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Variable for loop

If reverse == true I want to run one kind of loop, else I want to run another one.
Currently, here is an example of my code:
if (reverse) {
for (var i = 0; i < length; i++) {
...
}
} else {
for (var i = length; i >= 0; i--) {
...
}
}
The code inside is really big, and is quite the same. I could use a function, but this function would have so many params that is not a good choice.
So I've tried to do something like that:
var loopStart1 = startPageIndex;
if (!reverse) {
condition1 = function(i) {
return i < length;
}
increment1 = function(i) {
return ++i;
}
} else {
condition1 = function(i) {
return i >= 0;
}
increment1 = function(i) {
return i--;
}
}
mainLoop: for (var i = loopStart1; condition1(i); increment1(i)) {
But now I have an infinite loop.
Any idea on how to solve this issue?

Why not do it inline?! ;)
var start = startPageIndex;
for (var i = start; (reverse && i >= 0) || (!reverse && i < length); reverse ? --i : ++i) { }

Assuming the specific case is to traverse through an array either backwards or forwards, there are two simple ways to do that.
1) Just conditionally reverse the array before looping, or
2) Use a single loop but conditionally map the loop variable to something else. Something like...
for (var i = 0; i < length; i++) {
var j = i;
if (reverse) {
j = length - (i + 1);
}
// ...then use j instead of i
}

If you want to make it dynamic, I wouldn't use a for loop, but a do..while loop to be easier to customize.
Your main function would just have a simple reverse bool flag and you could just pass that.
Inside that function that you want to depend on the reverse flag, you can use the ternary expression in the condition (x ? y : z)
Makes it clearer to read. In theory you can do it in a for loop directly, using two ternary directions.
do {
//Your actions here, using i as counter
//Here you will do the counter direction change
if(reverse)
i--;
else
i++;
// Use ternary expression if (x ? y : z) in the while condition
// Reads as: If x is true, do y, else do z
// All in one line
} while(reverse ? i>=0 : i<n);
Ideally, in these situations I would consider using iterators.

How about a simple loop function,
Below I've created a simple function called myLoop, were you can pass the length, if it's reversed, and what callback you want for each loop iteration.
eg.
function myLoop(length, reverse, each) {
var index;
if (!reverse) {
for (index = 0; index < length; index ++) each(index);
} else {
for (index = length -1; index >= 0; index --) each(index);
}
}
function show(index) {
console.log("Loop index = " + index);
}
console.log("forward");
myLoop(5, false, show); //normal
console.log("revere");
myLoop(5, true, show); //reversed

I would opt for the same code, just change the array.
var array = ['one', 'two', 'three'];
var reversed = true;
let arrToUse = reversed ? array.reverse() : array;
for (var i = 0; i < arrToUse.length; i++) {
console.log(arrToUse[i]);
}

Check https://developer.mozilla.org/en-US/docs/Web/JavaScript/Guide/Functions
Primitive parameters (such as a number) are passed to functions by value; the value is passed to the function, but if the function changes the value of the parameter, this change is not reflected globally or in the calling function.
This is what happens when you call increment1(i), outer i in for loop is not changed. To fix it just set i from increment1(i)'s return value.
mainLoop: for (var i = loopStart1; condition1(i); i = increment1(i)) {

How to avoid my permutation algorithm of ERROR:heap out memory

Today I try to solve a problem on codewars，it requires me give the permutations of a given string.
Firstly,I try to use a recursion function looks like:
function permutate(str) {
var result = [];
if (str.length == 1) {
return [str]
} else {
var preResult = permutate(str.slice(1));
for (var j = 0; j < preResult.length; j++) {
for (var k = 0; k < preResult[j].length + 1; k++) {
var temp = preResult[j].slice(0, k) + str[0] + preResult[j].slice(k);
result.push(temp);
}
}
return result;
}
}
After I click the attemp button,the OJ tells me there is an error caused by heap out memory.Because my function called with a long string:"abcdefghijkl".
Secondly,I rewrite my function by using loop.just like:
function perm(str) {
let result = [],tempArr = [];
let subStr = str;
while (subStr.length !== 0) {
if (result.length === 0) {
result.push(str[0]);
} else {
for (let i = 0; i < result.length; i++) {
let item = result[i];
let itemLen = item.length;
for (let j = 0; j < itemLen+1; j++) {
let temp = item.slice(0, j) + subStr[0] + item.slice(j);
tempArr.push(temp);
}
}
result = tempArr;
tempArr = [];
}
subStr = subStr.slice(1);
}
return result;
}
It works when the given string is short.But still cause Error.
So,I want to know why cause this error and if there is a permutation algorithm can run in Node(v6.11.0) without memory error?
I searched a lot and tried many methods,but nothing works.So I ask my first question on stackoverflow,hoping you can give me some help.Thanks!

Try the module https://github.com/miguelmota/permutations, or even try to use the code from the module

In addition to previous answer as a possible try out, you can try to increase process max memory limit size, for example with node --max-old-space-size=8192 which is in bytes, the node process will run with extended 8GB memory limit.

How to analyse complexity of an Algorithm that contains hidden loops made by external function or method?

First of all, I would like to apologize in case my title is not concise as it should be, but my point is, if you take a look at the following code which is selection sort algorithm, it's obvious for someone to analyze its complexity.
module.exports = function (arr) {
var temp;
for (var i = 0; i < arr.length; i++) {
var iTh = i;
for (var j = i+1; j < arr.length; j++) {
if (arr[j] < arr[iTh]) {
iTh = j;
}
}
temp = arr[i];
arr[i] = arr[iTh];
arr[iTh] = temp;
}
return arr;
}
But what if an algorithm contains hidden loops which are provided by particular language's functions or methods. For instance these two functions are both reversing a string, and they have JavaScript methods which have complexity behind them too.
So! How can someone analyze the complexity of these two and pick the optimal one? Or they don't qualify to be algorithms?
First Reverse
exports.reverse1 = function (str) {
if (str == undefined || str.length) {
return 0;
}
let collector = [];
for (var i = str.length; i >= 0; i--) {
collector.push(str.charAt(i));
}
return collector.join("");
}
Second Reverse
exports.reverse2 = function (str) {
if (str == undefined || str === "") {
return 0;
}
return str.split("").reverse().join("");
}

How to do Nested For Loops in Functional Style

I'm in the process of learning functional programming, and completely getting rid of for loops has been a challenge sometimes, because they provide so much control and freedom. Below is an example of checking if a string is an isogram or not (no letters should be repeated). With nested for loops, it became an easy solution. Is there a way to do this the functional way with any high order functions or anything else? Any suggestion would be a huge help.
Code:
function isIsogram(string) {
let array = string.split('');
let condition = true;
for (let i = 0; i < string.length; i++) { //first loop picks character
for (j = i + 1; j < string.length; j++) { //second loop compares it rest
if (array[i].toLowerCase() == array[j].toLowerCase())
condition = false; //if repeat, the condition false
}
return condition;
}
}

You can use every or some together with a suitable string function:
function isIsogram(string) {
string = string.toLowerCase(); // case insensitive
return string.split('').every(function(character, index) {
return !string.includes(character, index+1);
});
}
Instead of includes you might also have utilised indexOf.

You can sort the String first and then apply every on it. It will stop the iteration as soon as two successive letters are the same:
Here is an improved implementation. Credit goes to #Xotic750:
function isIsogram(x) {
return Array.from(x.toLowerCase()).sort().every((y, i, xs) => i === 0
? true
: y !== xs[i - 1]);
}
console.log( isIsogram("consumptively") );
console.log( isIsogram("javascript") );
The implementation uses Array.prototype.every's second parameter, which represents the index of the current element (of the iteration). Please note that isIsogram solely depends on functions and their arguments.

Another example, like #Bergi but using some ES6 features for comparison.
function isIsogram(string) {
string = string.toLowerCase(); // case insensitive
for (let character of Array.from(string).entries()) {
if (string.includes(character[1], character[0] + 1)) {
return false;
}
}
return true;
}
console.log(isIsogram('abc'));
console.log(isIsogram('abca'));
How your ES3 style code could have looked (noting some of the issues pointed out in the comments)
function isIsogram(string) {
string = string.toLowerCase(); // case insensitive
var length = string.length;
for (var i = 0; i < length; i += 1) {
for (var j = i + 1; j < length; j += 1) {
if (string.charAt(i) === string.charAt(j)) {
return false;
}
}
}
return true;
}
console.log(isIsogram('abc'));
console.log(isIsogram('abca'));

How can I search for a value in an array and delete just one of those values if 2 or more is found?

I have an array
var aos = ["a","a","a","b","b","c","d","d"];
I want to know if I can remove just 1 item if it finds 2 or more of the same value in the array. So for instance if it finds
"a", "a"
it will remove one of those "a"
This is my current code:
var intDennis = 1;
for (var i = 0; i < aos.length; i++) {
while (aos[i] == aos[intDennis]) {
aos.splice(i, 1);
intDennis++;
console.log(aos[intDennis], aos[i]);
}
intDennis = 1;
}
NOTE: My array is sorted.

Edited after better understanding of OP use-case.
Updated solution and fiddle test to incorporate suggestion from pst in comments.
(Not for nothing, but this method does not require the original array be sorted.)
Try this...
var elements = [];
var temp = {};
for (i=0; i<aos.length; i++) {
temp[aos[i]] = (temp[aos[i]] || 0) + 1;
}
for (var x in temp) {
elements.push(x);
for (i=0; i<temp[x]-2; i++) {
elements.push(x);
}
}
Fiddle Test

Because you said you have a sorted array, you only need to remove the second time a element is found. You only need one for.
The splice() function returns the removed element so, just use it to not remove more elements of that kind.
This solution is more clean and efficient.
var aos = ["a","a","a","b","b","c","d","d"];
var lastRemoved = "";
for (var i = 1; i < aos.length; i++) {
if (aos[(i-1)] == aos[i] && lastRemoved != aos[i]) {
lastRemoved = aos.splice(i, 1);
}
}
Code tested and working. Result: ["a", "a", "b", "c", "d"]

I don't believe there's any better way to do this on an unsorted array than an approach with O(n^2) behaviour. Given ES5 array-builtins (supported in all modern browsers, though not in IE prior to IE9), the following works:
aos.filter(function(value, index, obj) { return obj.indexOf(value) === index; })

UPDATED ANSWER TO REMOVE ONLY 1 DUPLICATE:
Assuming that each object will resolve to a unique String, here's a potential solution. The first time the object is detected, it sets a counter for that object to one. If it finds that object again, it splices that element out and increments the associated counter. If it finds that element more times, it will leave it alone.
var elements = {};
for (var i = 0; i < aos.length; i++) {
if(elements[aos[i]]){
if(elements[aos[i]] == 1){
aos.splice(i,1);//splice the element out of the array
i--;//Decrement the counter to account for the reduced array
elements[aos[i]]++;//Increment the count for the object
}
} else {
elements[aos[i]] = 1;//Initialize the count for this object to 1;
}
}
Here's the test fiddle for this.

I would not mutate the input -- that is, don't use splice. This will simplify the problem a good deal. Using a new array object here may actually be more efficient. This approach utilizes the fact that the input array is sorted.
Consider: (jsfiddle demo)
var input = ["a","a","a","b","b","c","d","d"]
var result = []
for (var i = 0; i < input.length; i++) {
var elm = input[i]
if (input[i+1] === elm) {
// skip first element (we know next is dup.)
var j = i + 1
for (; input[j] === elm && j < input.length; j++) {
result.push(input[j])
}
i = j - 1
} else {
result.push(elm)
}
}
alert(result) // a,a,b,c,d
Happy coding.
Replace === with a custom equality, as desired. Note that it is the first item is omitted from the output, which may not always be "correct".

REVISED EXAMPLE
function removeDuplicate(arr) {
var i = 1;
while(i < arr.length) {
if(arr[i] == arr[i - 1]) {
arr.splice(i, 1);
}
while(arr[i] == arr[i - 1] && i < arr.length) {
i += 1;
}
i += 1;
}
return arr;
}
alert(removeDuplicate(["a","a","a","b","b","c","d","d"]));