0300 - 0399
0347 - Top K Frequent Elements (Medium)
Problem Link
https://leetcode.com/problems/top-k-frequent-elements/
Problem Statement
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 <= 10^5kis 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.
Approach 1: Priority Queue
When you need to solve something started with "Top K ...", priority queue may possibly be the solution.
We use priority queue to sort the input by their frequency and pop the top element when the queue size is greater than . What's left in the queue would be the top frequent elements and we just list them out in a reverse order.
class Solution {
public:
vector<int> topKFrequent(vector<int>& nums, int k) {
int n = nums.size();
if(n == k) return nums;
map<int, int> m;
// store the frequency of each number
for(auto x : nums) m[x]++;
// define a custom comparator
auto cmp = [&](int x, int y) {
return m[x] > m[y];
};
// init priority queue with a custom comparator
priority_queue<int, vector<int>, decltype(cmp)> pq(cmp);
// iterate each number
for(auto x : m) {
// add this number
pq.push(x.first);
// out of queue size, pop the one with min frequency out
if(pq.size() > k) {
pq.pop();
}
}
// construct ans
vector<int> ans(k);
for(int i = k - 1; i >= 0; i--) {
ans[i] = pq.top();
pq.pop();
}
return ans;
}
};