This is an O(n^2) solution because it would require passing through each number, and then at each number scanning for a count.

A better way to do this is to take advantage of the property of a majority element that it consists of more than half of the set's elements.

What do I mean by this? Let's take another example, [4, 2, 2, 3, 2].

2 is obviously the majority element in the example from a visual scan, but how can we tip the algorithm off to this? Well, suppose we sorted the array:

[2, 2, 2, 3, 4]

Notice that if we hone in on index 2, that we see another 2. This is expected-- if we sort an array of numbers, the majority element will represent at least 50% of the elements, so it will always end up at the Math.floor(nums.length / 2)th index.

Depending on the sorting algorithm, this technique is usually O(n * log n) and requires either O(1) space if the sort is in-place, or O(n) space if not.

Another Method

Yet another way to solve this is to use the speed of a hash map! Hash maps allow access/look-ups in constant time, and its keys are guaranteed to be unique.

Thus, we can iterate through [2, 2, 2, 3, 4], and store each unique value as a key, and its attached value being the count. Each time we pass it, we increment its count, and use that information to relay we've got a current majority candidate.

1function majorityElement(nums) {
2  const hash = {};
3  // we keep a local max variable to track what
4  // element is in the majority
5  let max = 0,
6    val;
7
8  // iterate through
9  for (let i = 0; i < nums.length; i++) {
10    // increment or initialize the num as a key
11    if (hash[nums[i]]) {
12      hash[nums[i]]++;
13    } else {
14      hash[nums[i]] = 1;
15    }
16
17    if (hash[nums[i]] > max) {
18      max = hash[nums[i]];
19      val = nums[i];
20    }
21  }
22
23  return val;
24}

The time complexity of this solution is O(n) as it requires just one pass through the array, but has a space complexity of O(n) since it needs to maintain an auxiliary hash table.