Let's go through the example array [6, 4, 4, 6, 4] step by step.

Step 1: Create an empty Map called counts. This Map will store the elements from the array as keys and their corresponding counts as values. Set the majorityCount variable to Math.ceil(nums.length / 2), which is 3 in this case.

Step 2: The first element we encounter is 6. Since the element doesn't exist in the counts, we add it to the counts map with a count of 1. The counts map is now {6: 1}. Since the count 1 is less than 3, we continue to the next element.

Step 3: The second element is 4. Since the element doesn't exist in the counts, we add it to the counts map with a count of 1. The counts map is now {6: 1, 4: 1}. Since the count 1 is less than 3, we continue to the next element.

Step 4: The third element is 4 again. Since the element already exists in the counts, we update the count of 4 to 2 (increment by 1). The counts map is now {6: 1, 4: 2}. Since the count 2 is less than 3, continue to the next element.

Step 5: The fourth element is 6. Since the element already exists in the counts, we update the count of 6 to 2 (increment by 1). The counts map is now {6: 2, 4: 2}. Once again, since the count 2 is less than 3, we continue to the next element.

Step 6: The fifth element is 4 again. Since the element already exists in the counts, we update the count of 4 to 3 (increment by 1). The counts map is now {6: 2, 4: 3}. Since the count for 4, 3 is equal to our majorityCount, 3, we have found the majority element, which is 4.

Solution Code:

The optimal linear solution using a Map.

function findMajority(nums) {
  const counts = new Map();
  const majorityCount = Math.ceil(nums.length / 2);

  for (let num of nums) {
    if (counts.has(num)) {
      counts.set(num, counts.get(num) + 1);
    } else {
      counts.set(num, 1);
    }

    if (counts.get(num) >= majorityCount) {
      return num;
    }
  }
}