703. Kth Largest Element in a Stream
1. Description
Design a class to find the kth largest element in a stream. Note that it is the kth largest element in the sorted order, not the kth distinct element.
Implement KthLargest class:
- KthLargest(int k, int[] nums) Initializes the object with the integer k and the stream of integers nums.
- int add(int val) Appends the integer val to the stream and returns the element representing the kth largest element in the stream.
2. Example
Example 1:
Input
[“KthLargest”, “add”, “add”, “add”, “add”, “add”]
[[3, [4, 5, 8, 2]], [3], [5], [10], [9], [4]]
Output
[null, 4, 5, 5, 8, 8]
Explanation
KthLargest kthLargest = new KthLargest(3, [4, 5, 8, 2]);
kthLargest.add(3); // return 4
kthLargest.add(5); // return 5
kthLargest.add(10); // return 5
kthLargest.add(9); // return 8
kthLargest.add(4); // return 8
3. Constraints
- 1 <= k <= $10^4$
- 0 <= nums.length <= $10^4$
- $-10^4$ <= nums[i] <= $10^4$
- $-10^4$ <= val <= $10^4$
- At most $10^4$ calls will be made to add.
- It is guaranteed that there will be at least k elements in the array when you search for the kth element.
4. Solutions
Heap(Priority Queue)
class KthLargest {
public:
KthLargest(int k, vector<int> &nums) {
// n = nums.size()
// Time complexity: O(nlogk)
// Space complexity: O(k)
k_ = k;
for (int i = 0; i < nums.size(); ++i) {
if (data_.size() < k) {
data_.push(nums[i]);
} else if (nums[i] > data_.top()) {
data_.pop();
data_.push(nums[i]);
}
}
}
int add(int val) {
// Time complexity: O(logk)
// Space complexity: O(1)
if (data_.size() < k_) {
data_.push(val);
return data_.top();
} else if (val < data_.top()) {
return data_.top();
} else {
data_.pop();
data_.push(val);
return data_.top();
}
}
private:
int k_;
priority_queue<int, vector<int>, greater<int>> data_;
};