Related
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)
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
I try to loop the 2d arrays, but the I variable is undefined or not iterable, why?
can anyone tell me ??
function sum (arr) {
var total = 0
for(let [a1,a2,a3] of arr){
for(let i of [a1,a2,a3]){
for(let j of i){
total += j
}
}
if(typeof a2 == "undefined" && typeof a3 == "undefined"){
a2 = [0]
a3 = [0]
}
}
};
console.log(sum([
[
[10, 10],
[15],
[1, 1]
],
[
[2, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11],
[4],
[9, 11]
],
[
[3, 5, 1],
[1, 5, 3],
[1]
],
[
[90]
]
]));
but when i sum another 2D array, it works, like this :
function sum (arr) {
var total = 0
for(let [a1,a2,a3] of arr){
for(let i of [a1,a2,a3]){
for(let j of i){
total += j
}
}
}
return total
}
console.log(sum([
[
[4, 5, 6],
[9, 1, 2, 10],
[9, 4, 3]
],
[
[4, 14, 31],
[9, 10, 18, 12, 20],
[1, 4, 90]
],
[
[2, 5, 10],
[3, 4, 5],
[2, 4, 5, 10]
]
]));
i try to loop 3 times for this 2d arrays, the first top code is each lengths are diffreen in array
and the last code is same,
Cause
let [a1,a2,a3] of [ [90] ])
will result in a2 and a3 being undefined, therefore in the following line it is:
for(const i of [90, undefined, undefined])
And at the second index it does:
for(let j of undefined)
which doesnt work.
You just need to move your if statement that checks if the value is undefined and assigns it to zero if it is ahead of the part of code that iterates over those values. You were getting this error because there wasn't anything there.
function sumTwo(arr) {
var total = 0
for(let [a1,a2,a3] of arr){
if(typeof a2 == "undefined" && typeof a3 == "undefined"){
a2 = [0]
a3 = [0]
}
for(let i of [a1,a2,a3]){
for(let j of i){
total += j
}
}
}
return total
};
console.log(sumTwo([
[
[10, 10],
[15],
[1, 1]
],
[
[2, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11],
[4],
[9, 11]
],
[
[3, 5, 1],
[1, 5, 3],
[1]
],
[
[90]
]
])); //prints 237
When you say
let [a1,a2,a3] of [ [90] ])
there is no a2 or a3 there...
My suggestion would be using the code before you get into the first for loop:
if(arr.length < 3){
for(let y = arr.length, y > 3, y++ ){
arr.push(0)
}
}
Cheers!
It's probably better to recursively reduce the array using concat until you have a flat array and then reduce that to the sum of it's numbers:
const arr = [
[[10, 10], [15], [1, 1]],
[[2, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [4], [9, 11]],
[[3, 5, 1], [1, 5, 3], [1]],
[[90]],
];
const flatten = (arr) => {
const recur = (result, item) =>
!Array.isArray(item)
? result.concat(item)
: result.concat(item.reduce(recur, []));
return arr.reduce(recur, []);
};
console.log(
flatten(arr).reduce((result, item) => result + item, 0),
);
I am trying to create a function that will get the items that cannot be seen on the 2nd or 3rd and upcoming arrays passed within the function.
Right now my function gets only the similar items. How can I make it get the difference (w/c are the items that doesn't exist to the 2nd and 3rd and proceeding arrays.
const callM = function(arrays) {
arrays = Array.prototype.slice.call(arguments);
let result = [];
for(let i = 1; i < arrays.length; i++){
for(let x = 0; x < arrays[i].length; x++){
if(arrays[0].includes(arrays[i][x])){
result.push(arrays[i][x]);
}
}
}
return result;
};
console.log(callM([1, 2, 3, 4, 5], [5, 2, 10])); // -> must be [1, 3, 4]
console.log(callM([1, 2, 3, 4, 5], [5, 2, 10], [7, 1, 8])); // -> must be [3,4]
The logic right now is a bit off as it gets the opposite. How do i fix this?
Also is there a way to do this using Higher Order functions such as reduce or filter?
Thanks!
I'd think about this differently. As the difference between two sets: array 0 and array 1...n
To get array 0, just shift it off the top
const arr0 = arrays.shift()
Ref: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/shift
This removes the first array from arrays
Next we combine the remaining arrays
const arrN = arrays.reduce(function(prev, curr) {
return prev.concat(curr)
})
Ref: http://www.jstips.co/en/javascript/flattening-multidimensional-arrays-in-javascript/
Unneeded, handled by includes as mentioned by #Phil
Next filter duplicates from arrN by comparing with itself
const unique = arrN.filter(function(elem, index, self) {
return index == self.indexOf(elem);
})
Ref: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/filter
Next filter with includes to find the difference (or union)
const diff = arr0.filter(function(item) {
return !arrN.includes(item))
}
Full snippet:
function callM(arrays) {
const arr0 = arrays.shift()
const arrN = arrays.reduce(function(prev, curr) {
return prev.concat(curr)
})
//const unique = arrN.filter(function(elem, index, self) {
// return index == self.indexOf(elem)
//})
return arr0.filter(function(item) {
return !arrN.includes(item)
})
}
console.log(callM([[1, 2, 3, 4, 5], [5, 2, 10]]))
console.log(callM([[1, 2, 3, 4, 5], [5, 2, 10], [7, 1, 8]]))
of course ES6 would be easier. ;)
const callM = (first, ...rest) => {
const arrays = [].concat(...rest)
return first.filter(item => !arrays.includes(item))
}
console.log(callM([1, 2, 3, 4, 5], [5, 2, 10]))
console.log(callM([1, 2, 3, 4, 5], [5, 2, 10], [7, 1, 8]))
A short solution for small and medium sized arrays:
// Return elements in array but not in filters:
function difference(array, ...filters) {
return array.filter(el => !filters.some(filter => filter.includes(el)));
}
// Example:
console.log(difference([1, 2, 3, 4, 5], [5, 2, 10])); // [1, 3, 4]
console.log(difference([1, 2, 3, 4, 5], [5, 1, 10], [7, 2, 8])); // [3, 4]
For large inputs, consider creating a Set from all filters and filtering in linear time using set.has(el).
In order to fix your implementation, you could label the outer for-loop and continue from there whenever a filter contains one of the array elements. Only when all filters pass without match, you push the array element into the result:
// Return elements in array but not in filters:
function difference(array, ...filters) {
const result = [];
loop: for (const el of array) {
for (const filter of filters) {
if (filter.includes(el)) continue loop;
}
result.push(el);
}
return result;
}
// Example:
console.log(difference([1, 2, 3, 4, 5], [5, 2, 10])); // [1, 3, 4]
console.log(difference([1, 2, 3, 4, 5], [5, 2, 10], [7, 1, 8])); // [3,4]
If you're willing to use Underscore, you can do this in one line of code:
console.log(_.difference([1, 2, 3, 4, 5], [5, 2, 10], [7, 1, 8]))
https://jsfiddle.net/o1zuaa6m/
You can use array#reduce to create object lookup of all the other array excluding the first array. Then use array#filter to get the values which are not present in the object lookup
var callM = (first, ...rest) => {
var combined = rest
.reduce((res,arr) => res.concat(arr))
.reduce((o, v) => {
o[v] = true;
return o;
},{});
return first
.filter(v => !combined[v]);
}
console.log(callM([1, 2, 3, 4, 5], [5, 2, 10])); // -> must be [1, 3, 4]
console.log(callM([1, 2, 3, 4, 5], [5, 2, 10], [7, 1, 8])); // -> must be [3,4]
The "proper" way to exclude values is usually to use a lookup hash set with the values to exclude:
const callM = (a, ...b) => (b = new Set(b.concat.apply(...b)), a.filter(v => !b.has(v)))
console.log(callM([1, 2, 3, 4, 5], [5, 2, 10])); // [1, 3, 4]
console.log(callM([1, 2, 3, 4, 5], [5, 2, 10], [7, 1, 8])); // [3, 4]
I have this object which its keys are guaranteed sorted and will be used for the operation. And each of its value is a 2d array.
var obj = {
"0": [
[0, 1], [0, 3], [0, 4]
],
"1": [
[1, 2], [1, 3]
],
"2": [
[2, 3], [2, 5]
],
"3": [
[3, 4], [3, 6]
],
"5": [
[5, 6]
],
"6": [
[6, 5]
]
}
I am trying to concatenate them and for each of its last value of the array is the next index of the object. So, my expected result is an array like this,
The pattern is, I have to find a way from 0 which is the first index of obj, to the last index which is 6 by using the values in each of it and linking its last array value to the next object. If that makes sense.
[0, 1, 2, 3, 4, 5, 6]
[0, 1, 2, 3, 6]
[0, 1, 2, 5, 6]
[0, 1, 3, 4, 5, 6]
[0, 1, 3, 4]
[0, 1, 3, 6]
[0, 3, 4, 5, 6]
[0, 3, 6]
[0, 4]
This is my code so far, as I don't know how to proceed further..
var result = [];
for (var key in obj) {
var myarr = obj[key];
for (var i = 0; i < myarr.length; i++) {
result.push(myarr[i])
}
}
Any idea or feedback is welcome.
Edit
One of the expected result was [0, 1, 2, 3, 4, 5, 6], here's the step by step explanation.
The obj key starts from 0 and ends in 6, I have to form a way from 0 to 6 with the arrays in its value.
Starts from obj[0], the first array returns [0, 1], save this to res. (res is now [0, 1])
The last value of array in res is 1, now find the next value in obj[1]
obj[1] has two arrays, and ends with 2 or 3.. So it's possible to append with both of them, so it can be [0, 1, 2] or [0, 1, 3]. In this case, get the first one which is [1, 2] and append the last value to res. (res is now [0, 1, 2]).
The last value of array in res is now 2, now find the next value in obj[2].
obj[2] has two arrays, and ends with 3, or 5.. It's possible to append with both of them, so it can be [0, 1, 2, 3] or [0, 1, 2, 5]. In this case, get the first one which is [2, 3] and append the last value to res. (res is now [0, 1, 2, 3])
The last value of array in res is now 3, now find the next value in obj[3].
Repeat step 4 or 6. (res is now [0, 1, 2, 3, 4]).
The last value of array in res is now 4, now find the next value in obj[4].
Repeat step 4 or 6. (res is now [0, 1, 2, 3, 4, 5]).
The last value of array in res is now 5, now find the next value in obj[5].
Now value 6 is found which should be the end of iteration if you look at the step 4. Repeat step 4 or 6. (res is now [0, 1, 2, 3, 4, 5, 6]).
Repeat from step 1, and form another way to do it, with no duplicates of [0, 1, 2, 3, 4, 5 ,6].
This is a proposal, with a single extra output, mentioned below.
[
[0, 1, 2, 3, 4, 5, 6],
[0, 1, 2, 3, 6],
[0, 1, 2, 5, 6],
[0, 1, 3, 4, 5, 6], /* extended from below */
[0, 1, 3, 4], /* original result */
[0, 1, 3, 6],
[0, 3, 4, 5, 6], /* extended from below */
[0, 3, 4], /* extra line, line should not be in result */
[0, 3, 6], /* but follows the same building rule than above */
[0, 4]
]
Basically this solution is building a tree with the given information about linked nodes.
If some nodes are not contiguous, a backtracking is made for the missing links, with the above function for nodes, checkNodes or with iterPath, to walk the actual collected nodes for missing items.
function getParts(value, path, nodes) {
function checkNodes(a) {
if (a[1] === value + 1) {
getParts(a[1], path.concat(a[1]), nodes);
return true;
}
}
function iterPath(k) {
return (object[k] || []).some(function (a) {
return path[path.length - 1] + 1 === a[1] || iterPath(a[1]);
});
}
value = value || 0;
path = path || [value];
nodes = nodes || [];
if (object[value]) {
object[value].forEach(function (a, i, aa) {
if (a[1] === lastKey) {
parts.push(path.concat(a[1]));
return;
}
getParts(a[1], path.concat(a[1]), nodes.concat(aa.slice(i + 1)));
});
return;
}
if (nodes.some(checkNodes)) {
return;
}
path.slice(1).some(iterPath) && getParts(path[path.length - 1] + 1, path.concat(path[path.length - 1] + 1), nodes);
parts.push(path);
}
var object = {
0: [[0, 1], [0, 3], [0, 4]],
1: [[1, 2], [1, 3]],
2: [[2, 3], [2, 5]],
3: [[3, 4], [3, 6]],
5: [[5, 6]],
6: [[6, 5]]
},
lastKey = 6,
parts = [];
getParts();
parts.forEach(function (a) { console.log(JSON.stringify(a)); });
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Well, I was sitting on this for some time now, and sharing across my take on the problem:
The input object can be considered as an adjacency list of a tree:
var obj={0:[[0,1],[0,3],[0,4]],1:[[1,2],[1,3]],2:[[2,3],[2,5]],3:[[3,4],[3,6]],5:[[5,6]],6:[[6,5]]};
and the following as the result required, which is in fact, as I see it, the list of all root-to-leaf paths of the tree:
[0,1,2,3,4]
[0,1,2,3,6]
[0,1,2,5,6]
[0,1,3,4]
[0,1,3,6]
[0,3,4]
[0,3,6]
[0,4]
a little different than the result set mentioned in the question which is the below:
[0,1,2,3,4,5,6]
[0,1,2,3,6]
[0,1,2,5,6]
[0,1,3,4,5,6]
[0,1,3,4]
[0,1,3,6]
[0,3,4,5,6]
[0,3,6]
[0,4]
The difference between the results is only the question whether 4 and 6 are leaf nodes
Solution:
So I assume that for our Tree here:
0 is the root node
4 and 6 are the leaf nodes
See code below - I created a tree first, and from that listed out all the root to leaf paths:
// removed "6:[[6,5]]" as 6 is a 'leaf' of the tree
var obj={0:[[0,1],[0,3],[0,4]],1:[[1,2],[1,3]],2:[[2,3],[2,5]],3:[[3,4],[3,6]],5:[[5,6]]};
var availableNodes = Object.keys(obj);
var tree = availableNodes.reduce(function(hash) {
return function(prev, curr) {
hash[curr] = hash[curr] || {};
hash[curr].children = hash[curr].children || [];
obj[curr].forEach(function(element) {
hash[element[1]] = hash[element[1]] || {};
hash[element[1]].children = hash[element[1]].children || [];
hash[curr].rootPath = hash[curr].rootPath || [];
hash[curr].children.push({value: element[1],children: hash[element[1]].children});
});
curr && prev.push({value: curr,children: hash[curr].children});
return prev;
};
}(Object.create(null)), []);
//console.log(JSON.stringify(tree));
var result = [];
function rootToLeafPaths(node, path) {
path.push(+node.value);
if (node.children.length === 0) {
result.push(Array.from(path));
path.pop();
} else {
node.children.forEach(function(element) {
rootToLeafPaths(element, path);
});
path.pop();
}
}
rootToLeafPaths(tree[0], []);
console.log(JSON.stringify(result));
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