Method majorityElement

java/0229-majority-element-ii.java:7–24

First solution utilizes a hashmap and then does the due diligience of adding the appropriate values that appear more than n/3 times Runtime O(n) : Space O(n)

(int[] nums)

     * Runtime O(n) : Space O(n)
     */
    public List<Integer> majorityElement(int[] nums) {
        List<Integer> res = new ArrayList<>();
        Map<Integer, Integer> map = new HashMap<>();
        for (int i = 0; i < nums.length; i++) {
            if (map.containsKey(nums[i])) 
                map.put(nums[i], map.get(nums[i]) + 1);
            else 
                map.put(nums[i], 1);
        }

        for (Map.Entry<Integer, Integer> entry: map.entrySet()) {
            int potentialCandidate = entry.getValue();
            if (potentialCandidate > nums.length / 3)
                res.add(entry.getKey());
        }

        return res;
    }


    /**