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Why is my for loop breaking earlier than expected?

I am trying to solve a problem on leetCode:
Given an unsorted integer array nums, return the smallest missing positive integer.
This is the code I came up with
var firstMissingPositive = function(nums) {
nums.sort();
let x = 1; //this is to compare the elements of nums
for (let num in nums) {
if (nums[num] <= 0) continue; //because anything less than 1 does not matter
else if (nums[num] != x) break; //if x is not present, x is the answer
else x++; // if x is present, x should increment and check for the next integer
}
return x;
};
This code works 106/173 testcases. It does not pass the following case, which looks very simple -
nums = [1,2,3,4,5,6,7,8,9,20];
The output I get is 3, whereas the expected output is 10.
I'm not looking for the right solution to the problem. I'm just curious why this seemingly simple test fails. I do not understand why my loop breaks at 3 when it passes 1 and 2. Please help!

Here's the root cause of your problem (mdn):
The sort() method sorts the elements of an array in place and returns
the sorted array. The default sort order is ascending, built upon
converting the elements into strings, then comparing their sequences
of UTF-16 code units values.
So what you get after sort is [1, 2, 20, 3, ...], as '20' string precedes '3' string. One possible way to fix this it to force sorting by numeric value:
nums.sort((a, b) => a - b);

Javascript sort method for arrays [duplicate]

How does the following code sort this array to be in numerical order?
var array=[25, 8, 7, 41]
array.sort(function(a,b){
return a - b
})
I know that if the result of the computation is...
Less than 0: "a" is sorted to be a lower index than "b".
Zero: "a" and "b" are considered equal, and no sorting is performed.
Greater than 0: "b" is sorted to be a lower index than "a".
Is the array sort callback function called many times during the course of the sort?
If so, I'd like to know which two numbers are passed into the function each time. I assumed it first took "25"(a) and "8"(b), followed by "7"(a) and "41"(b), so:
25(a) - 8(b) = 17 (greater than zero, so sort "b" to be a lower index than "a"): 8, 25
7(a) - 41(b) = -34 (less than zero, so sort "a" to be a lower index than "b": 7, 41
How are the two sets of numbers then sorted in relation to one another?
Please help a struggling newbie!

Is the array sort callback function called many times during the course of the sort?
Yes
If so, I'd like to know which two numbers are passed into the function each time
You could find out your self with:
array.sort((a,b) => {
console.log(`comparing ${a},${b}`);
return a > b ? 1
: a === b ? 0
: -1;
});
EDIT
This is the output I've got:
25,8
25,7
8,7
25,41

The JavaScript interpreter has some kind of sort algorithm implementation built into it. It calls the comparison function some number of times during the sorting operation. The number of times the comparison function gets called depends on the particular algorithm, the data to be sorted, and the order it is in prior to the sort.
Some sort algorithms perform poorly on already-sorted lists because it causes them to make far more comparisons than in the typical case. Others cope well with pre-sorted lists, but have other cases where they can be "tricked" into performing poorly.
There are many sorting algorithms in common use because no single algorithm is perfect for all purposes. The two most often used for generic sorting are Quicksort and merge sort. Quicksort is often the faster of the two, but merge sort has some nice properties that can make it a better overall choice. Merge sort is stable, while Quicksort is not. Both algorithms are parallelizable, but the way merge sort works makes a parallel implementation more efficient, all else being equal.
Your particular JavaScript interpreter may use one of those algorithms or something else entirely. The ECMAScript standard does not specify which algorithm a conforming implementation must use. It even explicitly disavows the need for stability.

Pairs of values are compared, one pair at a time. The pairs that are compared are an implementation detail--don't assume they will be the same on every browser. The callback can be anything (so you can sort strings or Roman numerals or anything else where you can come up with a function that returns 1,0,-1).
One thing to keep in mind with JavaScript's sort is that it is not guaranteed to be stable.

Deeply Knowledge
If the result is negative a is sorted before b.
If the result is positive b is sorted before a.
If the result is 0 no changes are done with the sort order of the two values.
NOTE:
This code is the view inside of the sort method step by step.
OUTPUT:
let arr = [90, 1, 20, 14, 3, 55];
var sortRes = [];
var copy = arr.slice(); //create duplicate array
var inc = 0; //inc meant increment
copy.sort((a, b) => {
sortRes[inc] = [ a, b, a-b ];
inc += 1;
return a - b;
});
var p = 0;
for (var i = 0; i < inc; i++) {
copy = arr.slice();
copy.sort((a, b) => {
p += 1;
if (p <= i ) {
return a - b;
}
else{
return false;
}
});
p = 0;
console.log(copy +' \t a: '+ sortRes[i][0] +' \tb: '+ sortRes[i][1] +'\tTotal: '+ sortRes[i][2]);
}

To help clarify the behavior of Array#sort and its comparator, consider this naive insertion sort taught in beginning programming courses:
const sort = arr => {
for (let i = 1; i < arr.length; i++) {
for (let j = i; j && arr[j-1] > arr[j]; j--) {
[arr[j], arr[j-1]] = [arr[j-1], arr[j]];
}
}
};
const array = [3, 0, 4, 5, 2, 2, 2, 1, 2, 2, 0];
sort(array);
console.log("" + array);
Ignoring the choice of insertion sort as the algorithm, focus on the hardcoded comparator: arr[j-1] > arr[j]. This has two problems relevant to the discussion:
The > operator is invoked on pairs of array elements but many things you might want to sort such as objects don't respond to > in a reasonable way (the same would be true if we used -).
Even if you are working with numbers, oftentimes you want some other arrangement than the ascending sort that's been baked-in here.
We can fix these problems by adding a comparefn argument which you're familiar with:
const sort = (arr, comparefn) => {
for (let i = 1; i < arr.length; i++) {
for (let j = i; j && comparefn(arr[j-1], arr[j]) > 0; j--) {
[arr[j], arr[j-1]] = [arr[j-1], arr[j]];
}
}
};
const array = [3, 0, 4, 5, 2, 2, 2, 1, 2, 2, 0];
sort(array, (a, b) => a - b);
console.log("" + array);
sort(array, (a, b) => b - a);
console.log("" + array);
const objArray = [{id: "c"}, {id: "a"}, {id: "d"}, {id: "b"}];
sort(objArray, (a, b) => a.id.localeCompare(b.id));
console.log(JSON.stringify(objArray, null, 2));
Now the naive sort routine is generalized. You can see exactly when this callback is invoked, answering your first set of concerns:
Is the array sort callback function called many times during the course of the sort? If so, I'd like to know which two numbers are passed into the function each time
Running the code below shows that, yes, the function is called many times and you can use console.log to see which numbers were passed in:
const sort = (arr, comparefn) => {
for (let i = 1; i < arr.length; i++) {
for (let j = i; j && comparefn(arr[j-1], arr[j]) > 0; j--) {
[arr[j], arr[j-1]] = [arr[j-1], arr[j]];
}
}
};
console.log("on our version:");
const array = [3, 0, 4, 5];
sort(array, (a, b) => console.log(a, b) || (a - b));
console.log("" + array);
console.log("on the builtin:");
console.log("" +
[3, 0, 4, 5].sort((a, b) => console.log(a, b) || (a - b))
);
You ask:
How are the two sets of numbers then sorted in relation to one another?
To be precise with terminology, a and b aren't sets of numbers--they're objects in the array (in your example, they're numbers).
The truth is, it doesn't matter how they're sorted because it's implementation-dependent. Had I used a different sort algorithm than insertion sort, the comparator would probably be invoked on different pairs of numbers, but at the end of the sort call, the invariant that matters to the JS programmer is that the result array is sorted according to the comparator, assuming the comparator returns values that adhere to the contract you stated (< 0 when a < b, 0 when a === b and > 0 when a > b).
In the same sense that I have the freedom to change my sort's implementation as long as I don't breach my specification, implementations of ECMAScript are free to choose the sort implementation within the confines of the language specification, so Array#sort will likely produce different comparator calls on different engines. One would not write code where the logic relies on some particular sequence of comparisons (nor should the comparator produce side effects in the first place).
For example, the V8 engine (at the time of writing) invokes Timsort when the array is larger than some precomputed number of elements and uses a binary insertion sort for small array chunks. However, it used to use quicksort which is unstable and would likely give a different sequence of arguments and calls to the comparator.
Since different sort implementations use the return value of the comparator function differently, this can lead to surprising behavior when the comparator doesn't adhere to the contract. See this thread for an example.

Is the array sort callback function called many times during the course of the sort?
Yes, that's exactly it. The callback is used to compare pairs of elements in the array as necessary to determine what order they should be in. That implementation of the comparison function is not atypical when dealing with a numeric sort. Details in the spec or on some other more readable sites.

Is the array sort callback function called many times during the course of the sort?
Since this is a comparison sort, given N items, the callback function should be invoked on average (N * Lg N) times for a fast sort like Quicksort. If the algorithm used is something like Bubble Sort, then the callback function will be invoked on average (N * N) times.
The minimum number of invocations for a comparison sort is (N-1) and that is only to detect an already sorted list (i.e. early out in Bubble Sort if no swaps occur).

Is the array sort callback function called many times during the course of the sort?
Yes
If so, I'd like to know which two numbers are passed into the function each time.
a: The first element for comparison.
b: The second element for comparison.
In the following example, a will be "2" and b will be "3" in the first iteration
How are the two sets of numbers then sorted in relation to one another?
Elements are sorted according to the return value of the compare function.
greater than 0: sort a after b
less than 0: sort a before b
equal to 0: keep original order of a and b
Here is an example
var arr = [3, 2, 1, 5, 4, 6, 7, 9, 8, 10];
console.log(arr.sort((a, b) => {
console.log(a - b, a, b);
//b-a if sorting in decending order
return a - b;
}));

Javascript find the highest value in 2d array based on array itself

I've a multidimensional array
arr = [[[1,1],[2,1],[3,1],[4,2],[5,2],[6,2],[7,3],[8,4]],
[[1,1],[2,1],[3,1],[4,3],[5,3],[6,4],[7,4],[8,5],[9,5],[10,5]]
];
and so on ... but the dimension of the arr is not fixed, is variable.
I've a variable that tell me where to point my attention
var number = 2;
So my goal is the look in any arr[i] and find the max 1st argument based on the 2nd argument, I try to explain better my self, in this particular case if number is 2 my expectation is to have from arr:
for the 1st array in arr -> 6 (because the second argument is 1,1,1,2,2,2,3 so I've to point at the last 2 and return the 1st argument)
for the 2nd array in arr -> 3 (because 2 is missing and the 1 is the last second argument)
I know is a little tricky
My first idea was to make a for loops where I delete all value over my number, then I can take the very last one, but I think I'm over-complicating all.
There is a better and fast way to achieve the same result?
J

You present lists (arrays) of pairs of numbers, where the pairs are sorted in ascending order, first by the second number, then by the first.
What you seem to ask for is: Given a number to search for among the second numbers, e.g. number = 2, find the last pair where the second number is less than or equal to this number, and return the corresponding first number in this pair.
You state that you could use for loops to solve the problem. A straightforward approach could be like the following snippet:
var arr = [[[1,1],[2,1],[3,1],[4,2],[5,2],[6,2],[7,3],[8,4]],
[[1,1],[2,1],[3,1],[4,3],[5,3],[6,4],[7,4],[8,5],[9,5],[10,5]]
];
var findNumber = 2;
var result = [];
for(var i = 0; i < arr.length; i++){
var maxIndex = -1;
for(var j = 0;
j < arr[i].length && arr[i][j][1] <= findNumber;
j++){
maxIndex = j;
}
result.push(arr[i][maxIndex][0]);
}
//gives the expected answers 6 and 3
console.log(result);
Then you ask:
There is a better and fast way to achieve the same result?
A solution involving .map and .reduce could be considered more elegant, like the following:
var arr = [[[1,1],[2,1],[3,1],[4,2],[5,2],[6,2],[7,3],[8,4]],
[[1,1],[2,1],[3,1],[4,3],[5,3],[6,4],[7,4],[8,5],[9,5],[10,5]]
];
var findNumber = 2;
var result = arr.map(function(val){
return val[val.reduce(function(acc, curr, index){
return curr[1] <= findNumber? index : acc;
}, -1)][0];
});
//gives the expected answers 6 and 3
console.log(result);
However, in terms of performance, for loops are likely to perform better (run faster) and are easy to comprehend.
In addition, you mention that
the dimension of the arr is not fixed
You would need to post some code examples on how the dimensionality of your data may vary before it would be possible to provide any answer that handles this aspect.
Update
To handle a single array of pairs, you do not need the outer loop or .map(). Putting the solution above into a reusable function:
function lookupFirstNumberFromSecond(secondNumberToFind, arr){
var j = 0, maxIndex = -1;
while(j < arr.length && arr[j][1] <= secondNumberToFind){
maxIndex = j++;
}
return arr[maxIndex][0];
}
//gives the expected answer 6
console.log(lookupFirstNumberFromSecond(
2,
[[1,1],[2,1],[3,1],[4,2],[5,2],[6,2],[7,3],[8,4]]
));
//gives the expected answer 3
console.log(lookupFirstNumberFromSecond(
2,
[[1,1],[2,1],[3,1],[4,3],[5,3],[6,4],[7,4],[8,5],[9,5],[10,5]]
));

I'm not entirely sure about what you are trying to achieve but I guess Array.reduce is a pretty elegant solution to get a single value out of an array.
e.g.
var number = 2;
[[1,4],[2,1],[3,1],[4,2],[5,2],[6,2],[7,3],[8,4]]
.reduce(function (a, b) {
return ((!a || b[0] > a[0]) && b[1] === number) ? b : a;
});

Not entirely sure what you're trying to solve either, but if you're trying to get the max value in a n dimensional array, then the most straightforward method is to solve this standardly in a recursive manner
function recurseMax(arr) {
if (Number.isInteger(arr)) {
return arr;
}
else {
answer = 0;
for (let i = 0; i < arr.length; i++) {
answer = answer > recurseMax(arr[i]) ? answer : recurseMax(arr[i]);
}
return answer;
}
}
console.log(recurseMax([1,[3, 5], [5, 6, 7, 10], [2, [3, [500]]]])); //Outputs 500
For each element, either is a number or another possible multidimensional element, so we recursively find its max. This avoids potential overhead from a reduce operation (though I'm not experienced enough to speak with confidence whether or not it is completely faster, not really sure of the optimizations V8 can do on reduce or a plain old recursion loop). Either way, the solution is fairly straightforward.
I am answering the question based on the assumption that you mean that the array can have a max dimension of n.

Sorting an array in javascript

var num_list = [1, 2, 3, 4];
function num_order(a, b) {return b-a; }
num_list.sort(num_order);
I've been through blogs and i have searched on this topic but to no avail. All describe this function to be sorting in descending order but none specify how does that happen..
For example, what values are stored in the parameters a and b and how are these values assigned.. Finally what results the function passes and how does sort method do to those values.. Any help would be appreciated..
Let me specify that there is a post similar to this but the answer in that post is not clear.. One of the user has provided a link here which makes it much clearer

The parameter you pass to the sort method is the comparison function. It will define the order the elements are sorted.
To see what values are being passed to the parameters a and b. Try this in your console:
var num_list = [1, 2, 3, 4];
num_list.sort(function(a, b) {
debugger;
return b-a;
})
MDN has good documentation on the compare function: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/sort

Have a look at any sort algorithm : at some point they need to compare two elements of the array.
The sort function of most Browsers is heapsort or quicksort, but i will take bubble sort as an example of a sort algorithm :
n = array size
repeat
swapped = false
for i= 0 to n-2
if array [i] > array [i+1] then
swap ( array [i] , array [i+1] )
swapped = true
end for
n = n -1
until swapped = false
we can easily rewrite the comparison to use a comparison function :
n = array size
repeat
swapped = false
for i= 0 to n-2
a = array [i]
b = array [i+1]
if compareFunction(a,b) > 0 then
swap ( array [i] , array [i+1] )
swapped = true
end for
n = n -1
until swapped = false
with :
compareFunction (a,b)
return a-b
So the comparison function is just a function that returns an integer that reflect the items order.
If two elements are the same => the function returns 0, first is bigger => returns >0, second is bigger => returns <0.
Using a comparison function allows to sort any kind of array (i.e. an array containing any kind of item), while still using the very same sort algorithm as the one used for integer sort.

How to create an array in JavaScript whose indexing starts at 1?

By default the indexing of every JavaScript array starts from 0. I want to create an array whose indexing starts from 1 instead.
I know, must be very trivial... Thanks for your help.

It isn't trivial. It's impossible. The best you could do is create an object using numeric properties starting at 1 but that's not the same thing.
Why exactly do you want it to start at 1? Either:
Start at 0 and adjust your indices as necessary; or
Start at 0 and just ignore index 0 (ie only use indices 1 and up).

A simple solution is to fill the zeroth item:
var map = [null, 'January', 'February', 'March'];
'First month : ' + map[1];
Semantically it would be better to use an object:
var map = {1:'January', 2:'February', 3:'March'};
'First month : ' + map[1];
Note these keys are not ints actually, object keys are always strings.
Also, we can't use dot notation for accessing. (MDN - Property Accessors)
I'd choose the first solution, which I think is less confusing.

Since this question also pops up for a Google search like "javascript start array at 1" I will give a different answer:
Arrays can be sliced. So you can get a sliced version of the Array like this:
var someArray = [0, 1, 2, 3];
someArray.slice(1);
[1, 2, 3]
someArray.slice(2, 4);
[2, 3]
Source: https://developer.mozilla.org/en/docs/Web/JavaScript/Reference/Global_Objects/Array/slice

You could use delete to remove the first element like so:
let arr = ['a','b','c'];
delete arr[0];
console.log(arr[0]);
console.log(arr[1]);
Or just not define it at all:
let arr = [,'b','c'];
console.log(arr[0]);
console.log(arr[1]);
If you want to make sure that you always get the first truthy element regardless of the index and have access to ES6 you can use:
arr.find(Boolean)

The question asks "How to create an array in JavaScript whose indexing starts at 1". The accepted answer states "It isn't trivial. It's impossible."
This is true, and should be understood for good reason. However, you can create an array and omit setting the first element, in which case it will still exist (hence the accepted answer being correct) but it'll be marked as empty for you.
let usernames = ['bob', 'sally', 'frank']
let myArray = [];
let arrayIndex = 1;
usernames.map(username => {
myArray[arrayIndex] = username;
arrayIndex++;
})
console.log(myArray);
Array(4) [ <1 empty slot>, "bob", "sally", "frank" ]
1: "bob"
2: "sally"
3: "frank"
​length: 4
Notice that the length is "4".
console.log(myArray[0]);
undefined
Using this, there's a quirk in our favour whereby using Object.keys() on an array doesn't return empty (undefined) elements. So with the above array:
console.log(Object.keys(myArray).length);
3
Note: This is arguably a little hacky so use it with caution.
As zero of something rarely exists in our world, doing this might be useful where you are only going to access pre-defined indexes. An example would be if you have pages of a book. There isn't a page 0 as that makes no sense. And if you are always access a value directly, e.g.
const currentPage = pages[1];
Then this is fine in my opinion, as long as the code shows intent. Some will argue that ignoring a valid array index is futile, and I don't fully disagree. However, it's also futile and very annoying when I want to get page 35 of a book and the array index is 34. Meh!
When you loop your (dodgy) array with map it ignores the 0 index you didn't want:
myArray.map((value, index) => {
console.log(index);
console.log(value);
})
1
bob
2
sally
3
frank
For general use however, you should use index 0, so when you loop some data and spit things out you're not going to get caught out by the first one being empty.

Okay, according to #cletus you couldn't do that because it's a built-in javascript feature but you could go slightly different way if you still want that. You could write your own index-dependent functions of Array (like reduce, map, forEach) to start with 1. It's not a difficult task but still ask yourself: why do I need that?
Array.prototype.mapWithIndexOne = function(func) {
const initial = []
for (let i = 1; i < this.length + 1; i++) {
initial.push(func(this[i - 1], i))
}
return initial
}
const array = ['First', 'Second', 'Third', 'Fourth', 'Fifth']
console.log(array.mapWithIndexOne((element, index) => `${element}-${index}`))
// => ["First-1", "Second-2", "Third-3", "Fourth-4", "Fifth-5"]
Codepen: https://codepen.io/anon/pen/rvbNZR?editors=0012

Using Array.map
[,1,2,3].map((v, i) => ++i)

Just wanted to point out that an index in c ish languages is also the offset from the first element. This allows all sorts of offset math where you don't have to subtract 1 before doing the math, only to add the 1 back later.
if you want a "1" array because the indexes are mapped to other values, that's the case for an enumeration or a hash.

First add this function to your javascript codes:
var oneArray = function(theArray)
{
theArray.splice(0,0,null);
return theArray
}
Now use it like this:
var myArray= oneArray(['My', 'name', 'is', 'Ram']);
alert(myArray[1]); << this line show you: My
See live demo

Just prepend a null:
a = [1, 2, 3, 4]
a.unshift(null)
a[3] // 3

Simple, just make two changes to the classic Javascript for loop.
var Array = ['a', 'b', 'c'];
for (var i = 1; i <= Array.length; i++) {
//"i" starts at 1 and ends
//after it equals "length"
console.log(i);
}