graph algorithm - Javascript Union Pairs Union Find -

i working on union finding. want group pairs of numbers based on whether 1 of indices shares number index of pair. so:

i have array of pairs such these:

pairs: [[1,3], [6,8], [3,8], [2,7]] 

whats best way group them in unions such this:

[ [ 1, 3, 8, 6 ], [ 2, 7 ] ] 

([1,3] , [3,8] go because share 3. group unites [6,8] because share 8. whats best way in javascript?

here other examples:

pairs: [[8,5], [10,8], [4,18], [20,12], [5,2], [17,2], [13,25],[29,12], [22,2], [17,11]]  [ [ 8, 5, 10, 2, 17, 22, 11 ],[ 4, 18 ],[ 20, 12, 29 ],[ 13, 25 ] ] 

edit here's method i'm using:

findunions = function(pairs, unions){    if (!unions){        unions = [pairs[0]];        pairs.shift();    }else{        if(pairs.length){            unions.push(pairs[0])            pairs.shift()        }    }      if (!pairs.length){         return unions     }     unite = true     while (unite && pairs.length){         unite = false         loop1:         (i in unions){             loop2:             var length = pairs.length;             (j=0;j<length;j++){                 if (unions[i].includes(pairs[j][0])){                     if (!unions[i].includes(pairs[j][1])){                         unions[i].push(pairs[j][1])                         pairs.splice(j, 1)                         j-=1;                         length-=1                         unite = true                     }else{                         pairs.splice(j, 1)                         j-=1                         length-=1                     }                 }else if (unions[i].includes(pairs[j][1])){                      unions[i].push(pairs[j][0])                      pairs.splice(j, 1)                      unite = true                     j-=1                     length-=1                 }             }         }     }     return findunions(pairs, unions) } 

method:

finalarray = [], positions = {};     array.length    j=i+1 array.length        find match between arr[i] , arr[j]        if match found           pos = postion mapped either or j in positions           add elements of arr[i] or arr[j] or both depending on pos. return finalarray 

in method keep storing positions of arrays adding finalarray in positions object , later can use object find suitable position add elements of matched arrays in finalarray.

function mergearrays(finalarray, pos, subarray) {  (var k = 0; k < subarray.length; k++) {      if (finalarray[pos].indexof(subarray[k]) < 0)          finalarray[pos].push(subarray[k]);  }    }    function unionarrays(arr) {  var finalarray = [arr[0]],      positions = {          0: 0      };  (var = 0; < arr.length; i++) {      (var j = + 1; j < arr.length; j++) {          (var k = 0; k < arr[i].length; k++) {              if (arr[j].indexof(arr[i][k]) >= 0) {                  if (i in positions) {                      mergearrays(finalarray, positions[i], arr[j]);                      positions[j] = positions[i];                  } else if (j in positions) {                      mergearrays(finalarray, positions[j], arr[i]);                      positions[i] = positions[j];                  } else {                      var pos = finalarray.length;                      finalarray.push([]);                      mergearrays(finalarray, pos, arr[i]);                      mergearrays(finalarray, pos, arr[j]);                      positions[i] = positions[j] = pos;                  }                  break;              }            }      }      if (!(i in positions)) {          finalarray.push(arr[i]);          positions[i] = finalarray.length - 1;      }  }  return finalarray;  }  console.log(unionarrays([[1,3], [6,8], [3,8], [2,7]]));  console.log(unionarrays([[8,5], [10,8], [4,18], [20,12], [5,2], [17,2], [13,25],[29,12], [22,2], [17,11]]));