I'm aware of the modular arithmetic approach to solve this question, I just took it upon myself to solve it in the more naive way, but I just created an infinite loop somewhere in the inner for loop and there is no way out (Please do not tell me to do it in the standard way, This is just a challenge that I made upon myself)
The function return the size of a maximal subset of S where the sum of any 2 numbers in s’ is not evenly divisible by k.
here is my code that causes an infinite loop, so be careful
function nonDivisibleSubset(arr, k) {
let maxElements = 0;
for (let i = 0; i < arr.length; i++) {
console.log(i);
// arr 1 to (arr.length - 1)
let result = [];
let noOfElements = 0;
for (let j = 0 + 1; j < arr.length; j++) {
console.log(`hi${j}`);
//arr 2 to arr.length
console.log(result.length);
if (
result.length === 0 &&
(arr[i] + arr[j]) % k !== 0
) {
console.log("hi2");
result.push(arr[i], arr[j]);
noOfElements += 2;
} else {
let isDivisible = false;
console.log(`result length: ${result.length}`);
for (let l = 0; l < result.length; l++) {
console.log(l);
if (i !== l && (arr[l] + arr[j]) % 4 === 0) {
isDivisible = true;
break;
}
if (!isDivisible) {
result.push(arr[l]);
console.log(result);
noOfElements += 1;
break; //removing this break will initialize the infinite loop
}
}
}
}
if (noOfElements > maxElements) {
maxElements = noOfElements;
}
}
return maxElements;
}
//initialise
nonDivisibleSubset([19, 10, 12, 10, 24, 25, 22], 4)
The resul array continues to increase in size as the repeat statement spins.
Here is an example of how you can do it:
for (let l = 0, size = result.length; l < size; l++) {
if (i !== l && (arr[l] + arr[j]) % 4 === 0) {
isDivisible = true;
break;
}
if (!isDivisible) {
result.push(arr[l]);
console.log(result);
console.log(result.length);
noOfElements += 1;
break; //removing this break will initialize the infinite loop
}
}
And,
if (i !== l && (arr[l] + arr[j]) % 4 === 0)
Isn't it k, not 4?
if (i !== l && (arr[l] + arr[j]) % k === 0)
I hope it helps.
Hello I am trying to refactor my function but it doesn't seem to be working properly. It works fine normally which is the commented code. Am I writing this wrong or is there a fundamental flaw in my code?
function isPrime(num) {
// if (num <= 1){
// return false
// } else {
// for(var i = 2; i < num; i++){
// if(num % i === 0) return false;
// }
// }
num <= 1 ? false : {
for(let i = 2; i < num; i++){
num % 1 === 0 ? false
}}
return true;
}
Thanks
As far as I can tell, no language having the ternary operator allows running a block of statements, which is what you are doing. The ternary operator is a shorthand version of:
let result;
if (<condition>)
result = resultTrue;
else
result = resultFalse;
Unless JavaScript stabs be in the back on this one, for is a statement that doesn't return anything, and therefore cannot be used the way you want.
The best you could do, I guess, would be:
function isPrime(num) {
if (num <= 1) {
return false;
}
for(var i = 2; i < num; i++) {
if(num % i === 0) return false;
}
}
The else part omitted since hitting the line after the if necessarily means that the condition for it was not met.
Technically what you have can be brought to life with an IIFE:
function isPrime(num) {
return num <= 1 ? false :
(n => {
for (let i = 2; i < n; i++) {
if (n % i === 0) return false;
}
return true;
})(num)
}
for (let i = 0; i < 10; i++) {
console.log(i, isPrime(i));
}
Now it works and there is a ternary operator inside.
Even the attempt without return can be done, just you need isPrime made into an arrow function:
let isPrime = num => num <= 1 ? false :
(n => {
for (let i = 2; i < n; i++) {
if (n % i === 0) return false;
}
return true;
})(num);
for (let i = 0; i < 10; i++) {
console.log(i, isPrime(i));
}
Tadaam, the num <= 1 ? false : part is literally there in the first line, without that nasty return.
But realistically this is not how isPrime should be implemented, your very first attempt is the best one so far.
If you want an actual improvement, use the fact that divisors come in pairs. If x is a divisor of num, then num/x is also divisor of num. So if you were looking for all divisors of num, you would find them in pairs as soon as x=num/x=sqrt(num), so it's enough to check numbers between 2 and Math.floor(Math.sqrt(num)), inclusive. As you preferably don't want JS to calculate this number in every iteration (when checking the condition), you could count downwards, so
for (let i = Math.floor(Math.sqrt(num)); i >= 2; i--)
I'm trying to get N ways of solves a N rook problem. The issue I am having is currently, I seem to get n*n solutions while it needs to be N! . Below is my code, I have written it in simple loops and functions, so it's quite long. Any help would be greatly appreciated
Note: Please ignore case for n = 2. I get some duplicates which I thought I would handle via JSON.stringify
var createMatrix = function (n) {
var newMatrix = new Array(n);
// build matrix
for (var i = 0; i < n; i++) {
newMatrix[i] = new Array(n);
}
for (var i = 0; i < n; i++) {
for (var j = 0; j < n; j++) {
newMatrix[i][j] = 0;
}
}
return newMatrix;
};
var newMatrix = createMatrix(n);
// based on rook position, greying out function
var collision = function (i, j) {
var col = i;
var row = j;
while (col < n) {
// set the row (i) to all 'a'
col++;
if (col < n) {
if (newMatrix[col][j] !== 1) {
newMatrix[col][j] = 'x';
}
}
}
while (row < n) {
// set columns (j) to all 'a'
row++;
if (row < n) {
if (newMatrix[i][row] !== 1) {
newMatrix[i][row] = 'x';
}
}
}
if (i > 0) {
col = i;
while (col !== 0) {
col--;
if (newMatrix[col][j] !== 1) {
newMatrix[col][j] = 'x';
}
}
}
if (j > 0) {
row = j;
while (row !== 0) {
row--;
if (newMatrix[i][row] !== 1) {
newMatrix[i][row] = 'x';
}
}
}
};
// checks position with 0 and sets it with Rook
var emptyPositionChecker = function (matrix) {
for (var i = 0; i < matrix.length; i++) {
for (var j = 0; j < matrix.length; j++) {
if (matrix[i][j] === 0) {
matrix[i][j] = 1;
collision(i, j);
return true;
}
}
}
return false;
};
// loop for every position on the board
loop1:
for (var i = 0; i < newMatrix.length; i++) {
var row = newMatrix[i];
for (var j = 0; j < newMatrix.length; j++) {
// pick a position for rook
newMatrix[i][j] = 1;
// grey out collison zones due to the above position
collision(i, j);
var hasEmpty = true;
while (hasEmpty) {
//call empty position checker
if (emptyPositionChecker(newMatrix)) {
continue;
} else {
//else we found a complete matrix, break
hasEmpty = false;
solutionCount++;
// reinitiaze new array to start all over
newMatrix = createMatrix(n);
break;
}
}
}
}
There seem to be two underlying problems.
The first is that several copies of the same position are being found.
If we consider the case of N=3 and we visualise the positions by making the first rook placed red, the second placed green and the third to be placed blue, we get these three boards:
They are identical positions but will count as 3 separate ones in the given Javascript.
For a 3x3 board there are also 2 other positions which have duplicates. The gets the count of unique positions to 9 - 2 - 1 -1 = 5. But we are expecting N! = 6 positions.
This brings us to the second problem which is that some positions are missed. In the case of N=3 this occurs once when i===j==1 - ie the mid point of the board.
This position is reached:
This position is not reached:
So now we have the number of positions that should be found as 9 - 2 - 1 - 1 +1;
There appears to be nothing wrong with the actual Javascript in as much as it is implementing the given algorithm. What is wrong is the algorithm which is both finding and counting duplicates and is missing some positions.
A common way of solving the N Rooks problem is to use a recursive method rather than an iterative one, and indeed iteration might very soon get totally out of hand if it's trying to evaluate every single position on a board of any size.
This question is probably best taken up on one of the other stackexchange sites where algorithms are discussed.
Ok guys, object developer newbie here. I try to do an animation of falling cube as explain here : Falling animation to fill a webpage
I have some algorithmic issues. I follow the model of a tetris game but I want multiple pixels falling at the same time. So I have a constructor with some methods to move my pixel.
But now I use my constructor to create an array of object like :
var a_player = [];
function addPlayer(pos){
var player = new Player(pos);
a_player.push(player);
}
addPlayer({x: 3, y: 3});
addPlayer({x: 0, y: 0});
And I want to use some public methods like a collide() method :
function collide(arena, player) {
const [m, o] = [player.matrix, player.pos];
for (let y = 0; y < m.length; ++y) {
for (let x = 0; x < m[y].length; ++x) {
if (m[y][x] !== 00 &&
(arena[y + o.y] &&
arena[y + o.y][x + o.x]) !== 0) {
return true;
}
}
}
return false;
}
But I don't know what's the best way to do it. I can use a "for" like
for (i = 0; i < a_player.length; i++){
console.log(a_player[i].pos);
}
but I have to apply it on all my methods, or I can duplicate my method by the number of player I have in my array (but in the ends I want more than 20k players...). So can you help me with that kind of problematic ?
I think this what you are looking for:
function collide(arena) {
const [m, o] = [this.matrix, this.pos];
for (let y = 0; y < m.length; ++y) {
for (let x = 0; x < m[y].length; ++x) {
if (m[y][x] !== 00 &&
(arena[y + o.y] &&
arena[y + o.y][x + o.x]) !== 0) {
return true;
}
}
}
return false;
}
Player.prototype.collide=collide;
for (i = 0; i < a_player.length; i++){
a_player[i].collide(arena)
}
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What's the best way to break from nested loops in Javascript?
//Write the links to the page.
for (var x = 0; x < Args.length; x++)
{
for (var Heading in Navigation.Headings)
{
for (var Item in Navigation.Headings[Heading])
{
if (Args[x] == Navigation.Headings[Heading][Item].Name)
{
document.write("<a href=\""
+ Navigation.Headings[Heading][Item].URL + "\">"
+ Navigation.Headings[Heading][Item].Name + "</a> : ");
break; // <---HERE, I need to break out of two loops.
}
}
}
}
Just like Perl,
loop1:
for (var i in set1) {
loop2:
for (var j in set2) {
loop3:
for (var k in set3) {
break loop2; // breaks out of loop3 and loop2
}
}
}
as defined in EMCA-262 section 12.12. [MDN Docs]
Unlike C, these labels can only be used for continue and break, as Javascript does not have goto.
Wrap that up in a function and then just return.
I'm a little late to the party but the following is a language-agnostic approach which doesn't use GOTO/labels or function wrapping:
for (var x = Set1.length; x > 0; x--)
{
for (var y = Set2.length; y > 0; y--)
{
for (var z = Set3.length; z > 0; z--)
{
z = y = -1; // terminates second loop
// z = y = x = -1; // terminate first loop
}
}
}
On the upside it flows naturally which should please the non-GOTO crowd. On the downside, the inner loop needs to complete the current iteration before terminating so it might not be applicable in some scenarios.
I realize this is a really old topic, but since my standard approach is not here yet, I thought I post it for the future googlers.
var a, b, abort = false;
for (a = 0; a < 10 && !abort; a++) {
for (b = 0; b < 10 && !abort; b++) {
if (condition) {
doSomeThing();
abort = true;
}
}
}
Quite simple:
var a = [1, 2, 3];
var b = [4, 5, 6];
var breakCheck1 = false;
for (var i in a) {
for (var j in b) {
breakCheck1 = true;
break;
}
if (breakCheck1) break;
}
Here are five ways to break out of nested loops in JavaScript:
1) Set parent(s) loop to the end
for (i = 0; i < 5; i++)
{
for (j = 0; j < 5; j++)
{
if (j === 2)
{
i = 5;
break;
}
}
}
2) Use label
exit_loops:
for (i = 0; i < 5; i++)
{
for (j = 0; j < 5; j++)
{
if (j === 2)
break exit_loops;
}
}
3) Use variable
var exit_loops = false;
for (i = 0; i < 5; i++)
{
for (j = 0; j < 5; j++)
{
if (j === 2)
{
exit_loops = true;
break;
}
}
if (exit_loops)
break;
}
4) Use self executing function
(function()
{
for (i = 0; i < 5; i++)
{
for (j = 0; j < 5; j++)
{
if (j === 2)
return;
}
}
})();
5) Use regular function
function nested_loops()
{
for (i = 0; i < 5; i++)
{
for (j = 0; j < 5; j++)
{
if (j === 2)
return;
}
}
}
nested_loops();
var str = "";
for (var x = 0; x < 3; x++) {
(function() { // here's an anonymous function
for (var y = 0; y < 3; y++) {
for (var z = 0; z < 3; z++) {
// you have access to 'x' because of closures
str += "x=" + x + " y=" + y + " z=" + z + "<br />";
if (x == z && z == 2) {
return;
}
}
}
})(); // here, you execute your anonymous function
}
How's that? :)
How about using no breaks at all, no abort flags, and no extra condition checks. This version just blasts the loop variables (makes them Number.MAX_VALUE) when the condition is met and forces all the loops to terminate elegantly.
// No breaks needed
for (var i = 0; i < 10; i++) {
for (var j = 0; j < 10; j++) {
if (condition) {
console.log("condition met");
i = j = Number.MAX_VALUE; // Blast the loop variables
}
}
}
There was a similar-ish answer for decrementing-type nested loops, but this works for incrementing-type nested loops without needing to consider each loop's termination value for simple loops.
Another example:
// No breaks needed
for (var i = 0; i < 89; i++) {
for (var j = 0; j < 1002; j++) {
for (var k = 0; k < 16; k++) {
for (var l = 0; l < 2382; l++) {
if (condition) {
console.log("condition met");
i = j = k = l = Number.MAX_VALUE; // Blast the loop variables
}
}
}
}
}
If you use Coffeescript, there is a convenient "do" keyword that makes it easier to define and immediately execute an anonymous function:
do ->
for a in first_loop
for b in second_loop
if condition(...)
return
...so you can simply use "return" to get out of the loops.
I thought I'd show a functional-programming approach. You can break out of nested Array.prototype.some() and/or Array.prototype.every() functions, as in my solutions. An added benefit of this approach is that Object.keys() enumerates only an object's own enumerable properties, whereas "a for-in loop enumerates properties in the prototype chain as well".
Close to the OP's solution:
Args.forEach(function (arg) {
// This guard is not necessary,
// since writing an empty string to document would not change it.
if (!getAnchorTag(arg))
return;
document.write(getAnchorTag(arg));
});
function getAnchorTag (name) {
var res = '';
Object.keys(Navigation.Headings).some(function (Heading) {
return Object.keys(Navigation.Headings[Heading]).some(function (Item) {
if (name == Navigation.Headings[Heading][Item].Name) {
res = ("<a href=\""
+ Navigation.Headings[Heading][Item].URL + "\">"
+ Navigation.Headings[Heading][Item].Name + "</a> : ");
return true;
}
});
});
return res;
}
Solution that reduces iterating over the Headings/Items:
var remainingArgs = Args.slice(0);
Object.keys(Navigation.Headings).some(function (Heading) {
return Object.keys(Navigation.Headings[Heading]).some(function (Item) {
var i = remainingArgs.indexOf(Navigation.Headings[Heading][Item].Name);
if (i === -1)
return;
document.write("<a href=\""
+ Navigation.Headings[Heading][Item].URL + "\">"
+ Navigation.Headings[Heading][Item].Name + "</a> : ");
remainingArgs.splice(i, 1);
if (remainingArgs.length === 0)
return true;
}
});
});
How about pushing loops to their end limits
for(var a=0; a<data_a.length; a++){
for(var b=0; b<data_b.length; b++){
for(var c=0; c<data_c.length; c++){
for(var d=0; d<data_d.length; d++){
a = data_a.length;
b = data_b.length;
c = data_b.length;
d = data_d.length;
}
}
}
}
Already mentioned previously by swilliams, but with an example below (Javascript):
// Function wrapping inner for loop
function CriteriaMatch(record, criteria) {
for (var k in criteria) {
if (!(k in record))
return false;
if (record[k] != criteria[k])
return false;
}
return true;
}
// Outer for loop implementing continue if inner for loop returns false
var result = [];
for (var i = 0; i < _table.length; i++) {
var r = _table[i];
if (!CriteriaMatch(r[i], criteria))
continue;
result.add(r);
}
There are many excellent solutions above.
IMO, if your break conditions are exceptions,
you can use try-catch:
try{
for (var i in set1) {
for (var j in set2) {
for (var k in set3) {
throw error;
}
}
}
}catch (error) {
}
Hmmm hi to the 10 years old party ?
Why not put some condition in your for ?
var condition = true
for (var i = 0 ; i < Args.length && condition ; i++) {
for (var j = 0 ; j < Args[i].length && condition ; j++) {
if (Args[i].obj[j] == "[condition]") {
condition = false
}
}
}
Like this you stop when you want
In my case, using Typescript, we can use some() which go through the array and stop when condition is met
So my code become like this :
Args.some((listObj) => {
return listObj.some((obj) => {
return !(obj == "[condition]")
})
})
Like this, the loop stopped right after the condition is met
Reminder : This code run in TypeScript
Assign the values which are in comparison condition
function test(){
for(var i=0;i<10;i++)
{
for(var j=0;j<10;j++)
{
if(somecondition)
{
//code to Break out of both loops here
i=10;
j=10;
}
}
}
//Continue from here
}
An example with for .. of, close to the example further up which checks for the abort condition:
test()
function test() {
var arr = [1, 2, 3,]
var abort = false;
for (var elem of arr) {
console.log(1, elem)
for (var elem2 of arr) {
if (elem2 == 2) abort = true;
if (!abort) {
console.log(2, elem2)
}
}
}
}
Condition 1 - outer loop - will always run
The top voted and accepted answer also works for this kind of for loop.
Result: the inner loop will run once as expected
1 1
2 1
1 2
1 3
XXX.Validation = function() {
var ok = false;
loop:
do {
for (...) {
while (...) {
if (...) {
break loop; // Exist the outermost do-while loop
}
if (...) {
continue; // skips current iteration in the while loop
}
}
}
if (...) {
break loop;
}
if (...) {
break loop;
}
if (...) {
break loop;
}
if (...) {
break loop;
}
ok = true;
break;
} while(true);
CleanupAndCallbackBeforeReturning(ok);
return ok;
};
the best way is -
1) Sort the both array which are used in first and second loop.
2) if item matched then break the inner loop and hold the index value.
3) when start next iteration start inner loop with hold index value.