I am building a function that count Rank Order Ballots and returns the winner. The rules of this are that if a candidate has a clear majority then the candidate wins the election.
If not, we remove all reference to that candidate and whichever ballots the candidate go are assigned to whoever came second
So for example if we have this
const sample = { "A,B,C": 4, "B,C,A": 3, "C,B,A": 2};
Since C has the least number of votes and noone has a majority, all votes C won are then assigned to B, giving B the majority.
This is what I have written:
function removeLowestVotedCandidate(ballots) {
let lowestVotes = Object.entries(ballots).reduce((largestVal, comparison) =>comparison[1] < largestVal[1] ? comparison
:largestVal)[0]
.split('')[0]
//remove lowest voted candidate from object
const newRankedOrderBallots = JSON.parse(
JSON.stringify(ballots)
.replaceAll(`${lowestVotes}`, '')
.replaceAll(/(^[,\s]+)/g, '')
)
//remove leading commas
return Object.fromEntries(Object.entries(newRankedOrderBallots).map(([key, value]) => [key.replace(/^,+/,''), value]))
}
// console.log(removeLowestVotedCandidate(sample))
function getRankedChoiceWinner(ballots) {
let sum = 0
let winnerFound = false
let stretchWin = 0;
let sumByChar = {};
let winner = []
let updatedBallot = ballots
while(winnerFound === false) {
//count overall votes
for(let votes of Object.values(updatedBallot)){
sum +=votes
}
//calculate what is required for a clear majority
stretchWin = Math.round(sum/2)
//count votes assigned to each candidate
for(const[key, val] of Object.entries(updatedBallot)) {
const char = key[0];
sumByChar[char] = (sumByChar[char] ?? 0) + val;
}
console.log('sumByChar is currently', sumByChar)
//check if any candidate has a clear majority
winner = Object.entries(sumByChar)
.filter(([, val]) => val >= stretchWin)
.map(([keys]) => keys)
console.log('winner is currently', winner)
if (winner.length === 1) {
winnerFound = true
} else {
updatedBallot = removeLowestVotedCandidate(updatedBallot)
console.log('we are inside else', updatedBallot)
}
}
return winner
}
However, I seem to be getting the wrong answer, I am getting A as opposed to B. This is what is happening with my console.logs
sumByChar is currently { A: 4, B: 3, C: 2 }
winner is currently []
we are inside else { 'A,B,': 4, 'B,,A': 3, 'B,A': 2 }
sumByChar is currently { A: 8, B: 6, C: 4 }
winner is currently []
we are inside else { 'A,,': 4, A: 2 }
sumByChar is currently { A: 12, B: 9, C: 6 }
winner is currently [ 'A' ]
[ 'A' ]
It seems sumByChar is not reseting to zero and instead
There are 2 issues:
Your sumByChar is created outside the loop and mutated inside the loop. Every time there's a new iteration, you add additional values to it. Create the object inside the loop instead, so you get the sum only for the current ballots, not the cumulative sum for all iterations so far.
Your sum variable is also declared outside the loop, and you're adding to it inside the loop. Declare it inside the loop instead.
Also, the input structure is pretty badly designed for something like this. I'd highly recommend restructuring it to make removing candidates easier - using JSON.stringify and a regex just to remove something is extremely suspicious.
const sample = {
"A,B,C": 4,
"B,C,A": 3,
"C,B,A": 2
};
function removeLowestVotedCandidate(ballots) {
let lowestVotes = Object.entries(ballots).reduce((largestVal, comparison) => comparison[1] < largestVal[1] ? comparison :
largestVal)[0]
.split('')[0]
//remove lowest voted candidate from object
const newRankedOrderBallots = JSON.parse(
JSON.stringify(ballots)
.replaceAll(`${lowestVotes}`, '')
.replaceAll(/(^[,\s]+)/g, '')
)
//remove leading commas
return Object.fromEntries(Object.entries(newRankedOrderBallots).map(([key, value]) => [key.replace(/^,+/, ''), value]))
}
// console.log(removeLowestVotedCandidate(sample))
function getRankedChoiceWinner(ballots) {
let winnerFound = false
let stretchWin = 0;
let winner = []
let updatedBallot = ballots
while (winnerFound === false) {
let sumByChar = {};
//count overall votes
let sum = 0
for (let votes of Object.values(updatedBallot)) {
sum += votes
}
//calculate what is required for a clear majority
stretchWin = Math.round(sum / 2)
//count votes assigned to each candidate
for (const [key, val] of Object.entries(updatedBallot)) {
const char = key[0];
sumByChar[char] = (sumByChar[char] ?? 0) + val;
}
// console.log('sumByChar is currently', sumByChar)
//check if any candidate has a clear majority
winner = Object.entries(sumByChar)
.filter(([, val]) => val >= stretchWin)
.map(([keys]) => keys);
console.log('winner is currently', winner)
if (winner.length === 1) {
winnerFound = true
} else {
updatedBallot = removeLowestVotedCandidate(updatedBallot)
// console.log('we are inside else', updatedBallot)
}
}
return winner
}
console.log(getRankedChoiceWinner(sample));
Or, refactored to look halfway decent IMO:
const sample = {
"A,B,C": 4,
"B,C,A": 3,
"C,B,A": 2
};
const getRankedChoiceWinner = badBallots => checkOneBallot(restructureBallots(badBallots));
const restructureBallots = badBallots => Object.entries(badBallots)
.map(([candidatesStr, votes]) => [candidatesStr.split(','), votes]);
const checkOneBallot = (ballots) => {
const sumVotes = ballots.reduce((a, b) => a + b[1], 0);
const sumByCandidate = {};
for (const [candidates, voteCount] of ballots) {
const candidate = candidates[0];
sumByCandidate[candidate] = (sumByCandidate[candidate] ?? 0) + voteCount;
}
const winningEntry = Object.entries(sumByCandidate).find(([, val]) => val >= sumVotes / 2);
if (winningEntry) return winningEntry[0][0];
return removeLowestAndRetry(ballots, sumByCandidate);
};
const removeLowestAndRetry = (ballots, sumByCandidate) => {
const lowestVal = Math.min(...Object.values(sumByCandidate));
const lowestCandidateEntry = Object.entries(sumByCandidate).reduce(
(a, entry) => entry[1] < a[1] ? entry : a,
['', Infinity]
);
const lowestCandidate = lowestCandidateEntry[0];
for (const ballot of ballots) {
ballot[0] = ballot[0].filter(candidate => candidate !== lowestCandidate);
}
return checkOneBallot(ballots);
};
console.log(getRankedChoiceWinner(sample));
I saw this interview question and gave a go. I got stuck. The interview question is:
Given a string
var s = "ilikealibaba";
and a dictionary
var d = ["i", "like", "ali", "liba", "baba", "alibaba"];
try to give the s with min space
The output may be
i like alibaba (2 spaces)
i like ali baba (3 spaces)
but pick no.1
I have some code, but got stuck in the printing.
If you have better way to do this question, let me know.
function isStartSub(part, s) {
var condi = s.startsWith(part);
return condi;
}
function getRestStr(part, s) {
var len = part.length;
var len1 = s.length;
var out = s.substring(len, len1);
return out;
}
function recPrint(arr) {
if(arr.length == 0) {
return '';
} else {
var str = arr.pop();
return str + recPrint(arr);
}
}
// NOTE: have trouble to print
// Or if you have better ways to do this interview question, please let me know
function myPrint(arr) {
return recPrint(arr);
}
function getMinArr(arr) {
var min = Number.MAX_SAFE_INTEGER;
var index = 0;
for(var i=0; i<arr.length; i++) {
var sub = arr[i];
if(sub.length < min) {
min = sub.length;
index = i;
} else {
}
}
return arr[index];
}
function rec(s, d, buf) {
// Base
if(s.length == 0) {
return;
} else {
}
for(var i=0; i<d.length; i++) {
var subBuf = [];
// baba
var part = d[i];
var condi = isStartSub(part, s);
if(condi) {
// rest string
var restStr = getRestStr(part, s);
rec(restStr, d, subBuf);
subBuf.unshift(part);
buf.unshift(subBuf);
} else {
}
} // end loop
}
function myfunc(s, d) {
var buf = [];
rec(s, d, buf);
console.log('-- test --');
console.dir(buf, {depth:null});
return myPrint(buf);
}
// Output will be
// 1. i like alibaba (with 2 spaces)
// 2. i like ali baba (with 3 spaces)
// we pick no.1, as it needs less spaces
var s = "ilikealibaba";
var d = ["i", "like", "ali", "liba", "baba", "alibaba"];
var out = myfunc(s, d);
console.log(out);
Basically, my output is, not sure how to print it....
[ [ 'i', [ 'like', [ 'alibaba' ], [ 'ali', [ 'baba' ] ] ] ] ]
This problem is best suited for a dynamic programming approach. The subproblem is, "what is the best way to create a prefix of s". Then, for a given prefix of s, we consider all words that match the end of the prefix, and choose the best one using the results from the earlier prefixes.
Here is an implementation:
var s = "ilikealibaba";
var arr = ["i", "like", "ali", "liba", "baba", "alibaba"];
var dp = []; // dp[i] is the optimal solution for s.substring(0, i)
dp.push("");
for (var i = 1; i <= s.length; i++) {
var best = null; // the best way so far for s.substring(0, i)
for (var j = 0; j < arr.length; j++) {
var word = arr[j];
// consider all words that appear at the end of the prefix
if (!s.substring(0, i).endsWith(word))
continue;
if (word.length == i) {
best = word; // using single word is optimal
break;
}
var prev = dp[i - word.length];
if (prev === null)
continue; // s.substring(i - word.length) can't be made at all
if (best === null || prev.length + word.length + 1 < best.length)
best = prev + " " + word;
}
dp.push(best);
}
console.log(dp[s.length]);
pkpnd's answer is along the right track. But word dictionaries tend to be quite large sets, and iterating over the entire dictionary at every character of the string is going to be inefficient. (Also, saving the entire sequence for each dp cell may consume a large amount of space.) Rather, we can frame the question, as we iterate over the string, as: given all the previous indexes of the string that had dictionary matches extending back (either to the start or to another match), which one is both a dictionary match when we include the current character, and has a smaller length in total. Generally:
f(i) = min(
f(j) + length(i - j) + (1 if j is after the start of the string)
)
for all j < i, where string[j] ended a dictionary match
and string[j+1..i] is in the dictionary
Since we only add another j when there is a match and a new match can only extend back to a previous match or to the start of the string, our data structure could be an array of tuples, (best index this match extends back to, total length up to here). We add another tuple if the current character can extend a dictionary match back to another record we already have. We can also optimize by exiting early from the backwards search once the matched substring would be greater than the longest word in the dictionary, and building the substring to compare against the dictionary as we iterate backwards.
JavaScript code:
function f(str, dict){
let m = [[-1, -1, -1]];
for (let i=0; i<str.length; i++){
let best = [null, null, Infinity];
let substr = '';
let _i = i;
for (let j=m.length-1; j>=0; j--){
let [idx, _j, _total] = m[j];
substr = str.substr(idx + 1, _i - idx) + substr;
_i = idx;
if (dict.has(substr)){
let total = _total + 1 + i - idx;
if (total < best[2])
best = [i, j, total];
}
}
if (best[0] !== null)
m.push(best);
}
return m;
}
var s = "ilikealibaba";
var d = new Set(["i", "like", "ali", "liba", "baba", "alibaba"]);
console.log(JSON.stringify(f(s,d)));
We can track back our result:
[[-1,-1,-1],[0,0,1],[4,1,6],[7,2,10],[11,2,14]]
[11, 2, 14] means a total length of 14,
where the previous index in m is 2 and the right index
of the substr is 11
=> follow it back to m[2] = [4, 1, 6]
this substr ended at index 4 (which means the
first was "alibaba"), and followed m[1]
=> [0, 0, 1], means this substr ended at index 1
so the previous one was "like"
And there you have it: "i like alibaba"
As you're asked to find a shortest answer probably Breadth-First Search would be a possible solution. Or you could look into A* Search.
Here is working example with A* (cause it's less bring to do than BFS :)), basically just copied from Wikipedia article. All the "turning string into a graph" magick happens in the getNeighbors function
https://jsfiddle.net/yLeps4v5/4/
var str = 'ilikealibaba'
var dictionary = ['i', 'like', 'ali', 'baba', 'alibaba']
var START = -1
var FINISH = str.length - 1
// Returns all the positions in the string that we can "jump" to from position i
function getNeighbors(i) {
const matchingWords = dictionary.filter(word => str.slice(i + 1, i + 1 + word.length) == word)
return matchingWords.map(word => i + word.length)
}
function aStar(start, goal) {
// The set of nodes already evaluated
const closedSet = {};
// The set of currently discovered nodes that are not evaluated yet.
// Initially, only the start node is known.
const openSet = [start];
// For each node, which node it can most efficiently be reached from.
// If a node can be reached from many nodes, cameFrom will eventually contain the
// most efficient previous step.
var cameFrom = {};
// For each node, the cost of getting from the start node to that node.
const gScore = dictionary.reduce((acc, word) => { acc[word] = Infinity; return acc }, {})
// The cost of going from start to start is zero.
gScore[start] = 0
while (openSet.length > 0) {
var current = openSet.shift()
if (current == goal) {
return reconstruct_path(cameFrom, current)
}
closedSet[current] = true;
getNeighbors(current).forEach(neighbor => {
if (closedSet[neighbor]) {
return // Ignore the neighbor which is already evaluated.
}
if (openSet.indexOf(neighbor) == -1) { // Discover a new node
openSet.push(neighbor)
}
// The distance from start to a neighbor
var tentative_gScore = gScore[current] + 1
if (tentative_gScore >= gScore[neighbor]) {
return // This is not a better path.
}
// This path is the best until now. Record it!
cameFrom[neighbor] = current
gScore[neighbor] = tentative_gScore
})
}
throw new Error('path not found')
}
function reconstruct_path(cameFrom, current) {
var answer = [];
while (cameFrom[current] || cameFrom[current] == 0) {
answer.push(str.slice(cameFrom[current] + 1, current + 1))
current = cameFrom[current];
}
return answer.reverse()
}
console.log(aStar(START, FINISH));
You could collect all possible combinations of the string by checking the starting string and render then the result.
If more than one result has the minimum length, all results are taken.
It might not work for extrema with string who just contains the same base string, like 'abcabc' and 'abc'. In this case I suggest to use the shortest string and update any part result by iterating for finding longer strings and replace if possible.
function getWords(string, array = []) {
words
.filter(w => string.startsWith(w))
.forEach(s => {
var rest = string.slice(s.length),
temp = array.concat(s);
if (rest) {
getWords(rest, temp);
} else {
result.push(temp);
}
});
}
var string = "ilikealibaba",
words = ["i", "like", "ali", "liba", "baba", "alibaba"],
result = [];
getWords(string);
console.log('all possible combinations:', result);
console.log('result:', result.reduce((r, a) => {
if (!r || r[0].length > a.length) {
return [a];
}
if (r[0].length === a.length) {
r.push(a);
}
return r;
}, undefined))
Use trie data structure
Construct a trie data structure based on the dictionary data
Search the sentence for all possible slices and build a solution tree
Deep traverse the solution tree and sort the final combinations
const sentence = 'ilikealibaba';
const words = ['i', 'like', 'ali', 'liba', 'baba', 'alibaba',];
class TrieNode {
constructor() { }
set(a) {
this[a] = this[a] || new TrieNode();
return this[a];
}
search(word, marks, depth = 1) {
word = Array.isArray(word) ? word : word.split('');
const a = word.shift();
if (this[a]) {
if (this[a]._) {
marks.push(depth);
}
this[a].search(word, marks, depth + 1);
} else {
return 0;
}
}
}
TrieNode.createTree = words => {
const root = new TrieNode();
words.forEach(word => {
let currentNode = root;
for (let i = 0; i < word.length; i++) {
currentNode = currentNode.set(word[i]);
}
currentNode.set('_');
});
return root;
};
const t = TrieNode.createTree(words);
function searchSentence(sentence) {
const marks = [];
t.search(sentence, marks);
const ret = {};
marks.map(mark => {
ret[mark] = searchSentence(sentence.slice(mark));
});
return ret;
}
const solutionTree = searchSentence(sentence);
function deepTraverse(tree, sentence, targetLen = sentence.length) {
const stack = [];
const sum = () => stack.reduce((acc, mark) => acc + mark, 0);
const ret = [];
(function traverse(tree) {
const keys = Object.keys(tree);
keys.forEach(key => {
stack.push(+key);
if (sum() === targetLen) {
const result = [];
let tempStr = sentence;
stack.forEach(mark => {
result.push(tempStr.slice(0, mark));
tempStr = tempStr.slice(mark);
});
ret.push(result);
}
if(tree[key]) {
traverse(tree[key]);
}
stack.pop();
});
})(tree);
return ret;
}
const solutions = deepTraverse(solutionTree, sentence);
solutions.sort((s1, s2) => s1.length - s2.length).forEach((s, i) => {
console.log(`${i + 1}. ${s.join(' ')} (${s.length - 1} spaces)`);
});
console.log('pick no.1');
What's the simplest, library-free code for implementing array intersections in javascript? I want to write
intersection([1,2,3], [2,3,4,5])
and get
[2, 3]
Use a combination of Array.prototype.filter and Array.prototype.includes:
const filteredArray = array1.filter(value => array2.includes(value));
For older browsers, with Array.prototype.indexOf and without an arrow function:
var filteredArray = array1.filter(function(n) {
return array2.indexOf(n) !== -1;
});
NB! Both .includes and .indexOf internally compares elements in the array by using ===, so if the array contains objects it will only compare object references (not their content). If you want to specify your own comparison logic, use Array.prototype.some instead.
Destructive seems simplest, especially if we can assume the input is sorted:
/* destructively finds the intersection of
* two arrays in a simple fashion.
*
* PARAMS
* a - first array, must already be sorted
* b - second array, must already be sorted
*
* NOTES
* State of input arrays is undefined when
* the function returns. They should be
* (prolly) be dumped.
*
* Should have O(n) operations, where n is
* n = MIN(a.length, b.length)
*/
function intersection_destructive(a, b)
{
var result = [];
while( a.length > 0 && b.length > 0 )
{
if (a[0] < b[0] ){ a.shift(); }
else if (a[0] > b[0] ){ b.shift(); }
else /* they're equal */
{
result.push(a.shift());
b.shift();
}
}
return result;
}
Non-destructive has to be a hair more complicated, since we’ve got to track indices:
/* finds the intersection of
* two arrays in a simple fashion.
*
* PARAMS
* a - first array, must already be sorted
* b - second array, must already be sorted
*
* NOTES
*
* Should have O(n) operations, where n is
* n = MIN(a.length(), b.length())
*/
function intersect_safe(a, b)
{
var ai=0, bi=0;
var result = [];
while( ai < a.length && bi < b.length )
{
if (a[ai] < b[bi] ){ ai++; }
else if (a[ai] > b[bi] ){ bi++; }
else /* they're equal */
{
result.push(a[ai]);
ai++;
bi++;
}
}
return result;
}
If your environment supports ECMAScript 6 Set, one simple and supposedly efficient (see specification link) way:
function intersect(a, b) {
var setA = new Set(a);
var setB = new Set(b);
var intersection = new Set([...setA].filter(x => setB.has(x)));
return Array.from(intersection);
}
Shorter, but less readable (also without creating the additional intersection Set):
function intersect(a, b) {
var setB = new Set(b);
return [...new Set(a)].filter(x => setB.has(x));
}
Note that when using sets you will only get distinct values, thus new Set([1, 2, 3, 3]).size evaluates to 3.
Using Underscore.js or lodash.js
_.intersection( [0,345,324] , [1,0,324] ) // gives [0,324]
// Return elements of array a that are also in b in linear time:
function intersect(a, b) {
return a.filter(Set.prototype.has, new Set(b));
}
// Example:
console.log(intersect([1,2,3], [2,3,4,5]));
I recommend above succinct solution which outperforms other implementations on large inputs. If performance on small inputs matters, check the alternatives below.
Alternatives and performance comparison:
See the following snippet for alternative implementations and check https://jsperf.com/array-intersection-comparison for performance comparisons.
function intersect_for(a, b) {
const result = [];
const alen = a.length;
const blen = b.length;
for (let i = 0; i < alen; ++i) {
const ai = a[i];
for (let j = 0; j < blen; ++j) {
if (ai === b[j]) {
result.push(ai);
break;
}
}
}
return result;
}
function intersect_filter_indexOf(a, b) {
return a.filter(el => b.indexOf(el) !== -1);
}
function intersect_filter_in(a, b) {
const map = b.reduce((map, el) => {map[el] = true; return map}, {});
return a.filter(el => el in map);
}
function intersect_for_in(a, b) {
const result = [];
const map = {};
for (let i = 0, length = b.length; i < length; ++i) {
map[b[i]] = true;
}
for (let i = 0, length = a.length; i < length; ++i) {
if (a[i] in map) result.push(a[i]);
}
return result;
}
function intersect_filter_includes(a, b) {
return a.filter(el => b.includes(el));
}
function intersect_filter_has_this(a, b) {
return a.filter(Set.prototype.has, new Set(b));
}
function intersect_filter_has_arrow(a, b) {
const set = new Set(b);
return a.filter(el => set.has(el));
}
function intersect_for_has(a, b) {
const result = [];
const set = new Set(b);
for (let i = 0, length = a.length; i < length; ++i) {
if (set.has(a[i])) result.push(a[i]);
}
return result;
}
Results in Firefox 53:
Ops/sec on large arrays (10,000 elements):
filter + has (this) 523 (this answer)
for + has 482
for-loop + in 279
filter + in 242
for-loops 24
filter + includes 14
filter + indexOf 10
Ops/sec on small arrays (100 elements):
for-loop + in 384,426
filter + in 192,066
for-loops 159,137
filter + includes 104,068
filter + indexOf 71,598
filter + has (this) 43,531 (this answer)
filter + has (arrow function) 35,588
My contribution in ES6 terms. In general it finds the intersection of an array with indefinite number of arrays provided as arguments.
Array.prototype.intersect = function(...a) {
return [this,...a].reduce((p,c) => p.filter(e => c.includes(e)));
}
var arrs = [[0,2,4,6,8],[4,5,6,7],[4,6]],
arr = [0,1,2,3,4,5,6,7,8,9];
document.write("<pre>" + JSON.stringify(arr.intersect(...arrs)) + "</pre>");
How about just using associative arrays?
function intersect(a, b) {
var d1 = {};
var d2 = {};
var results = [];
for (var i = 0; i < a.length; i++) {
d1[a[i]] = true;
}
for (var j = 0; j < b.length; j++) {
d2[b[j]] = true;
}
for (var k in d1) {
if (d2[k])
results.push(k);
}
return results;
}
edit:
// new version
function intersect(a, b) {
var d = {};
var results = [];
for (var i = 0; i < b.length; i++) {
d[b[i]] = true;
}
for (var j = 0; j < a.length; j++) {
if (d[a[j]])
results.push(a[j]);
}
return results;
}
The performance of #atk's implementation for sorted arrays of primitives can be improved by using .pop rather than .shift.
function intersect(array1, array2) {
var result = [];
// Don't destroy the original arrays
var a = array1.slice(0);
var b = array2.slice(0);
var aLast = a.length - 1;
var bLast = b.length - 1;
while (aLast >= 0 && bLast >= 0) {
if (a[aLast] > b[bLast] ) {
a.pop();
aLast--;
} else if (a[aLast] < b[bLast] ){
b.pop();
bLast--;
} else /* they're equal */ {
result.push(a.pop());
b.pop();
aLast--;
bLast--;
}
}
return result;
}
I created a benchmark using jsPerf. It's about three times faster to use .pop.
If you need to have it handle intersecting multiple arrays:
const intersect = (a1, a2, ...rest) => {
const a12 = a1.filter(value => a2.includes(value))
if (rest.length === 0) { return a12; }
return intersect(a12, ...rest);
};
console.log(intersect([1,2,3,4,5], [1,2], [1, 2, 3,4,5], [2, 10, 1]))
Sort it
check one by one from the index 0, create new array from that.
Something like this, Not tested well though.
function intersection(x,y){
x.sort();y.sort();
var i=j=0;ret=[];
while(i<x.length && j<y.length){
if(x[i]<y[j])i++;
else if(y[j]<x[i])j++;
else {
ret.push(x[i]);
i++,j++;
}
}
return ret;
}
alert(intersection([1,2,3], [2,3,4,5]));
PS:The algorithm only intended for Numbers and Normal Strings, intersection of arbitary object arrays may not work.
Using jQuery:
var a = [1,2,3];
var b = [2,3,4,5];
var c = $(b).not($(b).not(a));
alert(c);
A tiny tweak to the smallest one here (the filter/indexOf solution), namely creating an index of the values in one of the arrays using a JavaScript object, will reduce it from O(N*M) to "probably" linear time. source1 source2
function intersect(a, b) {
var aa = {};
a.forEach(function(v) { aa[v]=1; });
return b.filter(function(v) { return v in aa; });
}
This isn't the very simplest solution (it's more code than filter+indexOf), nor is it the very fastest (probably slower by a constant factor than intersect_safe()), but seems like a pretty good balance. It is on the very simple side, while providing good performance, and it doesn't require pre-sorted inputs.
For arrays containing only strings or numbers you can do something with sorting, as per some of the other answers. For the general case of arrays of arbitrary objects I don't think you can avoid doing it the long way. The following will give you the intersection of any number of arrays provided as parameters to arrayIntersection:
var arrayContains = Array.prototype.indexOf ?
function(arr, val) {
return arr.indexOf(val) > -1;
} :
function(arr, val) {
var i = arr.length;
while (i--) {
if (arr[i] === val) {
return true;
}
}
return false;
};
function arrayIntersection() {
var val, arrayCount, firstArray, i, j, intersection = [], missing;
var arrays = Array.prototype.slice.call(arguments); // Convert arguments into a real array
// Search for common values
firstArray = arrays.pop();
if (firstArray) {
j = firstArray.length;
arrayCount = arrays.length;
while (j--) {
val = firstArray[j];
missing = false;
// Check val is present in each remaining array
i = arrayCount;
while (!missing && i--) {
if ( !arrayContains(arrays[i], val) ) {
missing = true;
}
}
if (!missing) {
intersection.push(val);
}
}
}
return intersection;
}
arrayIntersection( [1, 2, 3, "a"], [1, "a", 2], ["a", 1] ); // Gives [1, "a"];
Simplest, fastest O(n) and shortest way:
function intersection (a, b) {
const setA = new Set(a);
return b.filter(value => setA.has(value));
}
console.log(intersection([1,2,3], [2,3,4,5]))
#nbarbosa has almost the same answer but he cast both arrays to Set and then back to array. There is no need for any extra casting.
Another indexed approach able to process any number of arrays at once:
// Calculate intersection of multiple array or object values.
function intersect (arrList) {
var arrLength = Object.keys(arrList).length;
// (Also accepts regular objects as input)
var index = {};
for (var i in arrList) {
for (var j in arrList[i]) {
var v = arrList[i][j];
if (index[v] === undefined) index[v] = 0;
index[v]++;
};
};
var retv = [];
for (var i in index) {
if (index[i] == arrLength) retv.push(i);
};
return retv;
};
It works only for values that can be evaluated as strings and you should pass them as an array like:
intersect ([arr1, arr2, arr3...]);
...but it transparently accepts objects as parameter or as any of the elements to be intersected (always returning array of common values). Examples:
intersect ({foo: [1, 2, 3, 4], bar: {a: 2, j:4}}); // [2, 4]
intersect ([{x: "hello", y: "world"}, ["hello", "user"]]); // ["hello"]
EDIT: I just noticed that this is, in a way, slightly buggy.
That is: I coded it thinking that input arrays cannot itself contain repetitions (as provided example doesn't).
But if input arrays happen to contain repetitions, that would produce wrong results. Example (using below implementation):
intersect ([[1, 3, 4, 6, 3], [1, 8, 99]]);
// Expected: [ '1' ]
// Actual: [ '1', '3' ]
Fortunately this is easy to fix by simply adding second level indexing. That is:
Change:
if (index[v] === undefined) index[v] = 0;
index[v]++;
by:
if (index[v] === undefined) index[v] = {};
index[v][i] = true; // Mark as present in i input.
...and:
if (index[i] == arrLength) retv.push(i);
by:
if (Object.keys(index[i]).length == arrLength) retv.push(i);
Complete example:
// Calculate intersection of multiple array or object values.
function intersect (arrList) {
var arrLength = Object.keys(arrList).length;
// (Also accepts regular objects as input)
var index = {};
for (var i in arrList) {
for (var j in arrList[i]) {
var v = arrList[i][j];
if (index[v] === undefined) index[v] = {};
index[v][i] = true; // Mark as present in i input.
};
};
var retv = [];
for (var i in index) {
if (Object.keys(index[i]).length == arrLength) retv.push(i);
};
return retv;
};
intersect ([[1, 3, 4, 6, 3], [1, 8, 99]]); // [ '1' ]
With some restrictions on your data, you can do it in linear time!
For positive integers: use an array mapping the values to a "seen/not seen" boolean.
function intersectIntegers(array1,array2) {
var seen=[],
result=[];
for (var i = 0; i < array1.length; i++) {
seen[array1[i]] = true;
}
for (var i = 0; i < array2.length; i++) {
if ( seen[array2[i]])
result.push(array2[i]);
}
return result;
}
There is a similar technique for objects: take a dummy key, set it to "true" for each element in array1, then look for this key in elements of array2. Clean up when you're done.
function intersectObjects(array1,array2) {
var result=[];
var key="tmpKey_intersect"
for (var i = 0; i < array1.length; i++) {
array1[i][key] = true;
}
for (var i = 0; i < array2.length; i++) {
if (array2[i][key])
result.push(array2[i]);
}
for (var i = 0; i < array1.length; i++) {
delete array1[i][key];
}
return result;
}
Of course you need to be sure the key didn't appear before, otherwise you'll be destroying your data...
function intersection(A,B){
var result = new Array();
for (i=0; i<A.length; i++) {
for (j=0; j<B.length; j++) {
if (A[i] == B[j] && $.inArray(A[i],result) == -1) {
result.push(A[i]);
}
}
}
return result;
}
For simplicity:
// Usage
const intersection = allLists
.reduce(intersect, allValues)
.reduce(removeDuplicates, []);
// Implementation
const intersect = (intersection, list) =>
intersection.filter(item =>
list.some(x => x === item));
const removeDuplicates = (uniques, item) =>
uniques.includes(item) ? uniques : uniques.concat(item);
// Example Data
const somePeople = [bob, doug, jill];
const otherPeople = [sarah, bob, jill];
const morePeople = [jack, jill];
const allPeople = [...somePeople, ...otherPeople, ...morePeople];
const allGroups = [somePeople, otherPeople, morePeople];
// Example Usage
const intersection = allGroups
.reduce(intersect, allPeople)
.reduce(removeDuplicates, []);
intersection; // [jill]
Benefits:
dirt simple
data-centric
works for arbitrary number of lists
works for arbitrary lengths of lists
works for arbitrary types of values
works for arbitrary sort order
retains shape (order of first appearance in any array)
exits early where possible
memory safe, short of tampering with Function / Array prototypes
Drawbacks:
higher memory usage
higher CPU usage
requires an understanding of reduce
requires understanding of data flow
You wouldn't want to use this for 3D engine or kernel work, but if you have problems getting this to run in an event-based app, your design has bigger problems.
I'll contribute with what has been working out best for me:
if (!Array.prototype.intersect){
Array.prototype.intersect = function (arr1) {
var r = [], o = {}, l = this.length, i, v;
for (i = 0; i < l; i++) {
o[this[i]] = true;
}
l = arr1.length;
for (i = 0; i < l; i++) {
v = arr1[i];
if (v in o) {
r.push(v);
}
}
return r;
};
}
A functional approach with ES2015
A functional approach must consider using only pure functions without side effects, each of which is only concerned with a single job.
These restrictions enhance the composability and reusability of the functions involved.
// small, reusable auxiliary functions
const createSet = xs => new Set(xs);
const filter = f => xs => xs.filter(apply(f));
const apply = f => x => f(x);
// intersection
const intersect = xs => ys => {
const zs = createSet(ys);
return filter(x => zs.has(x)
? true
: false
) (xs);
};
// mock data
const xs = [1,2,2,3,4,5];
const ys = [0,1,2,3,3,3,6,7,8,9];
// run it
console.log( intersect(xs) (ys) );
Please note that the native Set type is used, which has an advantageous
lookup performance.
Avoid duplicates
Obviously repeatedly occurring items from the first Array are preserved, while the second Array is de-duplicated. This may be or may be not the desired behavior. If you need a unique result just apply dedupe to the first argument:
// auxiliary functions
const apply = f => x => f(x);
const comp = f => g => x => f(g(x));
const afrom = apply(Array.from);
const createSet = xs => new Set(xs);
const filter = f => xs => xs.filter(apply(f));
// intersection
const intersect = xs => ys => {
const zs = createSet(ys);
return filter(x => zs.has(x)
? true
: false
) (xs);
};
// de-duplication
const dedupe = comp(afrom) (createSet);
// mock data
const xs = [1,2,2,3,4,5];
const ys = [0,1,2,3,3,3,6,7,8,9];
// unique result
console.log( intersect(dedupe(xs)) (ys) );
Compute the intersection of any number of Arrays
If you want to compute the intersection of an arbitrarily number of Arrays just compose intersect with foldl. Here is a convenience function:
// auxiliary functions
const apply = f => x => f(x);
const uncurry = f => (x, y) => f(x) (y);
const createSet = xs => new Set(xs);
const filter = f => xs => xs.filter(apply(f));
const foldl = f => acc => xs => xs.reduce(uncurry(f), acc);
// intersection
const intersect = xs => ys => {
const zs = createSet(ys);
return filter(x => zs.has(x)
? true
: false
) (xs);
};
// intersection of an arbitrarily number of Arrays
const intersectn = (head, ...tail) => foldl(intersect) (head) (tail);
// mock data
const xs = [1,2,2,3,4,5];
const ys = [0,1,2,3,3,3,6,7,8,9];
const zs = [0,1,2,3,4,5,6];
// run
console.log( intersectn(xs, ys, zs) );
.reduce to build a map, and .filter to find the intersection. delete within the .filter allows us to treat the second array as though it's a unique set.
function intersection (a, b) {
var seen = a.reduce(function (h, k) {
h[k] = true;
return h;
}, {});
return b.filter(function (k) {
var exists = seen[k];
delete seen[k];
return exists;
});
}
I find this approach pretty easy to reason about. It performs in constant time.
I have written an intesection function which can even detect intersection of array of objects based on particular property of those objects.
For instance,
if arr1 = [{id: 10}, {id: 20}]
and arr2 = [{id: 20}, {id: 25}]
and we want intersection based on the id property, then the output should be :
[{id: 20}]
As such, the function for the same (note: ES6 code) is :
const intersect = (arr1, arr2, accessors = [v => v, v => v]) => {
const [fn1, fn2] = accessors;
const set = new Set(arr2.map(v => fn2(v)));
return arr1.filter(value => set.has(fn1(value)));
};
and you can call the function as:
intersect(arr1, arr2, [elem => elem.id, elem => elem.id])
Also note: this function finds intersection considering the first array is the primary array and thus the intersection result will be that of the primary array.
This function avoids the N^2 problem, taking advantage of the power of dictionaries. Loops through each array only once, and a third and shorter loop to return the final result.
It also supports numbers, strings, and objects.
function array_intersect(array1, array2)
{
var mergedElems = {},
result = [];
// Returns a unique reference string for the type and value of the element
function generateStrKey(elem) {
var typeOfElem = typeof elem;
if (typeOfElem === 'object') {
typeOfElem += Object.prototype.toString.call(elem);
}
return [typeOfElem, elem.toString(), JSON.stringify(elem)].join('__');
}
array1.forEach(function(elem) {
var key = generateStrKey(elem);
if (!(key in mergedElems)) {
mergedElems[key] = {elem: elem, inArray2: false};
}
});
array2.forEach(function(elem) {
var key = generateStrKey(elem);
if (key in mergedElems) {
mergedElems[key].inArray2 = true;
}
});
Object.values(mergedElems).forEach(function(elem) {
if (elem.inArray2) {
result.push(elem.elem);
}
});
return result;
}
If there is a special case that cannot be solved, just by modifying the generateStrKey function, it could surely be solved. The trick of this function is that it uniquely represents each different data according to type and value.
This variant has some performance improvements. Avoid loops in case any array is empty. It also starts by walking through the shorter array first, so if it finds all the values of the first array in the second array, exits the loop.
function array_intersect(array1, array2)
{
var mergedElems = {},
result = [],
firstArray, secondArray,
firstN = 0,
secondN = 0;
function generateStrKey(elem) {
var typeOfElem = typeof elem;
if (typeOfElem === 'object') {
typeOfElem += Object.prototype.toString.call(elem);
}
return [typeOfElem, elem.toString(), JSON.stringify(elem)].join('__');
}
// Executes the loops only if both arrays have values
if (array1.length && array2.length)
{
// Begins with the shortest array to optimize the algorithm
if (array1.length < array2.length) {
firstArray = array1;
secondArray = array2;
} else {
firstArray = array2;
secondArray = array1;
}
firstArray.forEach(function(elem) {
var key = generateStrKey(elem);
if (!(key in mergedElems)) {
mergedElems[key] = {elem: elem, inArray2: false};
// Increases the counter of unique values in the first array
firstN++;
}
});
secondArray.some(function(elem) {
var key = generateStrKey(elem);
if (key in mergedElems) {
if (!mergedElems[key].inArray2) {
mergedElems[key].inArray2 = true;
// Increases the counter of matches
secondN++;
// If all elements of first array have coincidence, then exits the loop
return (secondN === firstN);
}
}
});
Object.values(mergedElems).forEach(function(elem) {
if (elem.inArray2) {
result.push(elem.elem);
}
});
}
return result;
}
Here is underscore.js implementation:
_.intersection = function(array) {
if (array == null) return [];
var result = [];
var argsLength = arguments.length;
for (var i = 0, length = array.length; i < length; i++) {
var item = array[i];
if (_.contains(result, item)) continue;
for (var j = 1; j < argsLength; j++) {
if (!_.contains(arguments[j], item)) break;
}
if (j === argsLength) result.push(item);
}
return result;
};
Source: http://underscorejs.org/docs/underscore.html#section-62
Create an Object using one array and loop through the second array to check if the value exists as key.
function intersection(arr1, arr2) {
var myObj = {};
var myArr = [];
for (var i = 0, len = arr1.length; i < len; i += 1) {
if(myObj[arr1[i]]) {
myObj[arr1[i]] += 1;
} else {
myObj[arr1[i]] = 1;
}
}
for (var j = 0, len = arr2.length; j < len; j += 1) {
if(myObj[arr2[j]] && myArr.indexOf(arr2[j]) === -1) {
myArr.push(arr2[j]);
}
}
return myArr;
}
I think using an object internally can help with computations and could be performant too.
// Approach maintains a count of each element and works for negative elements too
function intersect(a,b){
const A = {};
a.forEach((v)=>{A[v] ? ++A[v] : A[v] = 1});
const B = {};
b.forEach((v)=>{B[v] ? ++B[v] : B[v] = 1});
const C = {};
Object.entries(A).map((x)=>C[x[0]] = Math.min(x[1],B[x[0]]))
return Object.entries(C).map((x)=>Array(x[1]).fill(Number(x[0]))).flat();
}
const x = [1,1,-1,-1,0,0,2,2];
const y = [2,0,1,1,1,1,0,-1,-1,-1];
const result = intersect(x,y);
console.log(result); // (7) [0, 0, 1, 1, 2, -1, -1]
I am using map even object could be used.
//find intersection of 2 arrs
const intersections = (arr1,arr2) => {
let arrf = arr1.concat(arr2)
let map = new Map();
let union = [];
for(let i=0; i<arrf.length; i++){
if(map.get(arrf[i])){
map.set(arrf[i],false);
}else{
map.set(arrf[i],true);
}
}
map.forEach((v,k)=>{if(!v){union.push(k);}})
return union;
}
This is a proposed standard: With the currently stage 2 proposal https://github.com/tc39/proposal-set-methods, you could use
mySet.intersection(mySet2);
Until then, you could use Immutable.js's Set, which inspired that proposal
Immutable.Set(mySet).intersect(mySet2)
I extended tarulen's answer to work with any number of arrays. It also should work with non-integer values.
function intersect() {
const last = arguments.length - 1;
var seen={};
var result=[];
for (var i = 0; i < last; i++) {
for (var j = 0; j < arguments[i].length; j++) {
if (seen[arguments[i][j]]) {
seen[arguments[i][j]] += 1;
}
else if (!i) {
seen[arguments[i][j]] = 1;
}
}
}
for (var i = 0; i < arguments[last].length; i++) {
if ( seen[arguments[last][i]] === last)
result.push(arguments[last][i]);
}
return result;
}
If your arrays are sorted, this should run in O(n), where n is min( a.length, b.length )
function intersect_1d( a, b ){
var out=[], ai=0, bi=0, acurr, bcurr, last=Number.MIN_SAFE_INTEGER;
while( ( acurr=a[ai] )!==undefined && ( bcurr=b[bi] )!==undefined ){
if( acurr < bcurr){
if( last===acurr ){
out.push( acurr );
}
last=acurr;
ai++;
}
else if( acurr > bcurr){
if( last===bcurr ){
out.push( bcurr );
}
last=bcurr;
bi++;
}
else {
out.push( acurr );
last=acurr;
ai++;
bi++;
}
}
return out;
}