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
}
If I have two arrays as parameters how can I find the starting index where the second parameter occurs as a sub-array in the array given as the first parameter.
E.g.: [5,9,3,6,8], [3,6] should return 2.
Is there a function in JavaScript for this, or does it just loop through both of them and compare?
findArrayInArray = function(a, b) {
var ai = a.length
, bi = b.length;
for(var i=0; i<ai; i++) {
if (a[i] === b[0]) {
if(bi === 1) return i;
for(var x=1; x<bi; x++) {
if(a[i+x] === b[x]) {
if(x === bi-1) return i;
} else {
break;
}
}
}
}
}
var arr1 = [5,9,3,6,8];
var arr2 = [3,6];
console.log(findArrayInArray(arr1,arr2)); // 2
http://jsfiddle.net/ymC8y/3/
In direct answer to your question, there is no built in function in JS to look in an array for a sub-array.
You will have to do some sort of brute force looping search like this or use some external library function that already has array comparison logic. Here's what a brute force solution in plain JS looks like:
function findSubArrayIndex(master, sub) {
for (var m = 0; m < master.length - sub.length + 1; m++) {
for (var s = 0; s < sub.length; s++) {
if (master[m + s] !== sub[s]) {
break;
} else if (s === sub.length - 1) {
return m;
}
}
}
return -1;
}
Working demo: http://jsfiddle.net/jfriend00/mt8WG/
FYI, here's a somewhat performance optimized version of this function:
function findSubArrayIndex(master, sub) {
var subLen = sub.length, subFirst, m, mlen;
if (subLen > 1) {
subFirst = sub[0];
for (m = 0, mlen = master.length - subLen + 1; m < mlen; m++) {
if (master[m] === subFirst) {
for (var s = 1; s < subLen; s++) {
if (master[m + s] !== sub[s]) {
break;
} else if (s === subLen - 1) {
return m;
}
}
}
}
} else if (subLen === 1) {
subFirst = sub[0];
for (m = 0, mlen = master.length; m < mlen; m++) {
if (master[m] === subFirst) {
return m;
}
}
}
return -1;
}
Working demo: http://jsfiddle.net/jfriend00/CGPtX/
function index (a, b) {
var as = new String(a),
bs = new String(b),
matchIndex = as.indexOf(bs);
if (matchIndex === -1) {
return -1;
} else if (matchIndex === 0) {
return 0;
}
return as.substring(0, matchIndex + 1).match(/,/g).length;
}
console.log(index([5,9,3,6,8], [3, 6]));
Try this - You loop through both arrays and compare each element:
var arr1 = [5,9,3,6,8];
var arr2 = [3,6];
findArrayInArray = function(arr1, arr2) {
for(var i=0; i<arr1.length; i++) {
for(var j=0; j<arr2.length; j++){
if(arr1[i] === arr2[j]){
return i;
}
}
}
return false;
}
findArrayInArray(arr1, arr2);
I have a strange problem with my alghoritm, which work if array size less than 114468 and doesn't work if more than 114468. Browse with google chrome. Can't understand why =\ Here is the code:
Generate array:
var arr = [];
var res = [];
for (var i = 114467; i > 0; i--) {
arr.push([i - 1, i]);
}
Find first elem in array to sort:
for (var i = 0, j = arr.length; i < j && res.length == 0; i++) {
var found = false;
for (var m = 0; m < j; m++) {
if (i == m || arr[i][0] == arr[m][1] || arr[i][1] == arr[m][0]) {
found = true;
break;
}
if (!found) {
res.push(arr[m]);
arr.splice(m, 1);
}
}
}
Sorting:
do {
for (var i = 0, j = arr.length; i < j; i++) {
var resLength = res.length - 1;
if (arr[i][1] == res[resLength][0] || arr[i][0] == res[resLength][1]) {
res.push(arr[i]);
arr.splice(i, 1);
break;
}
}
} while (arr.length > 0);
On the step sorting it stops to work.
All code:
var t = function () {
var arr = [];
var res = [];
for (var i = 114467; i > 0; i--) {
arr.push([i - 1, i]);
}
var startsec = new Date().getSeconds();
var startmilsec = new Date().getMilliseconds();
document.write(startsec + '.' + startmilsec + '<br>');
for (var i = 0, j = arr.length; i < j && res.length == 0; i++) {
var found = false;
for (var m = 0; m < j; m++) {
if (i == m || arr[i][0] == arr[m][1] || arr[i][1] == arr[m][0]) {
found = true;
break;
}
if (!found) {
res.push(arr[m]);
arr.splice(m, 1);
}
}
}
do {
for (var i = 0, j = arr.length; i < j; i++) {
var resLength = res.length - 1;
if (arr[i][1] == res[resLength][0] || arr[i][0] == res[resLength][1]) {
res.push(arr[i]);
arr.splice(i, 1);
break;
}
}
} while (arr.length > 0);
var stopsec = new Date().getSeconds();
var stopmilsec = new Date().getMilliseconds();
document.write(stopsec + '.' + stopmilsec + '<br>');
var executionTime = (stopsec - startsec).toString() + "s" + (stopmilsec - startmilsec).toString() + "'ms";
document.write(executionTime + '<br>');
} ();
Do i get my memory limit?
Alright, I isolated the problem. It seems that splice(0,1) slows down astronomically when the array size increases from 114467 to 114468.
Using this custom benchmark:
var t;
function startBench(){t=new Date().getTime();}
function stopBench(){console.log(new Date().getTime()-t);}
var arr=[];
for (var i = 114467; i > 0; i--) {
arr.push([i - 1, i]);
}
var arr2=[];
for (var i = 114468; i > 0; i--) {
arr2.push([i - 1, i]);
}
startBench();
for(i=0;i<1000;i++){
arr.splice(0,1);
}
stopBench();
startBench();
for(i=0;i<1000;i++){
arr2.splice(0,1);
}
stopBench();
I get 3 ms for 114467 and 2740ms for 114468 on Chrome (1000 iterations), but 170 each on Firefox. Maybe you ought to be using a different way to remove elements? Using a variant of bubble sort may work better.
I've submitted a bug report on this. Looking at the reply, it seems to be a valid bug. Hopefully it'll be fixed.
What is a fast and simple implementation of interleave:
console.log( interleave([1,2,3,4,5,6] ,2) ); // [1,4,2,5,3,6]
console.log( interleave([1,2,3,4,5,6,7,8] ,2) ); // [1,5,2,6,3,7,4,8]
console.log( interleave([1,2,3,4,5,6] ,3) ); // [1,3,5,2,4,6]
console.log( interleave([1,2,3,4,5,6,7,8,9],3) ); // [1,4,7,2,5,8,3,6,9]
This mimics taking the array and splitting it into n equal parts, and then shifting items off the front of each partial array in sequence. (n=2 simulates a perfect halving and single shuffle of a deck of cards.)
I don't much care exactly what happens when the number of items in the array is not evenly divisible by n. Reasonable answers might either interleave the leftovers, or even "punt" and throw them all onto the end.
function interleave( deck, step ) {
var copyDeck = deck.slice(),
stop = Math.floor(copyDeck.length/step),
newDeck = [];
for (var i=0; i<step; i++) {
for (var j=0; j<stop; j++) {
newDeck[i + (j*step)] = copyDeck.shift();
}
}
if(copyDeck.length>0) {
newDeck = newDeck.concat(copyDeck);
}
return newDeck;
}
It could be done with a counter instead of shift()
function interleave( deck, step ) {
var len = deck.length,
stop = Math.floor(len/step),
newDeck = [],
cnt=0;
for (var i=0; i<step; i++) {
for (var j=0; j<stop; j++) {
newDeck[i + (j*step)] = deck[cnt++];
}
}
if(cnt<len) {
newDeck = newDeck.concat(deck.slice(cnt,len));
}
return newDeck;
}
And instead of appending the extras to the end, we can use ceil and exit when we run out
function interleave( deck, step ) {
var copyDeck = deck.slice(),
stop = Math.ceil(copyDeck.length/step),
newDeck = [];
for (var i=0; i<step; i++) {
for (var j=0; j<stop && copyDeck.length>0; j++) {
newDeck[i + (j*step)] = copyDeck.shift();
}
}
return newDeck;
}
can i has prize? :-D
function interleave(a, n) {
var i, d = a.length + 1, r = [];
for (i = 0; i < a.length; i++) {
r[i] = a[Math.floor(i * d / n % a.length)];
}
return r;
}
according to my tests r.push(... is faster than r[i] = ... so do with that as you like..
note this only works consistently with sets perfectly divisible by n, here is the most optimized version i can come up with:
function interleave(a, n) {
var i, d = (a.length + 1) / n, r = [a[0]];
for (i = 1; i < a.length; i++) {
r.push(a[Math.floor(i * d) % a.length]);
}
return r;
}
O(n-1), can anyone come up with a log version? to the mathmobile! [spinning mathman logo]
Without for loops (I've added some checkup for the equal blocks):
function interleave(arr, blocks)
{
var len = arr.length / blocks, ret = [], i = 0;
if (len % 1 != 0) return false;
while(arr.length>0)
{
ret.push(arr.splice(i, 1)[0]);
i += (len-1);
if (i>arr.length-1) {i = 0; len--;}
}
return ret;
}
alert(interleave([1,2,3,4,5,6,7,8], 2));
And playground :) http://jsfiddle.net/7tC9F/
how about functional with recursion:
function interleave(a, n) {
function f(a1, d) {
var next = a1.length && f(a1.slice(d), d);
a1.length = Math.min(a1.length, d);
return function(a2) {
if (!a1.length) {
return false;
}
a2.push(a1.shift());
if (next) {
next(a2);
}
return true;
};
}
var r = [], x = f(a, Math.ceil(a.length / n));
while (x(r)) {}
return r;
}
Phrogz was pretty close, but it didn't interleave properly. This is based on that effort:
function interleave(items, parts) {
var len = items.length;
var step = len/parts | 0;
var result = [];
for (var i=0, j; i<step; ++i) {
j = i
while (j < len) {
result.push(items[j]);
j += step;
}
}
return result;
}
interleave([0,1,2,3], 2); // 0,2,1,3
interleave([0,1,2,3,4,5,6,7,8,9,10,11], 2) // 0,6,1,7,2,8,3,9,4,10,5,11
interleave([0,1,2,3,4,5,6,7,8,9,10,11], 3) // 0,4,8,1,5,9,2,6,10,3,7,11
interleave([0,1,2,3,4,5,6,7,8,9,10,11], 4) // 0,3,6,9,1,4,7,10,2,5,8,11
interleave([0,1,2,3,4,5,6,7,8,9,10,11], 5) // 0,2,4,6,8,10,1,3,5,7,9,11
Since I've been pushed to add my own answer early (edited to fix bugs noted by RobG):
function interleave(items,parts){
var stride = Math.ceil( items.length / parts ) || 1;
var result = [], len=items.length;
for (var i=0;i<stride;++i){
for (var j=i;j<len;j+=stride){
result.push(items[j]);
}
}
return result;
}
try this one:
function interleave(deck, base){
var subdecks = [];
for(count = 0; count < base; count++){
subdecks[count] = [];
}
for(var count = 0, subdeck = 0; count < deck.length; count++){
subdecks[subdeck].push(deck[count]);
subdeck = subdeck == base - 1? 0 : subdeck + 1;
}
var newDeck = [];
for(count = 0; count < base; count++){
newDeck = newDeck.concat(subdecks[count]);
}
return newDeck;
}