Related
Input:
[1,2,[3,4,[5,6]]]
Output:
[[1,2],[3,4],[5,6]]
Here is the solution:
function convert(a,res=[]) {
const group = (arr) => {
res.push(arr.slice(0,2));
arr.map((v) => Array.isArray(v) && group(v));
}
group(a);
return res;
}
console.log(convert([1,2,[3,4]])); // [[1,2],[3,4]]
console.log(convert([1,2,[3,4,[5,6]]])); // [[1,2],[3,4],[5,6]]
console.log(convert([1,2,[3,4,[5,6,[7,8]]]])); // [[1,2],[3,4],[5,6],[7,8]];
Although the problem is solved, For learning purposes, I've always wonder how to solve this without the nesting function approach. I've tried to refactor the code as followed:
function convert(a,i=0,res=[]) {
return i >= a.length
? res
: convert(
a,
i+1,
Array.isArray(a[i]) ? [...res,a.slice(0,2)] : res
)
}
console.log(convert([1,2,[3,4]])); // [[1,2]]
console.log(convert([1,2,[3,4,[5,6]]])); // [[1,2]]
console.log(convert([1,2,[3,4,[5,6,[7,8]]]])); // [[1,2]]
As you can see, the result is less than ideal. I just couldn't quite wrap my head around it. Any feedback and pointer will be greatly appreciated :)
UPDATE:
Here is the solution which covers even more test cases:
function convert(a,res=[]) {
return !a.length
? res
: convert(
a.filter(Array.isArray).flat(),
[...res,a.filter((v) => !Array.isArray(v))]
);
}
console.log(convert([1,2,[3,4]])); // [[1,2],[3,4]]
console.log(convert([1,2,[3,4,[5,6]]])); // [[1,2],[3,4],[5,6]]
console.log(convert([1,2,[3,4,[5,6,[7,8]]]])); // [[1,2],[3,4],[5,6],[7,8]];
console.log(convert([1,2,[5,6,[9,10],7,8],3,4])); // [[1,2,3,4],[5,6,7,8],[9,10]]
console.log(convert([1,5,5,[5,[1,2,1,1],5,5],5,[5]])); // [[1,5,5,5],[5,5,5,5],[1,2,1,1]]
console.log(convert([1,[2],1,[[2]],1,[[[2]]],1,[[[[2]]]]])); // [[1,1,1,1],[2],[2],[2],[2]]
I'm not sure if this is what you mean, but you can use a straight recursive call, spreading each result into the previous.
const input = [1, 2, [3, 4, [5, 6]]];
function flatten(arr) {
const res = [[]];
for (const e of arr) {
Array.isArray(e) ? res.push(...flatten(e)) : res[0].push(e);
}
return res;
}
console.log(flatten(input));
This works even if the elements on any level are split by nested levels, and for variable numbers of elements on any given level.
const input = [1, 2, [4, [5, 6, [8, 9], 7]], 3];
// [[ 1, 2, 3 ], [ 4 ], [ 5, 6, 7 ], [ 8, 9 ]]
function flatten(arr) {
const res = [[]];
for (const e of arr) {
Array.isArray(e) ? res.push(...flatten(e)) : res[0].push(e);
}
return res;
}
console.log(flatten(input));
Edit
To accommodate the added conditions brought up in the comments, but maintain straight recursion without passing an accumulator to later calls, I might do something like the following.
function flatten(arr) {
const level = [], nested = [];
for (const e of arr) {
Array.isArray(e) ? nested.push(...e) : level.push(e);
}
return [level, ...(nested.length ? flatten(nested) : [])]
}
console.log(flatten([1, 2, [3, 4, [5, 6]]]));
// [[ 1, 2 ], [ 3, 4 ], [ 5, 6 ]]
console.log(flatten([1, 2, [4, [5, 6, [8, 9], 7]], 3]));
// [[ 1, 2, 3 ], [ 4 ], [ 5, 6, 7 ], [ 8, 9 ]]
console.log(flatten([1, [2], 1, [[2]], 1, [[[2]]], 1, [[[[2]]]]]));
// [[ 1, 1, 1, 1 ], [ 2 ], [ 2 ], [ 2 ], [ 2 ]]
console.log(flatten([1, 5, 5, [5, [1, 2, 1, 1], 5, 5], 5, [5]]));
// [[ 1, 5, 5, 5 ], [ 5, 5, 5, 5 ], [ 1, 2, 1, 1 ]]
I think your updated algorithm is fine.
It might be easier to express with a reusable utility function, though. I usually have a partition function lying around which accepts a predicate function and returns a function which splits an array into two sub-arrays, those for which the predicate returns true, and those for which it's false.
Using that, this becomes fairly simple:
const partition = (pred) => (xs) =>
xs .reduce (([t, f], x) => pred (x) ? [t .concat (x), f]: [t, f .concat (x)], [[], []])
const convert = (xs, [rest, first] = partition (Array .isArray) (xs)) =>
xs .length == 0 ? [] : [first, ...convert (rest)]
console.log (convert ([1, 2, [3, 4]])) //=> [[1, 2], [3, 4]]
console.log (convert ([1, 2, [3, 4, [5, 6]]])) //=> [[1, 2], [3, 4], [5, 6]]
console.log (convert ([1, 2, [3, 4, [5, 6, [7, 8]]]])) //=> [[1, 2], [3, 4], [5, 6], [7, 8]]
console.log (convert ([1, 2, [5, 6, [9, 10], 7, 8], 3, 4])) //=> [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10]]
console.log (convert ([1, 5, 5, [5, [1, 2, 1, 1], 5, 5], 5, [5]])) //=> [[1, 5, 5, 5], [5, 5, 5, 5], [1, 2, 1, 1]]
console.log (convert ([1, [2], 1, [[2]], 1, [[[2]]], 1, [[[[2]]]]])) //=> [[1, 1, 1, 1], [2], [2], [2], [2]]
.as-console-wrapper {max-height: 100% !important; top: 0}
We have an array of arrays like this:
const arrays = [
[0, 1, 2, 3, 4, 4, 4, 4],
[5, 6, 7, 8, 9, 10, 11, 11],
[2, 7, 10, 12],
[0, 7, 10, 14]
];
There may be duplicate elements in each array and that's fine.
But I'm after a proper solution to remove duplicate elements in each set comparing to lower sets!
So as we have a 0 in the first array and the last array, we should consider the 0 in last one a duplication and remove it...
the desired result would be:
[0, 1, 2, 3, 4, 4, 4, 4],
[5, 6, 7, 8, 9, 10, 11, 11],
[12],
[14]
It's a confusing issue for me please help...
You could collect the values in an object with index as value, and filter for values who are at the same index.
const
arrays = [[0, 1, 2, 3, 4, 4, 4, 4], [5, 6, 7, 8, 9, 10, 11, 11], [2, 7, 10, 12], [0, 7, 10, 14]],
seen = {},
result = arrays.map((array, i) => array.filter(v => (seen[v] ??= i) === i));
result.forEach(a => console.log(...a));
const arrays = [
[0, 1, 2, 3, 4, 4, 4, 4],
[4, 4, 5, 6, 7, 8, 9, 10, 11, 11],
[2, 7, 10, 12],
[0, 7, 10, 14]
]
let filtered = arrays.map((row, i) => {
// concat all previous arrays
let prev = [].concat(...arrays.slice(0, i))
// filter out duplicates from prev arrays
return row.filter(r => !prev.includes(r))
})
console.log(filtered)
We can do this using Array#reduce and maintain a seen Set, which will have all visited numbers from each array.
Once you iterate over an array you push all visited elements in the seen Set, then push a new array filtered by the elements not in the seen Set:
const arrays = [
[0, 1, 2, 3, 4, 4, 4, 4],
[5, 6, 7, 8, 9, 10, 11, 11],
[2, 7, 10, 12],
[0, 7, 10, 14]
];
const removeDupsInSibling = (arr) => {
let seen = new Set();
return arr.reduce((acc, a)=> {
const f = a.filter(v => !seen.has(v));
seen = new Set([...seen, ...a]);
acc.push(f);
return acc;
}, []);
}
console.log(removeDupsInSibling(arrays));
There are plenty of inefficient ways to do this, but if you want to do this in O(n), then we can make the observation that what we want to know is "which array a number is in". If we know that, we can run our algorithm in O(n):
for every element e in array at index i:
if index(e) == i:
this is fine
if index(e) < i:
remove this e
So let's just do literally that: we allocate an object to act as our lookup, and then we run through all elements:
const lookup = {};
const arrays = [
[0, 1, 2, 3, 4, 4, 4, 4],
[5, 6, 7, 8, 9, 10, 11, 11],
[2, 7, 10, 12],
[0, 7, 10, 14]
];
const reduced = arrays.map((array, index) => {
// run through the elements in reverse, so that we can
// safely remove bad elements without affecting the loop:
for(let i=array.length-1; i>=0; i--) {
let value = array[i];
let knownIndex = (lookup[value] ??= index);
if (knownIndex < index) {
// removing from "somewhere" in the array
// uses the splice function:
array.splice(i,1);
}
}
return array;
});
console.log(reduced);
For an alternative, where the loop+splice is taken care of using filter, see Nina's answer.
Simple, clean and high performance solution:
const arrays = [
[0, 1, 2, 3, 4, 4, 4, 4],
[5, 6, 7, 8, 9, 10, 11, 11],
[2, 7, 10, 12],
[0, 7, 10, 14]
];
const duplicates = {};
const answer = arrays.map( (array, level) => {
return array.filter( el => {
if ( duplicates[el] < level ) {
// return nothing; fine
} else {
duplicates[el] = level;
return el
}
})
});
console.log(JSON.stringify(answer))
here is on-liner and less-readable form:
const d = {}, arrays = [ [0, 1, 2, 3, 4, 4, 4, 4], [5, 6, 7, 8, 9, 10, 11, 11], [2, 7, 10, 12], [0, 7, 10, 14]];
const answer = arrays.map((a,l)=> a.filter(el=> d[el]<l ? 0 : (d[el]=l,el)));
console.log(JSON.stringify(answer))
const arrays = [
[0, 1, 2, 3, 4, 4, 4, 4],
[5, 6, 7, 8, 9, 10, 11, 11],
[2, 7, 10, 12],
[0, 7, 10, 14],
];
const output = arrays.reduce(
({ output, set }, current, i) => {
output[i] = current.filter((num) => !set.has(num));
[...new Set(output[i])].forEach((num) => set.add(num));
return { output, set };
},
{ output: [], set: new Set() }
).output;
console.log(output);
Gets the exact output you want:
[
[
0, 1, 2, 3,
4, 4, 4, 4
],
[
5, 6, 7, 8,
9, 10, 11, 11
],
[ 12 ],
[ 14 ]
]
Let say there are two array.
let MotherArray = [
[30, 1, 2, 3, 4, 5, 6],
[5, 6, 7, 8, 9],
[7, 8, 9],
];
let arraytoTest = [5,6];
What i want is that i want to return the array if
all the value inside the arraytoTest is included in the MotherArray[i]
I have tried
let MotherArray = [[30, 1, 2, 3, 4, 5, 6],[5, 6, 7, 8, 9],[7, 8, 9],];
let arraytoTest = [5, 6];
let result = MotherArray.includes(arraytoTest)
console.log(result);
But i don't think this is the correct method.
I also find the array.every() but i think my usage is not correct.
What I want is that I want to return MotherArray[0],MotherArray[1] which are [[30, 1, 2, 3, 4, 5, 6],[5, 6, 7, 8, 9]] in this particular example
since 5 and 6 are includes inside these 2 arrays.
You can combine array.filter() with array.every()
let MotherArray = [
[30, 1, 2, 3, 4, 5, 6],
[5, 6, 7, 8, 9],
[7, 8, 9],
];
let arraytoTest = [5,6];
let found = MotherArray.filter(childArray => arraytoTest.every(num => childArray.includes(num)));
console.log(found);
I think this is what you want; A combination of filter on the mother array and every for array you're testing.
let MotherArray = [
[30, 1, 2, 3, 4, 5, 6],
[5, 6, 7, 8, 9],
[7, 8, 9],
];
let arraytoTest = [5,6];
let result = MotherArray.filter(arr => arraytoTest.every(x => arr.indexOf(x)>-1));
console.log(result);
You can use filter and every like below.
Or you can use filter and some with negative condition like filter(x => !arraytoTest.some(y => !x.includes(y))). I think with some it would be efficient because
The some() method executes the callback function once for each element present in the array until it finds the one where callback returns a truthy value.
let MotherArray = [
[30, 1, 2, 3, 4, 5, 6],
[5, 6, 7, 8, 9],
[7, 8, 9],
];
let arraytoTest = [5,6];
let result = MotherArray.filter(x => arraytoTest.every(y => x.includes(y)));
console.log(result);
let result2 = MotherArray.filter(x => !arraytoTest.some(y => !x.includes(y)));
console.log(result2);
I have nested array data and I would like to extract all nested arrays to be siblings of their parent. I am pretty close, but I am getting an extra empty array in the results and I cannot figure out where it is coming from or how to get rid of it.
Note: I would really like to understand why this is happening and how to get rid of it in my function, and not just a .filter(arr => arr.length) on my results list.
This is my attempt so far:
var arrs = [
[1, 2, [3, 4], 5],
[6, [7, 8, 9, [10, 11]]],
[12, 13],
[[14, 15], [16, 17]],
[[1], 4, [1, 1], 4]
];
// Desired Output
// [
// [1, 2, 5],
// [3, 4],
// [6],
// [7, 8, 9],
// [10, 11],
// [12, 13],
// [14, 15],
// [16, 17],
// [4, 4]
// [1]
// [1, 1]
// ]
function extractArrays (arr) {
return arr.reduce((res, curr) => {
if (Array.isArray(curr)) {
res = res.concat(extractArrays(curr));
}
else {
res[0].push(curr);
}
return res;
}, [[]]);
}
console.log(extractArrays(arrs));
// Results:
// [
// [], <-- Where is this coming from?
// [ 1, 2, 5 ],
// [ 3, 4 ],
// [ 6 ],
// [ 7, 8, 9 ],
// [ 10, 11 ],
// [ 12, 13 ],
// [], <-- Also here
// [ 14, 15 ],
// [ 16, 17 ],
// [ 4, 4 ],
// [ 1 ],
// [ 1, 1 ]
// ]
.as-console-wrapper {
max-height: 100% !important;
}
Element like [[14, 15], [16, 17]] will introduce a [] after recursion. This should be handled by checking length.
var arrs = [
[1, 2, [3, 4], 5],
[6, [7, 8, 9, [10, 11]]],
[12, 13],
[[14, 15], [16, 17]],
[[1], 4, [1, 1], 4]
];
function extractArrays (arr, acc=[]) {
if (arr.length == 0 ) return acc;
let pure = arr.filter(elm => !Array.isArray(elm));
if (pure.length > 0) {
acc.push(pure);
}
acc.concat(arr.filter(elm => Array.isArray(elm)).map(elm => extractArrays(elm, acc)));
return acc;
}
console.log(extractArrays(arrs));
You can try the following code
var arrs = [
[1, 2, [3, 4], 5],
[6, [7, 8, 9, [10, 11]]],
[12, 13],
[
[14, 15],
[16, 17]
], // <-- added additional test case
[
[1], 4, [1, 1], 4
]
];
function extractArrays(arr) {
return arr.reduce((res, curr, i) => {
if (Array.isArray(curr)) {
res = res.concat(extractArrays(curr));
} else {
let index = 0;
for (let j = 0; j <= i; j++) {
if (!Array.isArray(arr[j])) {
res[index] ? res[index].push(curr) : res.push([curr]);
break;
} else {
index++;
}
}
}
return res;
}, []); // <-- no initial empty array inside here
}
console.log(extractArrays(arrs));
I just wanted to share my approach to this problem, I enjoyed trying to solve it, in my case I also passed an array to the extractArrays method, in order to make easier to capture and filter every array inside the arrs param.
let result = [];
extractArrays(arrs, result);
console.log(result);
function extractArrays(arr, result) {
let newResult = arr.reduce((acc, curr) => {
if (Array.isArray(curr)) {
extractArrays(curr, result);
} else {
acc.push(curr);
}
return acc;
}, []);
newResult.length && result.push(newResult);
}
You can check it when you return from function. stackblitz
function extractArray(arr) {
const res = arr.reduce((res, curr) => {
if(!Array.isArray(curr)){
return [[...res[0], curr], ...res.slice(1)]
}
return [...res, ...extractArray(curr)]
}, [[]]);
return res[0].length ? res : res.slice(1);
}
EDIT: More performant function (check stackblitz link)
function extractFaster(arr) {
let res = [0];
function recExtract(arr) {
let hasNonArrayElm = false;
let index = res.length -1;
arr.forEach(curr => {
if (!Array.isArray(curr)) {
hasNonArrayElm ? res[index].push(curr) : res.splice(index, 0, [curr]);
hasNonArrayElm = true;
return;
}
recExtract(curr);
});
}
recExtract(arr);
res.splice(-1, 1)
return res;
}
EDIT: The answer below the line is a great way to flatten arrays, but I suggested it because I misunderstood this question. I will leave it in case it benefits someone to know, but in order to keep an accurate record, I'll also update my answer to address the problem posed in the question.
The accepted answer seems sufficient enough, but I'll try my hand at it. I would use Array.reduce to cover everything with one swoop, and inside use Array.filter to separate the normal items from the array items, then use the spread ... operator on the nested arrays so everything gets pushed to the same level, after recursively calling the same extract function on all nested arrays. Honestly, the explanation might be harder to understand than the code, have a look:
const data = [
[1, 2, [3, 4], 5],
[6, [7, 8, 9, [10, 11]]],
[12, 13],
[[14, 15], [16, 17]],
[[1], 4, [1, 1], 4]
]
const extractChildArrays = arrs => arrs.reduce((acc, cur) => {
const nestedArrs = cur.filter(a => Array.isArray(a))
const normalItems = cur.filter(a => !Array.isArray(a))
acc.push(normalItems, ...extractChildArrays(nestedArrs))
return acc
}, [])
console.log(extractChildArrays(data))
UPDATE: Array.flat() is now accepted as part of the spec and it's supported by all modern browsers except Edge.
In ES6 there is actually an experimental array method called flat(). As of the writing of this answer, it's only compatible with Chrome, but a polyfill might be worth looking into, because it's SO EASY!
The first parameter of flat() is depth, so with the help of another answer, you can easily figure that out dynamically.
const data = arrs = [
[1, 2, [3, 4], 5],
[6, [7, 8, 9, [10, 11]]],
[12, 13],
[[14, 15], [16, 17]],
[[1], 4, [1, 1], 4]
]
const flattenArray = arr => {
const getDepth = a => Array.isArray(a) ?
1 + Math.max(...a.map(getDepth)) : 0
return arr.flat(getDepth(arr))
}
console.log(flattenArray(data))
I am trying to solve this freecodecamp algorithm question where I had to collect the difference of two or more arrays. I used map to get the difference of array but the problem is I only get two elements;
function sym(args) {
args = [].slice.call(arguments);
var newArr = args.map(function(el, index, arr){
console.log(arr.indexOf(arr[index]));
if(arr.indexOf(arr[index] === -1 )){
// console.log(arr[index]);
return args.push(arr[index]);
}
});
return newArr; // my newArr returns [3, 4] instead of [3,4,5]
}
console.log(sym([1, 2, 3], [5, 2, 1, 4]));
//sym([1, 2, 3], [5, 2, 1, 4]) should return [3, 4, 5]
//sym([3, 3, 3, 2, 5], [2, 1, 5, 7], [3, 4, 6, 6], [1, 2, 3], [5, 3, 9, 8], [1]) should return [1, 2, 4, 5, 6, 7, 8, 9]
I think we could do also this way since we want them to be ordered at the end.
For more detail about the original problem please consult this link: FreecodeCamp Link: Symmetric Difference
const sym = (...args) => {
// Merge all the different arrays and remove duplicate elements it means elements that are present both on two related arrays
let tab = args.reduce((a, b) => [
...a.filter(i => !b.includes(i)),
...b.filter(j => !a.includes(j))
], []);
// Then remove the rest of duplicated values and sort the obtained array
return Array.from(new Set(tab)).sort((a, b) => a - b);
}
console.log(sym([1, 2, 3, 3], [5, 2, 1, 4])); // [3, 4, 5]
console.log(sym([1, 1, 2, 5], [2, 2, 3, 5], [3, 4, 5, 5])); // [1, 4, 5]
console.log(sym([3, 3, 3, 2, 5], [2, 1, 5, 7], [3, 4, 6, 6], [1, 2, 3], [5, 3, 9, 8], [1])); // [1, 2, 4, 5, 6, 7, 8, 9]
The Set data structure is used here to remove duplicated values thanks to its characteristics.
Well your function is a little more complex than only selecting the unique values, cause you want to filter them out... and also accept multiple arrays. This should work.
var sym = (...arrays)=>{
//Concat Items
const allItems = arrays.reduce((a,c)=>a.concat(c), []);
// Identify repeated items
const repeatedItems = allItems.filter((v,i,a)=>a.indexOf(v) !== i);
// Filter repeated items out
const diff = allItems.filter(item=>repeatedItems.indexOf(item) < 0);
console.log(diff);
};
sym([1, 2, 3], [5, 2, 1, 4]); // [3,5,4]
I don't think your approach will work; you're supposed to create an array with elementos from both arrays, so a single .map won't do the job. Filtering through both arrays should work, although it will probably leave enough room for optimization.
my newArr returns [3, 4] instead of [3,4,5]
You are using map which will only return one value per iteration (which is why you are getting only 2 values) and in your case you are checking if the index is found or not (not the item)
You need to concatenate all the arrays and then remove those which are repeated
Concatenate
var newArr = args.reduce( ( a, c ) => a.concat( c ) , []);
Create a map by number of occurrences
var map = newArr.reduce( (a,c) => ( a[c] = (a[c] || 0) + 1, a ) , {});
Iterate and filter through those keys whose value is 1
var output = Object.keys( map ).filter( s => map[s] === 1 ).map( Number );
Demo
function sym(args)
{
args = [].slice.call(arguments);
var newArr = args.reduce( ( a, c ) => a.concat( c ) , []);
var map = newArr.reduce( (a,c) => ( a[c] = (a[c] || 0) + 1, a ) , {});
return Object.keys( map ).filter( s => map[s] === 1 ).map( Number );
}
console.log(sym([1, 2, 3], [5, 2, 1, 4]));
You could take an Object for counting the items and return only the items which have a count.
function sym(array) {
return array.reduce((a, b) => {
var count = {};
a.forEach(v => count[v] = (count[v] || 0) + 1);
b.forEach(v => count[v] = (count[v] || 0) - 1);
return Object.keys(count).map(Number).filter(k => count[k]);
});
}
console.log(sym([[3, 3, 3, 2, 5], [2, 1, 5, 7], [3, 4, 6, 6], [1, 2, 3], [5, 3, 9, 8], [1]]));