I have array of arrays, each array has unique items . I need to find and collect all duplicate items between all arrays to new array .
For this input : [[1,2,6,9],[3,2,7,5,12],[1,3]]
I need this output: [1,2,3]
Does anyone now what is the best way to do it ?
You could take a closure over an object for a count of the values.
const
data = [[1, 2, 6, 9], [3, 2, 7, 5, 12], [1, 3]],
result = data
.flat()
.filter(
(o => v => (o[v] = (o[v] || 0) + 1) === 2)
({})
);
console.log(result);You can try this, optimal with O(3N)=== O(N) complexity
let arr = [
[1, 2, 6, 9],
[3, 2, 7, 5, 12],
[1, 3]
];
let a = arr.flat()
let obj = {};
for (let i = 0; i < a.length; i++) {
if (obj[a[i]]) obj[a[i]]++;
else obj[a[i]] = 1
}
let repeated = [];
for (let i in obj) {
if (obj[i] > 1) {
repeated.push(i)
}
}
console.log(repeated)
Depending on what is meant by "duplicates" @Nina's solution can be modified to include only those cases where the value occurrs exactly twice. My slightly modified example snippet will exclude the value 1 as it occurs three times:
const data = [[1, 2, 6, 9], [3, 2, 7, 5, 12], [1, 3], [1, 7]], k=2, // no. of occurrences
result = data.reduce((o,c,i)=>(c.forEach(v=>o[v]=(o[v]||0)+1),o),[])
.map((v,i,a)=>(a[i] = v===k ? i : undefined ))
.filter(v=>v);
console.log(result);caveat: This will only work with numeric values (integers!).
But here is another version that will work with any kind of content:
var data = [[1, 2, 6, 9, "c"], [3, "a", 2, 7, 5, 12], [1, "c", 3], [1, 7]], k=2;
res=[...data.reduce((o,c,i)=>(c.forEach(v=>o.set(v,(o.get(v)||0)+1)),o),
new Map()).entries()]
.filter(([k,v])=>v===2)
.map(([k,v])=>k)
console.log(res)Update
If you want the addresses of the first location the duplicates were found you could do it like this:
var data = [[1, 2, 6, 9, "c"], [3, "a", 2, 7, 5, 12], [1, "c", 3], [1, 7]], n=2;
res=[...data.reduce((o,c,i)=>(c.forEach(v=>o.set(v,(o.get(v)||0)+1)),o),
new Map()).entries()]
.filter(([k,v])=>v===n)
.map(([k,v],j)=>[data.findIndex((d,i)=>(j=d.indexOf(k))>-1)+"_"+j, k])
console.log(Object.fromEntries(res))
{"0_2":1,"0_1":2,"1_2":3) in your original example? Or, for my sample data, the result would be {"0_1":2,"0_4":"c","1_0":3,"1_3":7) - check my edited answer.