Find if two arrays are repeated in array and then select them

You could take a Map with stringified arrays and count, then filter by count and restore the arrays.

var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]],
    result = Array
        .from(array.reduce(
            (map, array) =>
                (json => map.set(json, (map.get(json) || 0) + 1))
                (JSON.stringify(array)),
            new Map
         ))
        .filter(([, count]) => count > 2)
        .map(([json]) => JSON.parse(json));
        
console.log(result);
.as-console-wrapper { max-height: 100% !important; top: 0; }

Filter with a map at wanted count.

var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]],
    result = array.filter(
        (map => a => 
            (json =>
                (count => map.set(json, count) && !(2 - count))
                (1 + map.get(json) || 1)
            )
            (JSON.stringify(a))
        )
        (new Map)
    );
        
console.log(result);
.as-console-wrapper { max-height: 100% !important; top: 0; }

Unique!

var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]],
    result = array.filter(
        (s => a => (j => !s.has(j) && s.add(j))(JSON.stringify(a)))
        (new Set)
    );
        
console.log(result);
.as-console-wrapper { max-height: 100% !important; top: 0; }


Try this

array.filter(( r={}, a=>!(2-(r[a]=++r[a]|0)) ))

var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]];

var r= array.filter(( r={}, a=>!(2-(r[a]=++r[a]|0)) ))

console.log(JSON.stringify(r));

Time complexity O(n) (one array pass by filter function). Inspired by Nitish answer.

Explanation

The (r={}, a=>...) will return last expression after comma (which is a=>...) (e.g. (5,6)==6). In r={} we set once temporary object where we will store unique keys. In filter function a=>... in a we have current array element . In r[a] JS implicity cast a to string (e.g 1,17). Then in !(2-(r[a]=++r[a]|0)) we increase counter of occurrence element a and return true (as filter function value) if element a occurred 3 times. If r[a] is undefined the ++r[a] returns NaN, and further NaN|0=0 (also number|0=number). The r[a]= initialise first counter value, if we omit it the ++ will only set NaN to r[a] which is non-incrementable (so we need to put zero at init). If we remove 2- as result we get input array without duplicates - or alternatively we can also get this by a=>!(r[a]=a in r). If we change 2- to 1- we get array with duplicates only.


You can use Object.reduce, Object.entries for this like below

var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]];


let res = Object.entries(
            array.reduce((o, d) => {
              let key = d.join('-')
              o[key] = (o[key] || 0) + 1

              return o
          }, {}))
          .flatMap(([k, v]) => v > 2 ? [k.split('-').map(Number)] : [])
  
  
console.log(res)

OR may be just with Array.filters

var array = [[1, 17], [1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]];

let temp = {}
let res = array.filter(d => {
  let key = d.join('-')
  temp[key] = (temp[key] || 0) + 1
  
  return temp[key] == 3
})

console.log(res)