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Why is splicing an array in JavaScript removing the last few elements?

I'm currently working on an array function that converts subarrays of consecutive numbers into strings denoting that range of numbers — for example, this array...
[1, 2, 3, 6, 8, 10, 11, 12, 15, 18]
...would become this array:
["1-3", 6, 8, "10-12", 15, 18]
I've been able to develop a function that mostly works, but I've encountered a weird error where all the elements past the final range of numbers spliced into the array are completely deleted. For example, the test array above actually becomes this:
["1-3", 6, 8, "10-12"]
This is the code I've written so far. It's not super pretty yet, but as I mentioned above, it gets the job done right up until the very end:
let testArray = [1, 2, 3, 6, 8, 10, 11, 12, 15, 18];
for (i = 0; i < testArray.length; i++) {
let consecutives = [];
consecutives.push(testArray[i]);
let j = i + 1;
while (j < testArray.length) {
if (testArray[j] == (testArray[j - 1] + 1)) {
consecutives.push(testArray[j]);
j++;
} else {
break;
}
}
if (consecutives.length > 2) {
let range = String(testArray[i]) + "-" + String(testArray[j - 1]);
console.log(testArray);
console.log(testArray[i]);
console.log(testArray[j]);
testArray.splice(i, j, range);
}
}
console.log(testArray);
These are the console logs output by that code:
Array(10) [ 1, 2, 3, 6, 8, 10, 11, 12, 15, 18 ]
1
6
Array(8) [ "1-3", 6, 8, 10, 11, 12, 15, 18 ]
10
15
Array(4) [ "1-3", 6, 8, "10-12" ]
I initially figured this was caused by a mix-up with array indexes, but playing around with the index-1s hasn't fixed the problem yet. Has anyone else ever had a similar issue with JavaScript's splicing, and if so, how were you able to get it working?

The problem lies in one line of code:
testArray.splice(i, j, range);
According to the MDN, the second argument specifies how many elements in the array to delete.
deleteCount
An integer indicating the number of elements in the array to remove from start.
However, the code defines this argument as the index of the last array to remove from:
let j = i + 1;
The solution is to get the difference between i and j before passing it to splice:
testArray.splice(i, j - i, range);

When you do:
testArray.splice(i, j, range);
You are forgetting that j is the right limit index of the array that you want to erase, so you need to subtract i that is the left limit:
testArray.splice(i, j - i, range);
let testArray = [1, 2, 3, 6, 8, 10, 11, 12, 15, 18];
for (i = 0; i < testArray.length; i++) {
let consecutives = [];
consecutives.push(testArray[i]);
let j = i + 1;
while (j < testArray.length) {
if (testArray[j] == (testArray[j - 1] + 1)) {
consecutives.push(testArray[j]);
j++;
} else {
break;
}
}
if (consecutives.length > 2) { // doesn´t it should be > 1 ??
let range = String(testArray[i]) + "-" + String(testArray[j - 1]);
console.log(testArray);
console.log(testArray[i]);
console.log(testArray[j]);
testArray.splice(i, j - i, range);
}
}
console.log(testArray);

Reset counter when not in range and sum progression

I have an arrays of numbers, and specified range if sequence continues (range rule was met between two numbers) then i add value to result and increase counter by one, else i reset the counter and add nothing to result on this step. Better show in an example:
const numbers = [1, 4, 5, 6, 7, 33, 44, 46]; // they are always going to be from smallest to bigger
const progress = [0, 10, 20, 30, 40, 50, 60]; // 70, 80, etc
let res = 0;
for (let i = 1, j = 0; i < numbers.length; i++) {
const range = numbers[i] - numbers[i - 1];
if (range <= 5) {
j += 1;
res += progress[j];
} else {
j = 0;
}
}
res; // 110
Is there better way to approach this problem?

Well, by looking at your code & the explanation you gave, I think you have incremented 'j' before you added progress for 'j'. that portion should be like following...
if (range <= 5) {
res += progress[j];
j += 1;
}
You have asked for a better approach. But it would help if you specified from which perspective/scenario you are looking for a better approach.

you can do the same with reduce method
const numbers = [1, 4, 5, 6, 7, 33, 44, 46]; // they are always going to be from smallest to bigger
const progress = [0, 10, 20, 30, 40, 50, 60]; // 70, 80, etc
let resp = 0;
const result = numbers.reduce((acc, rec, i, arr) => {
if (rec - arr[i - 1] <= 5) {
resp += 1;
acc = acc + progress[resp];
return acc;
}
resp = 0;
return acc;
}, 0);
result;
You can read more about reduce here
Hope it answers your question.
Happy coding!

Identify longest period of variance in an array of integers

I'm trying to write a function that will identify the longest period of variance in an array of numbers. Variance begins when the previous number is higher than the current, and ends when the next number is the same as the current; however, if variance doesn't end, then it is assumed the variance began with the last two numbers.
For example: [10, 5, 3, 11, 8, 9, 9, 2, 10] The longest period of variance in this array is [5, 3, 11, 8, 9], or just 5 (the length). A variance ends when the following number is the same as the current, in this case, 9.
The function I've written works on this case; however, it doesn't when the entire array has variance, such as [10, 5, 10, 5, 10, 5, 10, 5, 10], which returns 8, when it should be 9.
In the case where the previous number is number is always lower, or always higher than the variance would be 2, because it never ended. For example [2, 4, 6, 8] and [8, 6, 4, 2].
I know the issue with the entire array variance can be solved by starting the for loop at 0, but then the other cases become invalid. Any help is greatly appreciated.
Without further ado, here is my code:
function findVariance(numbers) {
if ([0,1].includes(numbers.length)) return numbers.length;
const variance = [[0]];
let greater = numbers[1] > numbers[0];
let lesser = numbers[1] < numbers[0];
for (let i = 1; i < numbers.length; i++) {
let previous = variance.length - 1;
let previousVarianceGroup = variance[previous];
let previousVarianceGroupValue = previousVarianceGroup[previousVarianceGroup.length - 1];
if (greater) {
if (numbers[i] < numbers[previousVarianceGroupValue]) {
previousVarianceGroup.push(i);
greater = false;
lesser = true;
} else {
greater = numbers[i] < numbers[previousVarianceGroupValue];
lesser = numbers[i] < numbers[previousVarianceGroupValue];
variance.push([previousVarianceGroupValue, i]);
}
} else if (lesser) {
if (numbers[i] > numbers[previousVarianceGroupValue]) {
previousVarianceGroup.push(i);
greater = true;
lesser = false;
} else {
greater = numbers[i] > numbers[previousVarianceGroupValue];
lesser = numbers[i] > numbers[previousVarianceGroupValue];
variance.push([previousVarianceGroupValue, i]);
}
} else {
greater = numbers[i] > numbers[previousVarianceGroupValue];
lesser = numbers[i] < numbers[previousVarianceGroupValue];
variance.push([previousVarianceGroupValue, i]);
}
}
const result = [];
for (let i = 0; i < variance.length; i++) {
result[i] = variance[i].length;
}
result.sort();
return result[result.length - 1];
}
console.log(findVariance([10, 5, 3, 11, 8, 9, 9, 2, 10]));
console.log(findVariance([10, 5, 10, 5, 10, 5, 10, 5, 10]));
console.log(findVariance([2, 4, 6, 8]));

Here's what I got (trying to understand the question as best i could)
function calculateVariance(arr) {
// trivial cases
if (arr.length <= 1) { return arr.length; }
// store the difference between each pair of adjacent numbers
let diffs = [];
for (let i = 1; i < arr.length; i++) {
diffs.push(arr[i] - arr[i - 1]);
}
let max = 0;
// if the difference between two numbers is 0, they're the same.
// the base max variance encountered is 1, otherwise it's 2.
// the boolean zen here is that diffs[0] is falsy when it's 0, and truthy otherwise
let count = diffs[0] ? 2 : 1;
// go through the array of differences,
// and count how many in a row are alternating above/below zero.
for (i = 1; i < diffs.length; i++) {
if ((diffs[i] < 0 !== diffs[i - 1] < 0) && diffs[i] && diffs[i - 1]) {
count++;
} else {
max = Math.max(count, max);
// see above
count = diffs[i] ? 2 : 1;
}
}
// account for the maximum variance happening at the end
return Math.max(count, max);
}

You are overcomplicating things a bit, just increase a counter as long as an element equals the next one, reset on equal:
const counts = [];
let count = 0;
for(let i = 0; i < numbers.length - 1; i++) {
if(numbers[i] === numbers[i + 1]) {
counts.push(count);
count = 0;
} else {
count++;
}
}
counts.push(count);
return counts.sort()[counts.length - 1];

How to push multiples of a number to array?

How would one push multiples of a number to an array? For example, if the input is (6), I want to create an array that holds [6, 12, 18, 24, 30, 36, etc...]
The most intuitive method to me does not work.
for (var i = 0; i < 10; i++) {
firstArray.push(arr[0] *= 2);
}
This multiplies the number that comes before it by 2, causing an exponential growth. [14, 28, 56, 112, 224, 448, 896, 1792, etc.]
How would one achieve this?

Problem:
The problem in the code, as commented by Pranav is the use of multiplication by two in the for loop.
Using i iterator index can solve the problem.
firstArray.push(6 * (i + 1));
As i is starting from 0, i + 1 will give the number which is 1-based.
Another Approach:
First add the number
var num = 6,
arr = [num];
Then add the number which is double of the previous in the array.
for (var i = 1; i < 10; i++) {
arr.push(arr[i - 1] + num);
}
var arr = [6];
for (var i = 1; i < 10; i++) {
arr.push(arr[i - 1] + arr[0]);
}
console.log(arr);
The same thing can also be done in single line using for loop.
var arr = [];
for (let i = 0, num = 6; i < 10; i++, num += 6) {
arr.push(num);
}
console.log(arr);

You can use map:
function multiplyArrayElement(num) {
return num * 2;
}
numbers = [6, 12, 18, 24, 30, 36];
newArray = numbers.map(multiplyArrayElement);
https://jsfiddle.net/25c4ff6y/

It's cleaner to use Array.from. Just beware of its browser support.
Array.from({length: 10},(v,i) => (i + 1) * 6)

try this one
for (var i = 0; i < 10; i++) {
firstArray.push(arr[0] * (i+1));
}

var arr = [];
var x = 6; //Your desired input number
var z;
for(var i=1;i<10;i++){
z = (x*i);
arr.push(z);
}
console.log(arr);

"One line" solution with Array.fill and Array.map functions:
var num = 6;
var arr = new Array(10).fill(0).map(function(v, k){ return num *(k + 1); });
console.log(arr); // [6, 12, 18, 24, 30, 36, 42, 48, 54, 60]

Javascript - How Do I Check if 3 Numbers Are Consecutive and Return Starting Points?

If I have an array of [1, 2, 3, 5, 10, 9, 8, 9, 10, 11, 7] and wanted to find each case of 3 consecutive numbers (whether ascending or descending), how would I do that?
Second part would be then to alert an array with the index of each of these sequences.
For ex. the previous array would return [0,4,6,7].
So far I have this... which is a rough start
var arr = [1, 2, 3, 5, 10, 9, 8, 9, 10, 11, 7];
var results = [];
for (var i = 1; i < arr.length; i++) {
if ((arr[i] - arr[i-1] != 1) && (arr[i] - arr[i+1] != 1)) {
results.push(arr[i]);
}
}
alert(results);
Thanks for the help!
Thanks for the math.abs pointer. This is what I ended up doing:
var array = [1, 2, 3, 5, 10, 9, 8, 9, 10, 11, 7];
var indexes = [];
for(var i=0; i < array.length; i++) {
var diff = array[i+1] - array[i];
if(Math.abs(diff)==1 && array[i+1]+diff == array[i+2]) {
indexes.push(i);
}
}
alert(indexes);

It'd be interesting to know the context of this task as well... Anyway, here's my solution:
var arr = [1, 2, 3, 5, 10, 9, 8, 9, 10, 11, 7];
var results = [];
var limit = arr.length - 1;
var sequence = 0;
for (var i = 0; i < limit; ++i) {
var diff = arr[i+1] - arr[i];
if (sequence && sequence === diff) {
results.push(i-1);
continue;
}
sequence = (diff === 1 || diff === -1) // or ... Math.abs(diff) === 1
? diff
: 0;
}
console.log(results);
The idea is simple: we don't need to compare two neighbors twice. ) It's enough to raise a kind of sequence flag if this comparation starts a sequence, and lower it if no sequence is there.

This is a very literal approach to your question - I have only checked forwards numbers, but adding reverse would be done almost in the same way
var arr = [1, 2, 3, 4, 10, 9, 8, 9, 10, 11, 7];
var results = [];
for (var i = 0; i < arr.length; i++) {
// if next element is one more, and one after is two more
if (arr[i+1] == arr[i]+1 && arr[i+2] == arr[i]+2){
// store the index of matches
results.push(i);
// loop through next numbers, to prevent repeating longer sequences
while(arr[i]+1 == arr[i+1])
i++;
}
}
console.log(results);

You need to look closely at your expression in your if statement.
It currently says:
If the difference between the current element and previous element is not 1, and
If the difference between the current element and next element is not 1
then it's a result.
So, on the face of it, that's an incorrect logical statement to determine if the current element is in the middle of a consecutive set of three.
In addition, this doesn't account for an ascending or descending set of three either.
Try figuring out, in words, what the condition would look like and go from there.
Some things to consider
I suggest you start going through the list from i = 2
Research Math.abs

This is I think a simpler way to do it. First check the average of the left and right number is equal to the middle, then check that the absolute value of either neighbor is one.
var arr = [1, 2, 3, 5, 10, 9, 8, 9, 10, 11, 7];
var indexes = [];
for(var i=1; i < arr.length; i++) {
if((arr[i-1]+arr[i+1]) / 2 == arr[i] && Math.abs(arr[i]-arr[i-1]) == 1) {
indexes.push(i-1);
}
}
alert(indexes);

var arr = [1, 2, 3, 5, 10, 9, 8, 9, 10, 11, 7];
var results = [];
for (var i = 0; i < arr.length - 2; i++) {
if ((arr[i+1] - arr[i] === 1) && (arr[i+2] - arr[i+1] === 1)) {
results.push({
i:i,
mode:'up',
arr:[arr[i],arr[i+1],arr[i+2]
});
}
if ((arr[i+1] - arr[i] === -1) && (arr[i+2] - arr[i+1] === -1)) {
results.push({
i:i,
mode:'down',
arr:[arr[i],arr[i+1],arr[i+2]
});
}
}
alert(results);