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# 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<>();
}
}

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

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