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Majority Element III

Given an array of integers and a number k, the majority number is the number that occurs more than 1/k of the size of the array, find all the majority elements.

Example

Given [3,1,2,3,2,3,3,4,4,4] and k=3, return [3].

Challenge

O(n) time and O(k) extra space

This is a generalized problem of Majority Number and Majority Number II. The key observation here is that we try to find k different numbers to hide. And each number is at most hide once. So even if there is a O(k) inner loop, the total operations is still O(N) not O(N * k).

    public List<Integer> majorityElements(int[] nums, int k) {
        Map<Integer, Integer> cnt = new HashMap<>();
        for(int i = 0; i < nums.length; i++) {
            cnt.put(nums[i], cnt.getOrDefault(nums[i], 0) + 1);
            if(cnt.size() == k) {
                for(int key : cnt.keySet()) {
                    cnt.put(key, cnt.get(key) - 1);
                    if(cnt.get(key) == 0) {
                        cnt.remove(key);
                    }
                }
            }
        }
        List<Integer> ans = new ArrayList<>();
        for(int key : cnt.keySet()) {
            cnt.put(key, 0);
        }
        for(int i = 0; i < nums.length; i++) {
            if(cnt.containsKey(nums[i])) {
                cnt.put(nums[i], cnt.get(nums[i]) + 1);
            }
        }
        for(int key : cnt.keySet()) {
            if(cnt.get(key) > nums.length / k) {
                ans.add(key);
            }
        }
        return ans;
    }