0700 - 0799
0703 - Kth Largest Element in a Stream (Easy)
Problem Link
https://leetcode.com/problems/kth-largest-element-in-a-stream/
Problem Statement
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 integerkand the stream of integersnums.int add(int val)Appends the integervalto the stream and returns the element representing thekthlargest element in the stream.
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 8Constraints:
1 <= k <= 10^40 <= nums.length <= 10^4-10^4 <= nums[i] <= 10^4-10^4 <= val <= 10^4- At most
10^4calls will be made toadd. - It is guaranteed that there will be at least
kelements in the array when you search for thekthelement.
Approach 1: Priority Queue
We can use priority queue to handle the sort order and only maintain at most k element. Return to the top, which is the k-th element in a stream.
class KthLargest {
public:
KthLargest(int k, vector<int>& nums) {
// add val to priority queue
for (auto x : nums) pq.push(x);
// here the queue is sorted
// only keep at most k elements
// pop out first pq.size() - k elements
while (pq.size() > k) pq.pop();
K = k;
}
int add(int val) {
// add val to priority queue
pq.push(val);
// here the queue is sorted
// only keep at most k elements
// pop out first pq.size() - k elements
while (pq.size() > K) pq.pop();
return pq.top();
}
private:
int K;
// smaller first
priority_queue<int, vector<int>, greater<int>> pq;
};
/**
* Your KthLargest object will be instantiated and called as such:
* KthLargest* obj = new KthLargest(k, nums);
* int param_1 = obj->add(val);
*/