LeetCode-in-All
347. Top K Frequent Elements
Medium
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 <= 105kis 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.
Solution
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;
class Solution {
public:
vector<int> topKFrequent(vector<int>& nums, int k) {
sort(nums.begin(), nums.end());
// Min heap of <number, frequency>
priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> minHeap;
// Filter with min heap
int j = 0;
for (int i = 0; i <= nums.size(); ++i) {
if (i == nums.size() || nums[i] != nums[j]) {
minHeap.push({i - j, nums[j]});
if (minHeap.size() > k) {
minHeap.pop();
}
j = i;
}
}
// Convert to int array
vector<int> result(k);
for (int i = k - 1; i >= 0; --i) {
result[i] = minHeap.top().second;
minHeap.pop();
}
return result;
}
};