Top K Frequent Elements LeetCode Solution

Problem – Top K Frequent Elements

Given an integer array nums and an integer k, return the k most frequent elements. You may return the answer in any order.

Example 1:

Input: nums = [1,1,1,2,2,3], k = 2
Output: [1,2]

Example 2:

Input: nums = [1], k = 1
Output: [1]

Constraints:

  • 1 <= nums.length <= 105
  • -104 <= nums[i] <= 104
  • k is in the range [1, the number of unique elements in the array].
  • It is guaranteed that the answer is unique.

Follow up: Your algorithm’s time complexity must be better than O(n log n), where n is the array’s size.

Top K Frequent Elements LeetCode Solution in Java

public List<Integer> topKFrequent(int[] nums, int k) {

	List<Integer>[] bucket = new List[nums.length + 1];
	Map<Integer, Integer> frequencyMap = new HashMap<Integer, Integer>();

	for (int n : nums) {
		frequencyMap.put(n, frequencyMap.getOrDefault(n, 0) + 1);
	}

	for (int key : frequencyMap.keySet()) {
		int frequency = frequencyMap.get(key);
		if (bucket[frequency] == null) {
			bucket[frequency] = new ArrayList<>();
		}
		bucket[frequency].add(key);
	}

	List<Integer> res = new ArrayList<>();

	for (int pos = bucket.length - 1; pos >= 0 && res.size() < k; pos--) {
		if (bucket[pos] != null) {
			res.addAll(bucket[pos]);
		}
	}
	return res;
}

Top K Frequent Elements LeetCode Solution in C++

class Solution {
public:
    vector<int> topKFrequent(vector<int>& nums, int k) {
        unordered_map<int,int> map;
        for(int num : nums){
            map[num]++;
        }
        
        vector<int> res;
        // pair<first, second>: first is frequency,  second is number
        priority_queue<pair<int,int>> pq; 
        for(auto it = map.begin(); it != map.end(); it++){
            pq.push(make_pair(it->second, it->first));
            if(pq.size() > (int)map.size() - k){
                res.push_back(pq.top().second);
                pq.pop();
            }
        }
        return res;
    }
};

Top K Frequent Elements LeetCode Solution in Python

class Solution(object):
    def topKFrequent(self, nums, k):
        hs = {}
        frq = {}
        for i in xrange(0, len(nums)):
            if nums[i] not in hs:
                hs[nums[i]] = 1
            else:
                hs[nums[i]] += 1

        for z,v in hs.iteritems():
            if v not in frq:
                frq[v] = [z]
            else:
                frq[v].append(z)
        
        arr = []
        
        for x in xrange(len(nums), 0, -1):
            if x in frq:
                
                for i in frq[x]:
                    arr.append(i)

        return [arr[x] for x in xrange(0, k)]
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