Reputation: 3794
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.
Upvotes: 2
Views: 20543
Reputation: 109
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
}
Upvotes: 0
Reputation: 1
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>");
}
Upvotes: 0
Reputation: 11
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;
}
}
Upvotes: 0
Reputation: 21
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");
}
}
Upvotes: 0
Reputation: 61
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);
Upvotes: 0
Reputation: 31
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
Upvotes: 0
Reputation: 12155
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.
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;
}
Upvotes: 0
Reputation: 9
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);
}
Upvotes: 0
Reputation: 161
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");
Upvotes: 6
Reputation: 35
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));
Upvotes: 0
Reputation: 1433
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)
Upvotes: 0
Reputation: 3794
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>");
}
Upvotes: 1
Reputation: 844
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.
Upvotes: 2