在 JavaScript 中使用 filter() 在两个未排序数组中查找交集的大 O

Question

I just started learning Big O notation and I'm trying to understand the Big O of different functions to see which one is better.

I'm struggling to calculate the time and space complexity for the following code.

function findCommonElem(arr1, arr2) { 
  let result = arr1.filter(x => arr2.includes(x));
  console.log(result);
}

  findCommonElem(arr1, arr2);

From what I understand, common array methods like filter() usually have a big O of O(n) so in this case, it'd be O(m+n) depending on the length of each array. However, I could be super wrong.

Can someone explain, please? Thanks so much!

Bonus question: Compared to sorting the arrays then using a while loop for the same function, which one would be considered as "better"?

Answer 1

The above function has time complexity of O(M * N) .

But, can you make this solution more efficient? Yes. You can reduce it to O(M + N) .

TLDR - Use a hash table to achieve linear time complexity O(M + N)

Let's see.

---

Approach 1

Check every elements of array 1 with every element of array 2. (This is the approach you're using.)

 function findCommonElem(arr1, arr2) { return arr1.filter(x => arr2.includes(x)); } const arr1 = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]; const arr2 = [2, 4, 6, 8, 10, 12, 14, 16, 20]; console.log(findCommonElem(arr1, arr2)); // [2, 4, 6, 8, 10];

• Time complexity = O(M * N)

  • Every element from array 1 is checked with every element of array 2 at the worst case. So, it's M * N.

• Space compexity = O(M) or O(N)

  • At most, all of the elements from one of the either arrays could be in the intersection.

Approach 2

Use a hash map to linearize the nested loop. First, populate the hash map with array 1 elements. Then check through the array 2 using the map to find the intersection.

 function findCommonElem(arr1, arr2) { const map = new Map(); arr1.forEach(item => map.set(item, true)); return arr2.filter(item => map.has(item)); } const arr1 = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]; const arr2 = [2, 4, 6, 8, 10, 12, 14, 16, 20]; console.log(findCommonElem(arr1, arr2)); // [2, 4, 6, 8, 10];

The above function returns the same output. But here's the difference - the nested loop is reduced to two linear loops for the arrays. This means both arrays are traversed only once.

• Time complexity = O(M + N)

  • Array 1 is traversed once (M elements).
  • Array 2 is traversed once (N elements).
  • Checking if the map contains the element with map.has() takes constant time O(1) .
  • Total run time = M + N

• Space compexity = O(M) or O(N)

  • Space complexity is still the same here, because space needed for the new hash map is either O(M) or O(N) . And our intermediate array takes either O(M) or O(N) . Which is still the same.

Bonus: Don't know much about how hash maps work internally? Watch this .

Approach 3

Use set instead of map . The set data structure is best for this use case as you don't need the mapped value (the true value) in approach 2.

 function findCommonElem(arr1, arr2) { const set = new Set(arr1); return arr2.filter(item => set.has(item)); } const arr1 = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]; const arr2 = [2, 4, 6, 8, 10, 12, 14, 16, 20]; console.log(findCommonElem(arr1, arr2)); // [2, 4, 6, 8, 10];

This uses relatively lesser space but the algorithmic complexity of TC and SC are still same.

• Time complexity = O(M + N)
• Space compexity = O(M) or O(N)

Thanks to Nick Parsons for pointing this out.

Answer 2

let's say that arr1.length is n and arr2.length is m. so the filter function run the lambda function you wrote for each item in arr1 . The lambda function checks if an item is in the arr2 and in the worst case where it didn't find it the function ran on all the array so m times. therefore arr1.filter(x => arr2.includes(x)) run at the worst case O(n*m).

as for the space complexity, the filter function creates a new array and in the worst case that array size is as big as the original so the space complexity is O(n)

Answer 3

As it correctly mentioned, big O here would be equal to O(n*m) , here is an explanation for this:

1. arr1.filter complexity is O(n)
2. arr2.includes complexity is O(n) as well

However for each iteration in .filter you execute each time arr2.includes, which leads to O(n) * O(n) = O(n * m)

You may increase performance if you replace for example arr2 with JS Set. JS Set.has complexity is O(1) , so using Set instead of arr2 should help you to achieve O(n) complexity.

I believe I cannot answer on sorting question, cause I haven't understood clearly what you mean.