(allComb(arr));

You may do as follows;
var arr = [['red','blue','pink'],['dog','cat','bird'],['loud', 'quiet']],
res = arr.reduce((p,c) => p.reduce((r,x) => r.concat(c.map(y => x + " " + y)),[]));
console.log(res);

How to program Pascal's Triangle in Javascript - confusion re Arrays

I'm having a little trouble with my attempt at this problem. Code Below:
function pasc(n){
var result = [[1]];
for (var row = 1; row < n; row++){
for (var col = 1; col <= row; col++){
result[row][col] = result[row - 1][col] + result[row - 1][col - 1];
}
}
return result;
}
pasc(10)
for (var i = 0; i < result.length; i++){
document.write(result[i]+"<br>");
}
It seems the problem hinges on assigning values to an array using an expression like myArray[1][1] = "foo"
I'm confused about this because I can do this: var myArray = []; myArray[4] = "foo" which seems to suggest that an element can be created at an arbitrary position in a 1 dimensional array, but not with 2 dimensions.
Any help with clearing up my misconceptions appreciated.

The Pascal's Triangle can be printed using recursion
Below is the code snippet that works recursively.
We have a recursive function pascalRecursive(n, a) that works up till the number of rows are printed. Each row is a element of the 2-D array ('a' in this case)
var numRows = 10,
triangle,
start,
stop;
// N is the no. of rows/tiers
// a is the 2-D array consisting of the row content
function pascalRecursive(n, a) {
if (n < 2) return a;
var prevRow = a[a.length-1];
var curRow = [1];
for (var i = 1; i < prevRow.length; i++) {
curRow[i] = prevRow[i] + prevRow[i-1];
}
curRow.push(1);
a.push(curRow);
return pascalRecursive(n-1, a); // Call the function recursively
}
var triangle = pascalRecursive(numRows, [[1]]);
for(var i = 0; i < triangle.length; i++)
console.log(triangle[i]+"\n");

JavaScript doesn't have two-dimensional arrays. What it does have is arrays that happen to contain other arrays. So, yes, you can assign a value to any arbitrary position in an array, and the array will magically make itself big enough, filling in any gaps with 'undefined'... but you can't assign a value to any position in a sub-array that you haven't explicitly created yet. You have to assign sub-arrays to the positions of the first array before you can assign values to the positions of the sub-arrays.
Replacing
for (var row = 1; row < n; row++){
for (var col = 1; col <= row; col++){
with
for (var row = 1; row < n; row++){
result[row] = [];
for (var col = 1; col <= row; col++){
should do it. Assuming all of your indexing logic is correct, anyway. You've got some problems there, too, since your initial array only contains a single value, so result[row][col] = result[row - 1][col] + result[row - 1][col - 1]; is accessing at least one cell that has never been defined.

Thanks Logan R. Kearsley. I have now solved it:
function pasc(n){
var result = [];
result[0] = [1];
result[1] = [1,1];
for (var row = 2; row < n; row++){
result[row] = [1];
for (var col = 1; col <= row -1; col++){
result[row][col] = result[row-1][col] + result[row-1][col-1];
result[row].push(1);
}
}
return result;
}
for (var i = 0; i < pasc(10).length; i++){
document.write(pasc(10)[i]+"<br>");
console.log(pasc(10)[i]+"<br>");
}

you can create Pascal's triangle using below code:
function pascal(n) {
var arr = [];
if (n == 1) {
arr[0] = [];
arr[0][0] = 1;
} else if (n == 2) {
arr[0] = [];
arr[0][0] = 1;
arr[1] = [];
arr[1][0] = 1;
arr[1][1] = 1;
} else if (n > 2) {
arr[0] = [];
arr[1] = [];
arr[0][0] = 1;
arr[1][0] = 1;
arr[1][1] = 1;
for (i = 2; i < n; i++) {
arr[i] = [];
arr[i][0] = 1;
for (j = 1; j < i; j++) {
arr[i][j] = arr[i - 1][j - 1] + arr[i - 1][j];
}
arr[i][j] = 1;
}
}
console.log(arr);
for (i = 0; i < arr.length; i++) {
console.log(arr[i].join(' '))
}
}
function pascal(n) {
var arr = [];
if (n == 1) {
arr[0] = [];
arr[0][0] = 1;
} else if (n == 2) {
arr[0] = [];
arr[0][0] = 1;
arr[1] = [];
arr[1][0] = 1;
arr[1][1] = 1;
} else if (n > 2) {
arr[0] = [];
arr[1] = [];
arr[0][0] = 1;
arr[1][0] = 1;
arr[1][1] = 1;
for (i = 2; i < n; i++) {
arr[i] = [];
arr[i][0] = 1;
for (j = 1; j < i; j++) {
arr[i][j] = arr[i - 1][j - 1] + arr[i - 1][j];
}
arr[i][j] = 1;
}
}
console.log(arr);
for (i = 0; i < arr.length; i++) {
console.log(arr[i].join(' '))
}
}
pascal(5)

This function will calculate Pascal's Triangle for "n" number of rows. It will create an object that holds "n" number of arrays, which are created as needed in the second/inner for loop.
function getPascalsTriangle(n) {
var arr = {};
for(var row = 0; row < n; row++) {
arr[row] = [];
for(var col = 0; col < row+1; col++) {
if(col === 0 || col === row) {
arr[row][col] = 1;
} else {
arr[row][col] = arr[row-1][col-1] + arr[row-1][col];
}
}
}
return arr;
}
console.log(getPascalsTriangle(5));

Floyd triangle
You can try the following code for a Floyd triangle
var prevNumber=1,i,depth=10;
for(i=0;i<depth;i++){
tempStr = "";j=0;
while(j<= i){
tempStr = tempStr + " " + prevNumber;
j++;
prevNumber++;
}
console.log(tempStr);
}

You can create arbitrary 2d arrays and store it in there and return the correct Pascal.
JavaScript does not have a special syntax for creating multidimensional arrays. A common workaround is to create an array of arrays in nested loops.
source
Here is my version of the solution
function pascal(input) {
var result = [[1], [1,1]];
if (input < 0) {
return [];
}
if (input === 0) {
return result[0];
}
for(var j = result.length-1; j < input; j++) {
var newArray = [];
var firstItem = result[j][0];
var lastItem = result[j][result[j].length -1];
newArray.push(firstItem);
for (var i =1; i <= j; i++) {
console.log(result[j][i-1], result[j][i]);
newArray.push(sum(result[j][i-1], result[j][i]));
}
newArray.push(lastItem);
result.push(newArray);
}
return result[input];
}
function sum(one, two) {
return one + two;
}

Here is the code i created for pascal triangle in javascript
'use strict'
let noOfCoinFlipped = 5
let probabiltyOfnoOfHead = 2
var dataStorer = [];
for(let i=0;i<=noOfCoinFlipped;i++){
dataStorer[i]=[];
for(let j=0;j<=i;j++){
if(i==0){
dataStorer[i][j] = 1;
}
else{
let param1 = (j==0)?0:dataStorer[i-1][j-1];
let param2 = dataStorer[i-1][j]?dataStorer[i-1][j]:0;
dataStorer[i][j] = param1+param2;
}
}
}
let totalPoints = dataStorer[noOfCoinFlipped].reduce((s,n)=>{return s+n;})
let successPoints = dataStorer[noOfCoinFlipped][probabiltyOfnoOfHead];
console.log(successPoints*100/totalPoints)
Here is the link as well
http://rextester.com/TZX59990

This is my solve:
function pascalTri(n){
let arr=[];
let c=0;
for(let i=1;i<=n;i++){
arr.push(1);
let len=arr.length;
if(i>1){
if(i>2){
for(let j=1;j<=(i-2);j++){
let idx=(len-(2*i)+j+2+c);
let val=arr[idx]+arr[idx+1];
arr.push(val);
}
c++;
}
arr.push(1);
}
}
return arr;
}
let pascalArr=pascalTri(7);
console.log(pascalArr);

here is the pattern for n = 3
#
##
###
here is js code to print this.
function staircase(n) {
for(var i=0 ; i<n ; i++) {
for(var j=n-1 ; j>i ; j--)
process.stdout.write(" ");
for(var k=0 ; k<=i; k++) {
process.stdout.write("#");
}
process.stdout.write("\n");
}
}

class PascalTriangle {
constructor(n) {
this.n = n;
}
factoriel(m) {
let result = 1;
if (m === 0) {
return 1;
}
while (m > 0) {
result *= m;
m--;
}
return result;
}
fill() {
let arr = [];
for (let i = 0; i < this.n; i++) {
arr.push([]);
}
for (let i = 0; i < arr.length; i++) {
for (let j = 0; j <= i; j++) {
arr[i].push(this.factoriel(i) / (this.factoriel(j) * this.factoriel(i - j)));
}
}
return arr;
}
}

var m = prompt("enter number:");
var arrMain = new Array();
for (var i = 0; i < m; i++) {
arrMain[i] = [];
}
for (var i = 0; i < m; i++) {
if (i == 0) {
arrMain[i] = [1];
} else if (i == 1) {
(arrMain[i]) = [1, 1];
} else {
for (var j = 0; j <= i; j++) {
if (j == 0 || j == arrMain[i - 1].length) {
arrMain[i][j] = 1;
} else {
arrMain[i][j] = arrMain[i - 1][j] + arrMain[i - 1][j - 1];
}
}
}
document.write(arrMain[i] + "<br>");
}

This is my take on this problem by gaining access to the previous row.
const generate = numRows => {
const triangle = [[1]]
for (let i = 1; i < numRows; i++) {
// Previous row
const previous = triangle[i - 1]
// Current row
const current = new Array(i + 1).fill(1)
// Populate the current row with the previous
// row's values
for (let j = 1; j < i; j++) {
current[j] = previous[j - 1] + previous[j]
}
// Add to triangle result
triangle.push(current)
}
return triangle
}