347. Top K Frequent Elements

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Q:

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]

Example 3:

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

Output: [1,2]

Constraints:

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

A:



#include <iostream>
#include <vector>
#include <unordered_map>
#include <queue>
using namespace std;

vector<int> topKFrequent(vector<int>& nums, int k) {
    // Step 1: Count the frequency of each element
    unordered_map<int, int> freq_map;
    for (int num : nums) {
        freq_map[num]++;
    }

    // Step 2: Use a min-heap to keep the k most frequent elements
    // Priority queue stores pairs: (-frequency, element), we negate the frequency
    // to simulate a max-heap using a min-heap.
    priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> min_heap;

    for (const auto& entry : freq_map) {
        min_heap.push({entry.second, entry.first});
        // If heap size exceeds k, pop the least frequent element
        if (min_heap.size() > k) {
            min_heap.pop();
        }
    }

    // Step 3: Extract the top k elements from the heap
    vector<int> result;
    while (!min_heap.empty()) {
        result.push_back(min_heap.top().second);
        min_heap.pop();
    }

    return result;
}