Develop Reference

How to loop through all possible pairs of JSON object keys exactly once?

Say that I have a data structure of n elements and a function check(element1, element2) which performs some kind of checkup on two elements. I need to check exactly all possible pairs of elements. Using combinatorics it is easy to deduce that we need to perform exactly 'n choose 2' binomial coefficient iterations ( n*(n-1)/2 iterations)
So if my data structure is an array, the following nested loops would work:
for(let i = 0; i < elements.length; i++) {
for(let j = i + 1; j < elements.length; j++) {
check(elements[i], elements[j]);
}
}
This way we check the first element with all the others, the second element with elements 3 to n (since we already checked it with the first one), the third with elements 4 to n and so on and so forth. However if 'elements' was a JSON where the key to each element is not an integer, how can we achieve this effect? Obviously we can ensure that we perform all checkups with the following code:
for(var key1 in elements) {
for(var key2 in elements) {
if(key1 != key2) {
check(elements[key1], elements[key2]);
}
}
}
However obviously we are doing a lot of checkups more than once resulting in n^2 iterations.
What method can I use to achieve the same result as in the example with the array?

If you put all the keys you're going to be looping into an array using Object.keys() then you can use your standard for loop to "skip" over previously seen keys like so:
const keys = Object.keys(elements);
for(let i = 0; i < keys.length; i++) {
const key1 = keys[i];
for(let j = i + 1; j < keys.length; j++) {
const key2 = keys[j];
check(elements[key1], elements[key2]);
}
}

Perhaps you could get the list of keys in an array:
let elements = { a: 1, b: 2, c: 3 };
let keys = Object.keys(elements).sort(); // Sorts the keys array alphabetically, and thus simulating the numbers example and be sure you're not repeating "lower" order key after passing it
for(let i = 0; i < keys.length; i++) {
for(let j = i + 1; j < keys.length; j++) {
// check(elements[keys[i]], elements[keys[j]]);
console.log(elements[keys[i]], elements[keys[j]])
}
}
output:
1 2
1 3
2 3

How to shuffle characters in a vertical inverted pattern?

I have a string "ABCDEFGHIJKLMN" that I need to shuffle in a specific manner. To do that, I write the characters sequentially in columns bottom -> top and then left -> right (4 chars per column for example) until all characters are done. If the last column is not complete, then the empty spaces need to be on the bottom (this is very important). Like so:
D H L N
C G K M
B F J
A E I
The shuffle is accomplished by producing a new string reading the block of letters as we read text, in rows left -> right:
"DHLNCGKMBFJAEI"
The cases where the columns are not complete (word.size % column_height !=0) complicate things considerably.
I came up with a few solutions, but I'm not sure if there is a simpler (ie, shorter OR easier to read) and more elegant way of coding this problem. My solutions either have an ugly, separate block of code to handle the final incomplete column or seem way too complicated.
My question is, could it be done better?
If you don't want any spoilers and decide to try and figure it out for yourself, stop reading now. If you want to work from what I fiddled so far, then a working piece of code is
var result = "";
var str = "ABCDEFGHIJKLMN";
var nr_rows = 4;
var current_row = 4;
var columns = Math.floor(str.length / nr_rows);
var modulus_table = str.length % nr_rows;
var modulus_position = -1;
for (var i = 0; i < nr_rows; i++) {
for (var j = 0; j < columns; j++) {
result += str[current_row + j * nr_rows - 1];
}
if (modulus_table > 0) {
result += str[str.length + modulus_position];
modulus_table--;
modulus_position--;
}
current_row--;
}
console.log(result);
Moving on to arrays, the next example would loop through each character, placing it correctly in a matrix-like array, but it doesn't work. The array needs to be created another way. For another example of this issue, see How to create empty 2d array in javascript?. This would also need an ugly hack to fix the last characters on the last incomplete column aligning to the bottom instead of the top.
var result = [[],[]];
var str = "ABCDEFGHIJKLMN";
var nr_rows = 4;
var row = nr_rows - 1;
var column = 0;
for (var i = 0; i < str.length; i++) {
result[row][column] = str[i];
row--;
if (row < 0) {
row = nr_rows;
column++;
}
}
console.log(result);
This last method goes full matrix array, but it quickly becomes complicated, since it needs to loop through the array in 3 different directions. First, create a dummy array with the characters in the wrong place, but where the 'undefined' positions correspond to those that should be left empty. That is acomplished by populating the array 'rotated 90º' from the reading orientation.
Without this first step, the empty positions would be stacked at the bottom instead of the top.
A second pass is required to re-write the caracters in the correct places, skipping any holes in the matrix using the 'undefined' value. This check is made for every position and there is no separate block of code to handle an incomplete last line.
A third pass then reads every character in order to form the final shuffled string. All this seems way too complicated and confusing.
// matrix populated top->bottom and left->right
// with the characters in the wrong place
// but the undefined postions in the correct place of the empty positions
var matrix = [];
var str = "ABCDEFGHIJKLMN";
var rows = 4;
var columns = Math.ceil(str.length / rows);
var k = 0;
for (var i = 0; i < rows; i++) {
matrix[i] = [];
for (var j = columns - 1; j >= 0; j--) {
matrix[i][j] = str[k];
k++;
}
}
// populate the matrix with the chars in the correct place and the 'undefined' positions left empty
var k = 0;
for (var i = 0; i < rows; i++) {
for (var j = 0; j < columns; j++) {
if (matrix[i][j] != undefined) {
matrix[i][j] = str[k];
k++;
}
}
}
// read matrix in correct direction and send to string, skipping empty positions
var result = "";
for (var j = columns - 1; j >= 0; j--) {
for (var i = 0; i < rows; i++) {
if (matrix[i][j] != undefined) {
result += matrix[i][j];
}
}
}
console.log(result);

What if you just split/reverse the array into column groups, and convert to rows?
const result = str.match(/.{1,4}/g) // split string into groups of 4
.map(i => i.split('').reverse()) // reverse each group (bottom to top, and solves the last col issue)
.reduce((res, col) => { // reduce the groups into rows
col.forEach((c, i) => res[i] += c) // concat each string char to the right row
return res
}, ['','','','']) // initialise empty strings per row
.join('') // join the rows up
Fiddle here

If you wish to return a string, I don't see why any intermediate result should use an array when it doesn't have to. The following could use one less array, but it's convenient to use split and an array to control the while loop rather than mutate the string.
The idea is to fill the strings from the bottom up until the column is full, then keep adding from the bottom of each column until it runs out of characters to assign. The row to start filling from is based on how many characters are left and how many rows there are.
Rather than building strings, it could build arrays but then generating a string requires multiple joins.
It can also produce results where there are insufficient slots for all the characters, so a result using 9 characters from 10 or more using a 3x3 "matrix" (see last example).
function verticalShuffle(s, rows, cols) {
var result = [''];
s = s.split('');
while (s.length && result[0].length < cols) {
for (var i = (rows < s.length? rows : s.length) -1 ; i>=0; i--) {
if (!result[i]) result[i] = '';
result[i] += s.splice(0,1)[0] || '';
}
}
return result.join('');
}
var s = 'ABCDEFGHIJKLMN';
console.log(verticalShuffle(s, 4, 4)); // DHLNCGKMBFJAEI
console.log(verticalShuffle(s, 6, 3)); // FLNEKMDJCIBHAG
// Only use 9 characters
console.log(verticalShuffle(s, 3, 3)); // CFIBEHADG
This uses plain ed3 functionality that will run in any browser. I don't see the point of restricting it to ECMAScript 2015 or later hosts.

If interpret Question correctly, you can use for loop, String.prototype.slice() to to populate arrays with characters of string. Use Array.prototype.pop() within recursive function to get last element of array until each array .length is 0.
To create array
[
["D","H","L","N"],
["C","G","K","M"],
["B","F","J"],
["A","E","I"]
]
from vertically inverted string you can use for loop, String.prototype.slice() to set array of arrays containing elements having .length 4, or 3 once .length of parent array is 2, having been set with two arrays containing four elements
var str = "ABCDEFGHIJKLMN";
function fnVerticalInvert(str, arr, res) {
if (!arr && !res) {
arr = []; res = "";
}
if (str) {
for (var i = 0; i < str.length; i += 4) {
arr.push([].slice.call(str.slice(i, i + 4)));
}
}
for (var i = 0; i < arr.length; i++) {
if (arr[i].length) {
res += arr[i].pop()
}
}
if (arr.some(function(curr) {return curr.length}))
return fnVerticalInvert(null, arr, res);
for (var i = 0, l = 4, j = 0, n = l - 1, k; i < l; i++, j += l) {
if (i === l / 2) k = j;
arr[i] = [].slice.call(res.slice(!k ? j : k, !k ? j + l : k + n));
if (k) k += n;
}
return {str: res, arr:arr};
};
var res = fnVerticalInvert(str);
console.log(res);

Iterate through an array and concatenate another array all the values before next iteration of first array?

I'm trying to build a list of Urls. The structure is like this:
http://somedomain.com/game_CATEGORY?page=NUMBER.
I have an array of game categories, ranging from action games category to word games category.
I have an array of numbers, 1 through 20.
I have pieces of the url saved as strings.
I've been trying for a day to combine them in this way:
cats = ["action","adventure","arcade","board","card","casino","casual","educational","family","music","puzzle","racing","role_playing","simulation","sports","strategy","trivia","word"],
nums = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20],
urlString1 = "http://example.com/game_",
urlString2 = "?page=",
madeUrl1 = [],
x = 1, // counter for page numbers
madeUrl2 = [];
for (var i = 0; i < cats.length; i++) {
madeUrl1.push(urlString1+cats[i]+urlString2);
};
for (var i = 0; i < madeUrl1.length; i++) {
madeUrl2.push(madeUrl1[i]+x);
x++;
};
console.log(madeUrl2);
This gets me partially there. But its printing out one number per category. I need each category printout to have ALL 20 numbers added, then move on to the next category.

You'd need to nest another for loop inside your second one. Something like:
for (var i = 0; i < madeUrl1.length; i++) {
for (int j = 0; j < nums.length; j++) {
madeUrl2.push(madeUrl1[i]+nums[j]);
}
};
That way you're iterating through the base URLs you prepared in madeUrl1, and then for each of those you're iterating through each number you have in the array.
If the numbers are simply sequential from 1 to 20, you don't even need the nums array:
for (var i = 0; i < madeUrl1.length; i++) {
for (var x = 1; x <= 20; x++) {
madeUrl2.push(madeUrl1[i]+x);
}
};
And the whole thing could be accomplished with a single nested for loop:
for (var i = 0; i < cats.length; i++) {
for (var x = 1; x <= 20; x++) {
madeUrl1.push(urlString1+cats[i]+urlString2+x);
}
};

You can use the code below:
cats = ["action","adventure","arcade","board","card","casino","casual","educational","family","music","puzzle","racing","role_playing","simulation","sports","strategy","trivia","word"],
nums = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20],
urlString1 = "http://example.com/game_",
urlString2 = "?page=",
madeUrl1 = [],
x = 1;
for (var i = 0; i < cats.length; i++) {
for (var j = 0; j < nums.length; j++) {
madeUrl1.push(urlString1+cats[i]+urlString2+nums[j]);
x++;
};
};
console.log(madeUrl1);
What we did here, is first nesting our loops. E.g., it will first loop through the first array, and when it arrives at it first item, in this case a category, it will run the nested loop 20 times, appending each number to the page. After done, it continues to the second category and so on.

finding an array in another Multidimensional array in javascript

I'm writing a simple snakes game using JavaScript and HTML5 canvas.
I have a Multidimensional array that hold snake block like this:
snake=[[1,1],[1,2]];
and set it on arrayMap using (snake.indexOf([i],[j])!=-1) then draw arrayMap on canvas.
for (var i = 0; i < blocksHeightCount; i++) {
for (var j = 0; j < blocksWidthCount; j++) {
if ((snake.indexOf(i,j)!=-1)||
(walls.indexOf(i,j)!=-1)||
(foods.indexOf(i,j)!=-1)) {
arrayMap[i][j]=1;
} else {
arrayMap[i][j]=0;
}
}
}
for (var i = 0; i < blocksHeightCount; i++) {
for (var j = 0; j < blocksWidthCount; j++) {
Block = arrayMap[i][j];
if (Block!=0){
ctx.fillStyle = (Block != 9) ? colors[Block]
: "#bdc3c7";
ctx.fillRect(j * cubeWidth, i * cubeHeight
, cubeWidth-.4,cubeHeight-.4);
}
}
}
the problem is indexOf isn't working when I set array on it!
It works fine when I set indexOf("i,j") but i need it to be array.
please help, thx

First solution : using Array.map
Each element of your arrays snake, walls and foods is an array with 2 elements. So to check if an [x,y] exists in one of the arrays you need a simple way to
compare two arrays [x1, y1] and [x2, y2]. Comparing arrays directly using the operator == will compare their references and not values (Thanks #Elena for the remarque). A way to compare values
would be to affect a hash to each array and compare hashes. By hash I mean a number which is unique for each array of type [x, y]. That could be x * blocksWidthCount + y
in your case and the code will be :
function getHash(x, y){
return x * blocksWidthCount + y;
}
var blockHashes = snake.concat(walls).concat(foods).map(function(cell) {
return getHash(cell[0], cell[1]);
}); // hashes of all blocks in one array
for (var i = 0; i < blocksHeightCount; i++) {
for (var j = 0; j < blocksWidthCount; j++) {
if (blockHashes.indexOf(getHash(i, j)) != -1) {
arrayMap[i][j]=1;
} else {
arrayMap[i][j]=0;
}
}
}
Second Solution Changing the way we see things
Instead of looping over all cells and verifying if every single cell is a block which gives a complexity of O(N * M) (N number of cells and M number of blocks).
We can do better simply by supposing that there is no block and then loop over blocks and mark them as blocks which is in O(N + M) !
function markBlock(cell){
arrayMap[cell[0]][cell[1]] = 1;
}
for (var i = 0; i < blocksHeightCount; i++)
for (var j = 0; j < blocksWidthCount; j++)
arrayMap[i][j] = 0;
snake.forEach(markBlock);
walls.forEach(markBlock);
foods.forEach(markBlock);

Radix Sort in JavaScript

I've come up with the following but it predictably doesn't work.
var t = new Array(a.length);
var r = 4;
var b = 64;
var count = new Array(1<<r);
var pref = new Array(1<<r);
var groups = Math.ceil(b / r);
var mask = (1 << r) - 1;
var shift = 0;
for(var c = 0; c < groups; c++)
{
shift += r;
for(var j = 0; j < count.length; j++)
{
count[j] = 0;
}
for(var i = 0; i < a.length; i++)
{
count[ (a[i] >> shift) & mask ]++;
}
pref[0] = 0;
for(var i = 0; i < count.length; i++)
{
pref[i] = pref[i-1] + count[i-1];
}
for(var i = 0; i < a.length; i++)
{
t[ pref[ (a[i] >> shift) & mask ]++ ] = a[i];
}
for(var i = 0; i < a.length; i++)
{
a[i] = t[i];
}
// a is sorted?
}

This loop does basically the same thing, in a more Javascript-y way:
for (var div = 1, radix = 16; div < 65536 * 65536; div *= radix) {
var piles = [];
for (var i = 0; i < a.length; ++i) {
var p = Math.floor(a[i] / div) % radix;
(piles[p] || (piles[p] = [])).push(a[i]);
}
for (var i = 0, ai = 0; i < piles.length; ++i) {
if (!piles[i]) continue;
for (var pi = 0; pi < piles[i].length; ++pi)
a[ai++] = piles[i][pi];
}
}
Instead of doing it like a C programmer might, this loop builds a list of lists, one list for each possible 4-bit value. I avoid bit-shift operators because this is Javascript and while they do work, things get funny when numbers get large.
Starting with the low 4 bits of each value in "a", the code copies that element of "a" to the end of one of the "piles", that being the one corresponding to the 4-bit value. It then gathers up the piles and rebuilds "a" starting from all of the values whose low 4 bits were 0, then 1, etc. (Clearly there'll be some gaps, so those are skipped.) At the end of each iteration of the overall loop, the divisor is multiplied by the radix, so that the next set of 4 bits will be examined.
Once the divisor has exhausted the available range of integers, it's done.
Note that this will only work for positive numbers. Doing this with negative numbers gets a little weird; it might be easier to strip out the negative numbers into a separate array, flip the sign, sort, then reverse. Sort the positive numbers, and then finally glue the reversed negative numbers (flipping the signs again) to the front of the sorted positive numbers.

this
for(var i = 0; i < count.length; i++)
{
pref[i] = pref[i-1] + count[i-1];
}
is a problem because on the first iteration, i is zero and so pref[ 0 - 1 ] is not going to work very well.
I don't have a reference for radix sorts handy to determine what you should actually be doing here.