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How would I check if an array in another array contains x?

I am coding chess and am wondering how I would check if an array in another array contains x.
It would work something like this:
if(array contains array containing x)

You can use flat() to monodimensionalize the array and includes() to check.
let x = [0, 1, 2, [3, 4]];
if(x.flat(Infinity).includes(3)) {
console.log("yes");
}
EDIT: Using Infinity as a parameter, as said in the comments, for multi-level arrays.

If I understood your question you can do something like this
const contains = (data, x) => data.some(d => d.includes(x))
const data1 = [
[1],
[2, 3, 4],
[5]
]
console.log(contains(data1, 1), contains(data1,2), contains(data1, 7))
if you want a recursive approach you can do something like this
const contains = (data, x) => {
if(typeof data !== 'object'){
return data === x
}
return data.some(d => contains(d, x))
}
const data1 = [
[
[1]
],
[2],
[[3, [4]]]
]
console.log(contains(data1, 1), contains(data1, 2),contains(data1, 3),contains(data1, 5))

Get item that is present in all arrays (Javascript)

I have an object containing a bunch of arrays like this:
{
names: [0, 1, 2],
gender: [2, 5, 1],
boolean: [7, 2, 1, 6]
}
How can I get the value which is present in all arrays, in this case, 1?

A way of achieving this is by getting the distinct values for all the key values and then looping through the values, checking the value exists for each key's array.
const obj = {
names: [0, 1, 2],
gender: [2, 5, 1],
boolean: [7, 2, 1, 6]
}
// [1] Get the object's keys
const keys = Object.keys(obj);
// [2] Get distinct values by combining all the key arrays.
const distinctValues = [...new Set(keys.reduce((all, curr) => {
return all.concat(obj[curr]);
}, []))];
// [3] Filter the distinct values by checking each value against
// the key's array
const presentInAll = distinctValues.filter((value) => {
return keys.every((key) => obj[key].includes(value));
});
console.log(presentInAll);

I thought I'd add another solution that does not make any temporary copies of the original data which can be useful if the data is larger or memory management is important.
const data = {
names: [0, 1, 2],
gender: [2, 5, 1],
boolean: [7, 2, 1, 6]
};
function findCommon(...arrays) {
// sort arrays by length so we optimize and iterate first by the shortest array
arrays.sort((a, b) => {
return a.length - b.length;
});
let results = new Set();
// for each item in the first array
for (let item of arrays[0]) {
// look in other arrays for this value
let found = true;
for (let i = 1; i < arrays.length; i++) {
if (!arrays[i].includes(item)) {
found = false;
break;
}
}
if (found) {
results.add(item);
}
}
return results;
}
let results = findCommon(data.names, data.gender, data.boolean);
console.log(Array.from(results));

flatten an array without using .flat();

how can i flatten an array without using flat(). by 1 level?
so far i have this
function flatten(array) {
let flattened = [];
for (let i = 0; i < array.length; i++) {
const current = array[i];
for (let j = 0; i < current.length; j++) {
flattened.push(current[j])
}
}
return flattened
}
console.log(flatten([['foo', 'bar'], ['baz', 'qux']]));
// -> ["foo", "bar", "baz", "qux"]
flatten([[1], [2], 3, 4, [5]]);
// -> [1, 2, 3, 4, 5]
flatten([false, [true, [false]], [true]]);
// -> [false, true, [false], true]
flatten([]);
// -> []
and its crashing my memory

I hope this helps
var twoDimension = [[1], [2], 3, 4, [5]];
var plano = twoDimension.reduce((acc, el) => acc.concat(el), []);
console.log(plano);

You could use Array.reduce and the spread syntax:
function flatten(array) {
return array.reduce(
(accumulator, item) => {
// if `item` is an `array`,
// use the `spread syntax` to
// append items of the array into
// the `accumulator` array
if (Array.isArray(item)) {
return [...accumulator, ...item];
}
// otherwise, return items in the
// accumulator plus the new item
return [...accumulator, item];
}
, []); // initial value of `accumulator`
}
console.log(flatten([['foo', 'bar'], ['baz', 'qux']]));
// -> ["foo", "bar", "baz", "qux"]
console.log(flatten([[1], [2], 3, 4, [5]]));
// -> [1, 2, 3, 4, 5]
console.log(flatten([false, [true, [false]], [true]]));
// -> [false, true, [false], true]
console.log(flatten([]));
// -> []
References:
Array.reduce - MDN
Spread syntax - MDN

To flatten by a single level only, Array#concat() can be leveraged. It accepts any amount of arguments, so an array can be spread into the function call:
[].concat(...arr)
This avoids any explicit loops. JavaScript handles everything:
function flatten(arr) {
return [].concat(...arr);
}
console.log(flatten([['foo', 'bar'], ['baz', 'qux']]));
// -> ["foo", "bar", "baz", "qux"]
console.log(flatten([[1], [2], 3, 4, [5]]));
// -> [1, 2, 3, 4, 5]
console.log(flatten([false, [true, [false]], [true]]));
// -> [false, true, [false], true]
console.log(flatten([]));
// -> []

You can use the following method if your array have primitive data type and want to flat it completely:
arr.toString().split(',');

you can use the reducer of javascript as an alternative to flat().
const arr = [1, 2, [3, 4]];
arr.reduce((acc, val) => acc.concat(val), []);
// [1, 2, 3, 4]
or you can use decomposition syntax
const flattened = arr => [].concat(...arr);
For more details, go to Mozilla MDN

You have an error here:
for (let j = 0; i < current.length; j++) {
// ^ wrong variable, should be j
And you need to check if the value is not an array, then just push the current value and continue the loop.
function flatten(array) {
let flattened = [];
for (let i = 0; i < array.length; i++) {
const current = array[i];
if (!Array.isArray(current)) {
flattened.push(current);
continue;
}
for (let j = 0; j < current.length; j++) {
flattened.push(current[j])
}
}
return flattened
}
console.log(flatten([['foo', 'bar'], ['baz', 'qux']]));
// -> ["foo", "bar", "baz", "qux"]
console.log(flatten([[1], [2], 3, 4, [5]]));
// -> [1, 2, 3, 4, 5]
console.log(flatten([false, [true, [false]], [true]]));
// -> [false, true, [false], true]
console.log(flatten([]));
// -> []

You had a typo where in your innermost loop you set i to 0 instead of j. The only other thing you needed to do was check to see if each element in the outer array was scalar (not an array) and push it to the returned array if so.
function flatten(arr) {
let flat = []
for (let i=0; i < arr.length; i++) {
const cur = arr[i]
if(!Array.isArray(cur)){
flat.push(cur)
}else{
for (let j=0; j < cur.length; j++) {
flat.push(cur[j])
}
}
}
return flat
}
console.log(flatten([['foo','bar'],['baz','qux']]))
console.log(flatten([[1],[2],3,4,[5]]))
console.log(flatten([false,[true,[false]],[true]]))
console.log(flatten([]))

Well you can use spread operator with reduce.
function flatten(array) {
return array.reduce((a,v) => [...a, ...(Array.isArray(v) ? v : [v])], []);
}
console.log(flatten([['foo', 'bar'], 'baz', 'qux']))

Following could be used as a general implementation of Array.prototype.flat()
function flattenArray(arr, depth = 1) {
if (!Array.isArray(arr)) {
return [arr];
}
return depth > 0
? arr.reduce(
(acc, curr) =>
acc.concat(
Array.isArray(curr) ? flattenArray(curr, depth - 1) : curr
),
[]
)
: arr.slice();
}
const a = [1, 2, 3, 4];
const b = "a";
const c = [1, [2, 3], 4];
const d = [1, [2, [3, 4], 5], 6];
const e = [1, [2, [3, [4, [5], [6]], 7], 8], 9];
console.log(flattenArray(a, Infinity));
console.log(flattenArray(b, Infinity));
console.log(flattenArray(c, Infinity));
console.log(flattenArray(d, Infinity));
console.log(flattenArray(e, Infinity));

There is another interesting way to do it.
Stringify the array
remove all array start symbol ([) and array end symbol(])
Add array start symbol at the beginning and array end symbol at the end.
Now parse the resulting string
const arr2 = [0, 1, 2, [5, [10, [3, 4]]]]
const arr2 = [0, 1, 2, [5, [10, [3, 4]]]]
console.log( JSON.parse('['+ JSON.stringify(arr2).replace(/\[/g, ' ').replace(/\]/g, ' ') + ']'))

Suppose given flatten number list without using the flat function is:
let array = [2,3,[5,2,[6,[3, [4, 5, [5, 1, 3]]]],1,1],9];
//let array= [2,3,[5,2,[6,[3, [4, 5, [5, {"key":"value"}, 3]]]],1,1],9];
//achieve above commented nested array condition using second approach.
The best answer already given by #Mahipal that would be first approach i.e.
array.toString().split(',')
with number array conversion
array.toString().split(',').map(n => +n)
another approach would be using function recursion without toString()
function flatter(arr) {
if (!Array.isArray(arr) && (!isNaN(arr) || typeof arr === "object")) {
return arr;
}
return arr.reduce((a, b) => {
a.push(...[].concat(flatter(b)));
return a;
}, [])
}
flatter(array);
and output is:
[ 2, 3, 5, 2, 6, 3, 4, 5, 5, 1, 3, 1, 1, 9 ]
Hope this would help many ones.

//Using Recursion
const arr = [1, 2, 3, 4, [5, 6, [6, 7], 7, 8]]
let arr2 = [];
function flat(arr) {
arr.forEach(element => {
if (typeof (element) == 'object') {
flat(element);
} else {
arr2.push(element);
}
});
}
flat(arr);
console.log(arr2);

var multiDimensionArray = [["a"],["b","c"],["d"]]; //array of arrays
var flatArray = Array.prototype.concat.apply([], multiDimensionArray); //flatten array of arrays
console.log(flatArray);

This worked for me:
function myFlattern(arr) {
let check;
do{
check=false;
for(let i=0;i<arr.length;i++)
{
if(Array.isArray(arr[i]))
{
check=true;
let x=arr[i];
arr.splice(i,1,...x);
}
}
}while(check)
return arr;
}

A possible alternative would be without using flat():
var arr = [['object1', 'object2'],['object1'],['object1','object2','object3']];
var flattened = [].concat.apply([],arr);

You can use this to forget about the depth of nesting:
let multiArr = [1, [1, 2, [3, 4]], [2, 4, [45, 98]]];
while (multiArr.find((elem) => Array.isArray(elem))) {
multiArr = [].concat.apply([], multiArr);
}
console.log(multiArr);

Is there any direct method which returns the sum of lengths of arrays in the array of objects?

I have an array of objects and I want to find the sum of length of arrays of a certain property(key).
I have this array of objects like
var myArray =
[{
"a" : 1,
"b" : another Array
},
{
"c" : 2,
"b" : another Array
}
.....
]
Is there any way to simplify this below process?
var lengthOfEachObject = myArray.map(function(Obj){
if(Obj.b){
return Obj.b.length;
}
else{
return 0;
}
});
lengthofEachObject.reduce(function(x,y){
return x+y;
})
Answers can also include use of external libraries.

You can use .reduce without .map, this way you can get the total sum by only iterating once over your array. Furthermore, you can use destructing assignment instead of your if statements to set a default value for your b.length if it doesn't exist.
See working example below:
const arr = [{a: 1, b: [1, 2, 3, 4, 5] }, {c: 2, b: [1, 2, 3]}, {e: 3}],
total = arr.reduce((acc, {b={length:0}}) => acc + b.length, 0);
console.log(total);

You can use lodash's _.sumBy():
var myArray = [{"a":1,"b":[1,2,3]},{"c":2,"b":[4,5,6,7,8]},{"c":2}]
var result = _.sumBy(myArray, 'b.length')
console.log(result)
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.min.js"></script>

You could pull out the length with a closure over the wanted key and map this value.
const
length = k => ({ [k]: { length = 0 } = {} }) => length,
add = (a, b) => a + b,
array = [{ a: 1, b: [1, 2, 3, 4] }, { c: 2, b: [1, 2, 3] }];
console.log(array.map(length('b')).reduce(add));
console.log(array.map(length('a')).reduce(add));

How to merge 2D number array based on the first items of inner sub array?

Given 2d-array [ [ 2, 12 ], [ 3, 12 ], [ 3, 15 ], [ 5, 15 ] ].
I would like to check whether the first element of subarray is equal to the neighbour's first element. And if they are equal, I would like to sum their second elements and replace these 2 with the new array.
So the output must be: [ [ 2, 12 ], [ 3, 27 ], [ 5, 15 ] ].
Thank you!
I've tried the regular for loop, but it throws the TypeError saying it can't access the value of undefined
for (let i = 0; i < arr.length; i++) {
if (arr[i][0] === arr[i+1][0]) console.log( arr[i] );
}
TypeError: Cannot read property '0' of undefined
at evalmachine.<anonymous>:5:29
at Script.runInContext (vm.js:74:29)
at Object.runInContext (vm.js:182:6)
at evaluate (/run_dir/repl.js:133:14)
at ReadStream.<anonymous> (/run_dir/repl.js:116:5)
at ReadStream.emit (events.js:180:13)
at addChunk (_stream_readable.js:274:12)
at readableAddChunk (_stream_readable.js:261:11)
at ReadStream.Readable.push (_stream_readable.js:218:10)
at fs.read (fs.js:2124:12)

You can do something like this using reduce
Check if the accumulator's last item has the same key as the current item in context. If yes, sum the value parts. Else, push the item to accumulator
const data = [[2, 12], [3, 12], [3, 15], [5, 15]];
const newArray = data.reduce((acc, a, i) => {
if (i > 0 && a[0] === data[i-1][0])
acc[acc.length-1][1] += a[1]
else
acc.push(a);
return acc
}, [])
console.log(newArray)
The above logic using ternary and comma operators
const data = [[2, 12], [3, 12], [3, 15], [5, 15]];
const newArray = data.reduce((acc, a, i) => (
i > 0 && a[0] === data[i-1][0]
? acc[acc.length-1][1] += a[1]
: acc.push(a), acc), [])
console.log(newArray)

You coud reduce the array and take a helper variable for the last pushed element of the result set;
var array = [[2, 12], [3, 12], [3, 15], [5, 15]],
grouped = array.reduce((last => (r, a) => {
if (last && last[0] === a[0]) {
last[1] += a[1];
} else {
r.push(last = a);
}
return r;
})(), []);
console.log(grouped);