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Deleting arrays of same elements in 2 dimensional array in Javascript

I am wondering how you would go about deleting arrays that contain the same elements in a 2 dimensional array.
For example:
let 2dArr = [ [1, 2, 3],
[3, 2, 1],
[2, 4, 5],
[4, 5, 2],
[4, 3, 1] ];
This array would delete the second and fourth elements, returning the 2d array:
returnedArr = [ [1, 2, 3],
[2, 4, 5],
[4, 3, 1] ];
How exactly could this be done, preserving the 2d array? I could only think to loop through elements, comparing elements via a sort, and deleting them as you go along, but this would result in an indexing error if an element is deleted.

1) You can easily achieve the result using reduce and Set as:
let twodArr = [
[1, 2, 3],
[3, 2, 1],
[2, 4, 5],
[4, 5, 2],
[4, 3, 1],
];
const set = new Set();
const result = twodArr.reduce((acc, curr) => {
const key = [...curr].sort((a, b) => a - b).join();
if (!set.has(key)) {
set.add(key);
acc.push(curr);
}
return acc;
}, []);
console.log(result);
2) You can also use filter as:
let twodArr = [
[1, 2, 3],
[3, 2, 1],
[2, 4, 5],
[4, 5, 2],
[4, 3, 1],
];
const set = new Set();
const result = twodArr.filter((curr) => {
const key = [...curr].sort((a, b) => a - b).join();
return !set.has(key) ? (set.add(key), true) : false;
});
console.log(result);

const seen = []
const res = array.filter((item) => {
let key = item.sort().join()
if(!seen.includes(key)){
seen.push(key)
return item
}
})
console.log(res)

You can use hash map
let arr = [ [1, 2, 3], [3, 2, 1],[2, 4, 5],[4, 5, 2],[4, 3, 1] ];
let obj = {}
let final = []
for(let i=0; i<arr.length; i++){
// create a key
let sorted = [...arr[i]].sort((a,b)=> a- b).join`,`
// check if this is not present in our hash map
// add value to final out and update hash map accordingly
if(!obj[sorted]){
obj[sorted] = true
final.push(arr[i])
}
}
console.log(final)

Using Array.prototype.filter() and a Set as thisArg
let arr = [ [1, 2, 3],
[3, 2, 1],
[2, 4, 5],
[4, 5, 2],
[4, 3, 1] ];
let res = arr.filter(function(e){
const sorted = [...e].sort((a,b) => a-b).join('|');
return this.has(sorted) ? false : this.add(sorted)
},new Set)
console.log(res)

picking a unique set from an array of arrays

I attempted to ask a more complicated of this before but I couldn't explain it well so I am trying again with a simplified use case.
I will have an array of arrays like the following
var allData = [[1,2,3,4,5],[1,2,3,4,5],[1,2,3,4,5],[1,2,3,4,5],[1,2,3,4,5]]
I need to select 1 element from each array so that I get a unique set like [2,4,1,3,5] easy to do in this case as each array has all values. However this will rarely be the case. Instead I may have
var allData = [[1,2,4],[1,2],[1,2],[2,4,5],[1,2,3,5]]
In this case I couldn't pick 1 or 2 from the first array as that would prevent the 2nd and 3rd from having a unique combination. So something like [4,2,1,5,3] or [4,1,2,5,3] would be the only two possible answers for this combination.
The only way I see to do this is to just go through every combination but these will get fairly large so it doesn't seem reasonable as this happens real time. There are going to be at least 7 arrays, possibly 14 and distantly possible to have 31 so going through every combination would be fairly rough.
The 2nd part is if there is some way to "know" you have the best possible option. Say if there was some way I would know that having a single duplicate is my best case scenario. Even if I have to brute force it if I encounter a 1 duplication solution I would know to stop.
One easy way to get a very simple of this is to just subtract the number of possible choices from the number of elements but this is the correct answer in only the simplest of cases. Is there some type of library or anything to help solve these types of problems? It is a bit beyond my math abilities.
Here is something I have tried but it is too slow for larger sets and can fail. It works sometimes for the 2nd case I presented but only on luck
const allData = [[1,2,4],[1,2],[1,2],[2,4,5],[1,2,3,5]]
var selectedData = []
for (var i in allData){
console.log("length",allData[i].length)
var j = 0
while(j < allData[i].length){
console.log("chekcing",allData[i][j])
if (selectedData.includes(allData[i][j])){
console.log("removing item")
allData[i].splice(j,1)
}
else{j++}
}
var uniqueIds = Object.keys(allData[i])
console.log(uniqueIds)
var randId = Math.floor(Math.random() * uniqueIds.length)
console.log(randId)
selectedData.push(allData[i][randId])
console.log("selectedData",selectedData)
}

You can start with a fairly simple backtracking algorithm:
function pick(bins, n = 0, res = {}) {
if (n === bins.length) {
return res
}
for (let x of bins[n]) {
if (!res[x]) {
res[x] = n + 1
let found = pick(bins, n + 1, res)
if (found)
return found
res[x] = 0
}
}
}
//
let a = [[1, 2, 4], [1, 2], [1, 2], [2, 4, 5], [1, 2, 3, 4]]
console.log(pick(a))
This returns a mapping item => bin index + 1, which is easy to convert back to an array if needed.
This should perform relatively well for N < 10, for more/larger bins you can think of some optimizations, for example, avoid the worst case scenario by sorting bins from smallest to longest, or, depending on the nature of elements, represent bins as bitmasks.

You could count all elements and take various comparison with same indices.
function x([...data]) {
while (data.some(Array.isArray)) {
const
counts = data.reduce((r, a, i) => {
if (Array.isArray(a)) a.forEach(v => (r[JSON.stringify(v)] = r[JSON.stringify(v)] || []).push(i));
return r;
}, {}),
entries = Object.entries(counts),
update = ([k, v]) => {
if (v.length === 1) {
data[v[0]] = JSON.parse(k);
return true;
}
};
if (entries.some(update)) continue;
const grouped = entries.reduce((r, [, a]) => {
const key = JSON.stringify(a);
r[key] = (r[key] || 0) + 1;
return r;
}, {});
Object.entries(grouped).forEach(([json, length]) => {
const indices = JSON.parse(json);
if (indices.length === length) {
let j = 0;
indices.forEach(i => data[i] = data[i][j++]);
return;
}
if (length === 1) {
const value = JSON.parse(entries.find(([_, a]) => JSON.stringify(a) === json)[0]);
indices.forEach(i => data[i] = data[i].filter(v => v !== value));
data[indices[0]] = value;
}
});
}
return data;
}
console.log(...x([[1, 2, 3, 4, 5], [1, 2, 3, 4, 5], [1, 2, 3, 4, 5], [1, 2, 3, 4, 5], [1, 2, 3, 4, 5]]));
console.log(...x([[1, 2, 4], [1, 2], [1, 2], [2, 4, 5], [1, 2, 3, 5]]));
console.log(...x([[1, 2, 4], [1, 2], [1, 2], [2, 4, 5], [1, 2, 3, 5], [6, 7, 8, 9], [6, 7, 8, 9], [6, 7, 8, 10], [6, 7, 8, 10], [6, 7, 8, 10]]));

Here is an implementation based around counting occurrences across the arrays.
It first creates a map indexed by value counting the number of inner arrays each value occurs in. It then sorts by inner array length to prioritize shorter arrays, and then iterates over each inner array, sorting by occurrence and selecting the first non-duplicate with the lowest count, or, if there are no unique values, the element with the lowest count.
const
occurrencesAcrossArrays = (arr) =>
arr
.reduce((a, _arr) => {
[...new Set(_arr)].forEach(n => {
a[n] = a[n] || 0;
a[n] += 1;
});
return a;
}, {}),
generateCombination = (arr) => {
const dist = occurrencesAcrossArrays(arr)
return arr
.sort((a, b) => a.length - b.length)
.reduce((a, _arr) => {
_arr.sort((a, b) => dist[a] - dist[b]);
let m = _arr.find(n => !a.includes(n));
if (m !== undefined) {
a.push(m);
} else {
a.push(_arr[0]);
}
return a;
}, []);
};
console.log(generateCombination([[1, 2, 3, 4, 5], [1, 2, 3, 4, 5], [1, 2, 3, 4, 5], [1, 2, 3, 4, 5], [1, 2, 3, 4, 5]]).toString());
console.log(generateCombination([[1, 2, 4], [1, 2], [1], [2, 4, 5], [1, 2, 3, 5]]).toString());
console.log(generateCombination([[1, 2, 4], [1, 2], [1, 2], [2, 4, 5], [1, 2, 3, 5], [6, 7, 8, 9], [6, 7, 8, 9], [6, 7, 8, 10], [6, 7, 8, 10], [6, 7, 8, 10]]).toString());
Edit
In response to your comment – The situation seems to be emerging because the values all have the same occurrence count and are sequential.
This can be solved by keeping a running count of each value in the result array, and sorting each inner array by both by this running occurrence count as well as the original distribution count.This adds complexity to the sort, but allows you to simply access the first element in the array (the element with the lowest rate of occurrence in the result with the lowest occurrence count across all arrays).
const
occurrencesAcrossArrays = (arr) =>
arr
.reduce((a, _arr) => {
[...new Set(_arr)].forEach(n => {
a[n] = a[n] || 0;
a[n] += 1;
});
return a;
}, {}),
generateCombination = (arr) => {
const dist = occurrencesAcrossArrays(arr)
return arr
.sort((a, b) => a.length - b.length)
.reduce((acc, _arr) => {
_arr.sort((a, b) => (acc.occurrences[a] || 0) - (acc.occurrences[b] || 0) || dist[a] - dist[b]);
let m = _arr[0]
acc.occurrences[m] = acc.occurrences[m] || 0;
acc.occurrences[m] += 1;
acc.result.push(m);
return acc;
}, { result: [], occurrences: {} })
.result; // return the .result property of the accumulator
};
console.log(generateCombination([[2, 3, 4, 5, 6], [2, 3, 4, 5, 6], [2, 3, 4, 5, 6], [2, 3, 4, 5, 6], [2, 3, 4, 5, 6], [2, 3, 4, 5, 6], [2, 3, 4, 5, 6]]).toString());
// 2,3,4,5,6,2,3
console.log(generateCombination([[1, 2, 3, 4, 5], [1, 2, 3, 4, 5], [1, 2, 3, 4, 5], [1, 2, 3, 4, 5], [1, 2, 3, 4, 5]]).toString());
// 1,2,3,4,5
console.log(generateCombination([[1, 2, 4], [1, 2], [1], [2, 4, 5], [1, 2, 3, 5]]).toString());
// 1,2,4,5,3
console.log(generateCombination([[1, 2, 4], [1, 2], [1, 2], [2, 4, 5], [1, 2, 3, 5], [6, 7, 8, 9], [6, 7, 8, 9], [6, 7, 8, 10], [6, 7, 8, 10], [6, 7, 8, 10]]).toString());
//1,2,4,5,3,9,6,10,7,8
console.log(generateCombination([[1], [2, 3,], [3, 4, 5], [3, 4, 5, 6], [2, 3, 4, 5, 6, 7]]).toString());
// 1,2,4,6,7
A note on .reduce()
If you're having trouble getting your head around .reduce() you can rewrite all the instances of it in this example using .forEach() and declaring accumulator variables outside of the loop. (This will not always be the case, depending on how you manipulate the accumulator value within a reduce() call).
Example below:
const occurrencesAcrossArrays = (arr) => {
const occurrences = {};
arr.forEach(_arr => {
[...new Set(_arr)].forEach(n => {
occurrences[n] = occurrences[n] || 0;
occurrences[n] += 1;
});
});
return occurrences;
};
const generateCombination = (arr) => {
const dist = occurrencesAcrossArrays(arr);
const result = [];
const occurrences = {};
arr.sort((a, b) => a.length - b.length);
arr.forEach(_arr => {
_arr.sort((a, b) => (occurrences[a] || 0) - (occurrences[b] || 0) || dist[a] - dist[b]);
let m = _arr[0]
occurrences[m] = occurrences[m] || 0;
occurrences[m] += 1;
result.push(m);
});
return result;
};
console.log(generateCombination([[2, 3, 4, 5, 6], [2, 3, 4, 5, 6], [2, 3, 4, 5, 6], [2, 3, 4, 5, 6], [2, 3, 4, 5, 6], [2, 3, 4, 5, 6], [2, 3, 4, 5, 6]]).toString());
// 2,3,4,5,6,2,3
console.log(generateCombination([[1, 2, 3, 4, 5], [1, 2, 3, 4, 5], [1, 2, 3, 4, 5], [1, 2, 3, 4, 5], [1, 2, 3, 4, 5]]).toString());
// 1,2,3,4,5
console.log(generateCombination([[1, 2, 4], [1, 2], [1], [2, 4, 5], [1, 2, 3, 5]]).toString());
// 1,2,4,5,3
console.log(generateCombination([[1, 2, 4], [1, 2], [1, 2], [2, 4, 5], [1, 2, 3, 5], [6, 7, 8, 9], [6, 7, 8, 9], [6, 7, 8, 10], [6, 7, 8, 10], [6, 7, 8, 10]]).toString());
//1,2,4,5,3,9,6,10,7,8
console.log(generateCombination([[1], [2, 3,], [3, 4, 5], [3, 4, 5, 6], [2, 3, 4, 5, 6, 7]]).toString());
// 1,2,4,6,7

You could solve this problem using a MILP-model. Here is one implementation in MiniZinc (data has been extended to seven days):
int: Days = 7;
int: Items = 5;
set of int: DAY = 1..Days;
set of int: ITEM = 1..Items;
array[DAY, ITEM] of 0..1: A = % whether item k is allowed on day i
[| 1, 1, 0, 1, 0
| 1, 1, 0, 0, 0
| 1, 1, 0, 0, 0
| 0, 1, 0, 1, 1
| 1, 1, 0, 0, 0
| 0, 1, 0, 1, 1
| 1, 1, 1, 0, 1 |];
array[DAY, ITEM] of var 0..1: x; % 1 if item selected k on day i, otherwise 0
array[DAY, DAY, ITEM] of var 0..1: w; % 1 if item k selected on both day i and day j, otherwise 0
% exactly one item per day
constraint forall(i in DAY)
(sum(k in ITEM)(x[i, k]) = 1);
% linking variables x and w
constraint forall(i, j in DAY, k in ITEM where i < j)
(w[i, j, k] <= x[i, k] /\ w[i, j, k] <= x[j, k] /\ w[i, j, k] >= x[i, k] + x[j, k] - 1);
% try to minimize duplicates and if there are duplicates put them as far apart as possible
var int: obj = sum(i, j in DAY, k in ITEM where i < j)(((Days - (j - i))^2)*w[i, j, k]);
solve minimize obj;
output
["obj="] ++ [show(obj)] ++
["\nitem="] ++ [show([sum(k in ITEM)(k*x[i, k]) | i in DAY])];
Running gives:
obj=8
item=[2, 1, 5, 4, 3, 2, 1]
The following package looks promising for a JavaScript implementation: https://www.npmjs.com/package/javascript-lp-solver

Array algorithm: How to reform this nested array use javascript?

I have a table like this:
The colored blocks have coordinates as an array:
[
[1, 1],
[2, 1],
[2, 4],
[2, 5],
[3, 2],
[3, 4],
[4, 4],
[4, 5],
[4, 6],
[4, 7],
[5, 3],
[6, 3],
[6, 4],
[6, 5]
]
Now I want to group the neighboring blocks (horizontal and vertical) to independent child array.
The output estimated is like:
[
[
[1, 1],
[2, 1]
],
[
[2, 4],
[2, 5],
[3, 4],
[4, 4],
[4, 5],
[4, 6],
[4, 7]
],
[
[3, 2]
],
[
[5, 3],
[6, 3],
[6, 4],
[6, 5]
]
]
How to use a function to do this?
Edit: I tried to iterate each value in the input array and compare to [1, 1], if one of the coordinate is the same, push them to an new array and delete in the input array, and use recursion to do this again. But I stuck by a problem...as I should group [2, 5] and [4, 5] but cannot group [4, 4] and [6, 4]

You could filter the grouped items and build a new group with the own indices.
const
data = [[1, 1], [2, 1], [2, 4], [2, 5], [3, 2], [3, 4], [4, 4], [4, 5], [4, 6], [4, 7], [5, 3], [6, 3], [6, 4], [6, 5]],
offsets = [[-1, 0], [1, 0], [0, -1], [0, 1]],
groups = data.reduce((r, [i, j]) => {
const
own = [];
temp = r.filter(group => {
const found = group.some(g => offsets.some(o => i + o[0] === g[0] && j + o[1] === g[1]));
if (!found) return true;
own.push(...group);
});
return [...temp, [...own, [i, j]]];
}, []);
console.log(groups);
.as-console-wrapper { max-height: 100% !important; top: 0; }

It seems to me that you are having just a logic problem, not a JavaScript, since you know how to nest the array, is just having a problem to find the right logic to group neighbors.
You can't just compare one part of the coordinate, since the other can me really distant. You need to compare both. If you want to find the neighbor coordinate you need to have one part of the coordinate equal and the another part exactly with 1 distance away.
var arr =
[
[1, 1],
[2, 1],
[2, 4],
[2, 5],
[3, 2],
[3, 4],
[4, 4],
[4, 5],
[4, 6],
[4, 7],
[5, 3],
[6, 3],
[6, 4],
[6, 5]
];
var groupedArr = new Array();
for(i = 0; i < arr.length; i++)
{
var grouped = false;
for(j = 0; j < groupedArr.length; j++)
{
for(k = 0; k < groupedArr[j].length; k++)
{
if((arr[i][0] == groupedArr[j][k][0] && Math.abs(arr[i][1] - groupedArr[j][k][1]) == 1)
|| (arr[i][1] == groupedArr[j][k][1] && Math.abs(arr[i][0] - groupedArr[j][k][0]) == 1))
{
groupedArr[j].push(arr[i]);
grouped = true;
break;
}
}
if(grouped)
break;
}
if(!grouped)
{
var newGroup = new Array();
newGroup.push(arr[i]);
groupedArr.push(newGroup);
}
}
console.log(groupedArr);

looks like clustering problem. But we can solve with this simple code for small data.
ax = [[1, 1], [2, 1], [2, 4], [2, 5], [3, 2], [3, 4], [4, 4], [4, 5], [4, 6], [4, 7], [5, 3], [6, 3], [6, 4], [6, 5]];
console.log(JSON.stringify(reForm(ax)));
function reForm(a) {
const ret = [];
while (a.length > 0) {
var sub = [a.shift()];
var i=0;
while (i<sub.length){
var j=0
while(j<a.length){
if((a[j][0]==sub[i][0] && Math.abs(a[j][1]-sub[i][1])==1) ||
(a[j][1]==sub[i][1] && Math.abs(a[j][0]-sub[i][0])==1))
sub.push(...a.splice(j,1));
else j++
}
i++;
}
ret.push(sub);
}
return ret;
}

You can do this by using recursion
var array =[
[1, 1],
[2, 1],
[2, 4],
[2, 5],
[3, 2],
[3, 4],
[4, 4],
[4, 5],
[4, 6],
[4, 7],
[5, 3],
[6, 3],
[6, 4],
[6, 5]
]
var returnval = new Array();
while(array.length>0)
{
var temp = new Array();
var item = array[0];
array.splice(0, 1);
temp.push(item);
findnext(item,temp);
returnval.push(temp);
}
console.log(returnval);
function findnext(item, temp)
{
for(var i=0;i<array.length;i++)
{
if((array[i][0]==item[0] && Math.abs(array[i][1]-item[1])==1) || (array[i][1]==item[1] && Math.abs(array[i][0]-item[0])==1))
{
temp.push(array[i]);
item1 = array[i];
array.splice(i,1);
findnext(item1, temp);
}
}
}

How do I check if elements exist in an array using .some()?

I am trying to see if a particular pair of numbers exist in an array. I know that .some() works effectively when it's a simple array, but I can't seem to get it to work in nested arrays.
Thank you in advance.
const array = [[1, 0], [2, 0], [3, 1], [4, 3], [5, 2]];
// checks whether an element exists
const exists = (i) => i == [3, 1] || [1, 3];
console.log(array.some(exists));
// expected output: true

You need to check every value because of different arrays, you have different object references, and you misused the logical OR operator. This does not include a comparison with another value.
const array = [[1, 0], [2, 0], [3, 1], [4, 3], [5, 2]];
const exists = ([a, b]) => a === 3 && b === 1 || a === 1 && b === 3;
console.log(array.some(exists));

Here is a general solution.
const array = [[1, 0], [2, 0], [3, 1], [4, 3], [5, 2]];
const expected = [[1, 3], [3, 1]];
const exists = (i) => expected.some((j) => {
return i.every((k, n) => j[n] === k);
});
console.log(array.some(exists));
This could be wrapped in a function to make it reusable.
const array = [[1, 0], [2, 0], [3, 1], [4, 3], [5, 2]];
const expected = [[1, 3], [3, 1]];
const containsAny = (array, expected) => {
return array.some((i) => expected.some((j) => {
return i.every((k, n) => j[n] === k);
}))
};
console.log(containsAny(array, expected));

You could make a custom function out of it which makes it reusable.
This is a simple example. By flattening the array you can check if values are the same.
Got my inspiration on https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/some
const array = [[1, 0], [2, 0], [3, 1], [4, 3], [5, 2]];
const checkArrayFor = (needle, haystack) => {
const flattenNeedle = needle.join(':');
const flattenArray = haystack.map(item => item.join(':'));
return flattenArray.includes(flattenNeedle);
};
console.log("Check for [1, 0]", checkArrayFor([1, 0], array));
console.log("Check for [1, 1]", checkArrayFor([1, 1], array));

Comparing arrays in arrays

I'm looking for a way to make comparisons between arrays in arrays.
let small = [[1, 3], [2, 2], [2, 3], [3, 0]];
let large = [[1, 0], [1, 1], [2, 0], [2, 2], [2, 5], [3, 0], [3, 2]];
For example, I would like to be able to find out how many of the arrays in small are found in large. Some function that, given the arrays above as arguments, would return 2, since [2, 2] and [3, 0] from small are found in large.
How would you go about doing that?

You can convert one of the arrays into a Set of hashes, and than filter the 2nd array using the set:
const small = [[1, 3], [2, 2], [2, 3], [3, 0]];
const large = [[1, 0], [1, 1], [2, 0], [2, 2], [2, 5], [3, 0], [3, 2]];
const containsCount = (arr1, arr2, hashFn) => {
const arr1Hash = new Set(arr1.map(hashFn));
return arr2.filter(s => arr1Hash.has(hashFn(s))).length;
}
const result = containsCount(small, large, ([a, b]) => `${a}-${b}`);
console.log(result);

You can try something like:
let small = [[1, 3], [2, 2], [2, 3], [3, 0]];
let large = [[1, 0], [1, 1], [2, 0], [2, 2], [2, 5], [3, 0], [3, 2]];
let z = zeta(small, large);
console.log(z);
function zeta(a, b) {
let join = m => m.join();
let x = a.map(join);
let y = b.map(join);
return x.reduce((n, m) => (y.indexOf(m)>0) ? ++n : n, 0);
}
I hope this helps!

Use every and some to compare the arrays with each other.
If you want to get an array containing the subarrays that match, use filter:
let result = small.filter(arr =>
large.some(otherArr =>
otherArr.length === arr.length && otherArr.every((item, i) => item === arr[i])
)
);
Which filters the subarray from small that some subarray from large has the same length and the same elements/items.
Demo:
let small = [[1, 3], [2, 2], [2, 3], [3, 0]];
let large = [[1, 0], [1, 1], [2, 0], [2, 2], [2, 5], [3, 0], [3, 2]];
let result = small.filter(arr =>
large.some(otherArr =>
otherArr.length === arr.length && otherArr.every((item, i) => item === arr[i])
)
);
console.log(result);
And if you want just a count, then use reduce instead of filter to count the mathched items (this makes use of the fact that the numeric value of true is 1 and that of false is 0):
let count = small.reduce((counter, arr) =>
counter + large.some(otherArr =>
otherArr.length === arr.length && otherArr.every((item, i) => item === arr[i])
)
, 0);
Demo:
let small = [[1, 3], [2, 2], [2, 3], [3, 0]];
let large = [[1, 0], [1, 1], [2, 0], [2, 2], [2, 5], [3, 0], [3, 2]];
let count = small.reduce((counter, arr) =>
counter + large.some(otherArr =>
otherArr.length === arr.length && otherArr.every((item, i) => item === arr[i])
)
, 0);
console.log(count);
Note: If the subarrays contain only numbers, the code could be simplified to use Array#toString instead of every and length comparaison:
let result = small.filter(arr => large.some(otherArr => "" + otherArr === "" + arr));
Which casts both arrays into strings and compares the two strings instead. This can be used with the reduce as well.

Create two nesting map() function oute for small and inner for large then use JSON.stringify()
let small = [[1, 3], [2, 2], [2, 3], [3, 0]];
let large = [[1, 0], [1, 1], [2, 0], [2, 2], [2, 5], [3, 0], [3, 2]];
var same=[];
small.map(function(element){
large.map(function(element2){
if(JSON.stringify(element)==JSON.stringify(element2)){
same.push(element);
}
});
});
console.log(same);

let small = [[1, 3], [2, 2], [2, 3], [3, 0]];
let large = [[1, 0], [1, 1], [2, 0], [2, 2], [2, 5], [3, 0], [3, 2]];
let largeArrayStringForm = large.map(item => item.toString())
let matchingItems = small.filter(item => largeArrayStringForm.includes(item.toString()))
console.log(`matchCount: ${matchingItems.length}`)