buckets Archives - Huahua's Tech Road https://zxi.mytechroad.com/blog/tag/buckets/ Thu, 30 Aug 2018 20:33:47 +0000 en hourly 1 https://wordpress.org/?v=6.0.8 https://zxi.mytechroad.com/blog/wp-content/uploads/2017/09/cropped-photo-32x32.jpg buckets Archives - Huahua's Tech Road https://zxi.mytechroad.com/blog/tag/buckets/ 32 32 花花酱 LeetCode 347. Top K Frequent Elements https://zxi.mytechroad.com/blog/hashtable/leetcode-347-top-k-frequent-elements/ https://zxi.mytechroad.com/blog/hashtable/leetcode-347-top-k-frequent-elements/#respond Sun, 22 Oct 2017 07:05:08 +0000 http://zxi.mytechroad.com/blog/?p=669 Problem: Given a non-empty array of integers, return the k most frequent elements. For example, Given [1,1,1,2,2,3] and k = 2, return [1,2]. Note: You…

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Problem:

Given a non-empty array of integers, return the k most frequent elements.

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

Note:
You may assume k is always valid, 1 ≤ k ≤ number of unique elements.
Your algorithm’s time complexity must be better than O(n log n), where n is the array’s size.



Idea:

 

Solution 2: Priority queue / max heap

Time complexity: O(n) + O(nlogk)

Space complexity: O(n)

C++

// Author: Huahuas
// Running time: 16 ms
class Solution {
public:
  vector<int> topKFrequent(vector<int>& nums, int k) {
    unordered_map<int, int> count;        
    for (const int num : nums)
      ++count[num];
    priority_queue<pair<int, int>> q;
    for (const auto& pair : count) {
      q.emplace(-pair.second, pair.first);
      if (q.size() > k) q.pop();
    }
    vector<int> ans;
    for (int i = 0; i < k; ++i) {
      ans.push_back(q.top().second);
      q.pop();
    }
    return ans;
  }
};

Java

// Author: Huahua
// Running time: 56 ms
class Solution {
  public List<Integer> topKFrequent(int[] nums, int k) {    
    Map<Integer, Integer> counts = new HashMap<>();    
    for (int num : nums)
      counts.put(num, counts.getOrDefault(num, 0) + 1);    
    
    PriorityQueue<int[]> queue = new PriorityQueue<>((x, y) -> x[0] - y[0]);
    for (Integer num : counts.keySet()) {
      queue.offer(new int[]{counts.get(num), num});
      if (queue.size() > k) queue.poll();
    }
        
    List<Integer> ans = new ArrayList<>();
    for (int i = 0; i < k; ++i)
      ans.add(queue.poll()[1]);
    return ans;    
  }
}

Python

"""
Author: Huahua
Running time: 56 ms
"""
class Solution:
  def topKFrequent(self, nums, k):
    counts = collections.Counter(nums)
    h = []
    for num in counts.keys():
      heapq.heappush(h, (counts[num], num))
      if len(h) > k: heapq.heappop(h)
    return [heapq.heappop(h)[1] for i in range(k)]

Solution 3: Bucket Sort

Time complexity: O(n)

Space complexity: O(n)

C++/hashmap

class Solution {
public:
    vector<int> topKFrequent(vector<int>& nums, int k) {
        unordered_map<int, int> count;
        int max_freq = 1;
        for (const int num : nums)
            max_freq = max(max_freq, ++count[num]);
        map<int, vector<int>> buckets;
        for (const auto& kv : count)
            buckets[kv.second].push_back(kv.first);
        vector<int> ans;
        for (int i = max_freq; i >= 1; --i) {
            auto it = buckets.find(i);
            if (it == buckets.end()) continue;
            ans.insert(ans.end(), it->second.begin(), it->second.end());
            if (ans.size() == k) return ans;
        }
        return ans;
    }
};

C++/array

class Solution {
public:
    vector<int> topKFrequent(vector<int>& nums, int k) {
        unordered_map<int, int> count;
        for (const int num : nums)
            ++count[num];
        vector<vector<int>> buckets(nums.size() + 1);
        for (const auto& kv : count)
            buckets[kv.second].push_back(kv.first);
        vector<int> ans;
        for (auto it = buckets.rbegin(); it != buckets.rend(); ++it) {
            const vector<int>& keys = *it;
            if (keys.empty()) continue;
            ans.insert(ans.begin(), keys.begin(), keys.end());
            if (ans.size() == k) return ans;
        }
        return ans;
    }
};

Java

// Author: Huahua
// Running time: 23 ms
class Solution {
  public List<Integer> topKFrequent(int[] nums, int k) {
    List[] buckets = new List[nums.length + 1];
    Map<Integer, Integer> counts = new HashMap<>();

    for (int num : nums)      
      counts.put(num, counts.getOrDefault(num, 0) + 1);    
        
    for (int num : counts.keySet()) {
      int count = counts.get(num);
      if (buckets[count] == null)
        buckets[count] = new ArrayList<Integer>();
      buckets[count].add(num);
    }
    
    List<Integer> ans = new ArrayList<>();
    for (int i = buckets.length - 1; i > 0 && ans.size() < k; --i) {
      if (buckets[i] != null) ans.addAll(buckets[i]);      
    }
    return ans;    
  }
}

Python

"""
Author: Huahua
Running time: 64 ms
"""
class Solution:
  def topKFrequent(self, nums, k):
    counts = collections.Counter(nums)
    buckets = [[] for _ in range(len(nums) + 1)]    
    for num in counts.keys():
      buckets[counts[num]].append(num)
    ans = []
    for i in range(len(nums), 0, -1):      
      ans += buckets[i]      
      if len(ans) == k: return ans
    return ans

Related Problems

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