如何获取数组之间的差异并返回它们

Question

Im practicing my interviewing skills and i got this problem to solve:

"Write a function that processes two arrays of integers. Each array will have only distinct numbers (no repeats of the same integer in the same array), and the arrays are not sorted. find out which numbers are in array 1, but not array 2, and which numbers are in array 2, but not in array 1"

im trying to polish my ES6+ skills too so i know that a good way to loop through them is using array.map and also if i want the difference i need Array.prototype.filter() This way, i will get an array containing all the elements of arr1 that are not in arr2 and vice-versa

symmetric or non symmetric you ask?

i'll go with the fastest please

thanks!

Answer 1

Below two implementations.

The first one favors readability and is suitable for normal-sized arrays.

You can't go asymptotically faster than the second method (complexity in O(n*log(n)) ), although you can go faster for smaller arrays and reduce linearly the number of instructions executed.

 const arr1 = [1,2,9,8,6,7]; const arr2 = [8,6,3,4,1,0]; // most readable console.log( // arr1 without arr2 arr1.filter(x => arr2.indexOf(x)==-1), // arr1 and arr2 arr1.filter(x => arr2.indexOf(x)!=-1), // arr2 without arr1 arr2.filter(x => arr1.indexOf(x)==-1) ); // fastest - O(nln(n)) (() => { const count = // concatenate - O(n) [...arr1, ...arr1, ...arr2] // sort - O(nln(n)) .sort() // count - O(n) .reduce((acc, cur) => { acc[cur] = (acc[cur] || 0) +1; return acc; }, {}); // separate - O(n) const a1 = []; const a12 = []; const a2 = []; for (let i in count) { switch(count[i]) { case 1: a2.push(+i); break; case 2: a1.push(+i); break; case 3: a12.push(+i); break; } } console.log( a1, // arr1 without arr2 a12, // arr1 and arr2 a2 // arr2 without arr1 ); })();