# Kth Largest Element in a Stream

## 1: U-nderstand

Understand what the interviewer is asking for by using test cases and questions about the problem.

• Established a set (2-3) of test cases to verify their own solution later.
• Established a set (1-2) of edge cases to verify their solution handles complexities.
• Have fully understood the problem and have no clarifying questions.
• Have you verified any Time/Space Constraints for this problem?
• Could there be less than k elements in the array?
• No, it is guaranteed that there will be at least k elements in the array when you search for the kth element.
• What is the time and space complexity?
• Can you come up with an algorithm such that the add function has a O(logk) time complexity?
``````HAPPY CASE
Input
[[3, [4, 5, 8, 2]], , , , , ]
Output
[null, 4, 5, 5, 8, 8]

Explanation
KthLargest kthLargest = new KthLargest(3, [4, 5, 8, 2]);

Input
[[3,[4,5,8,2]],,,,,]
Output
[null,5,8,10,10,10]

Explanation
KthLargest kthLargest = new KthLargest(3, [4, 5, 8, 2]);

EDGE CASE (Multiple Spaces)
Input
KthLargest kthLargest = new KthLargest(1, [4, 5, 8, 2]);
Output
[null,10,15,15,15,15]

## 2: M-atch

Match what this problem looks like to known categories of problems, e.g. Linked List or Dynamic Programming, and strategies or patterns in those categories.

For Array/Strings, common solution patterns include:

• Sort
• Does sorting help us achieve what we need in order to solve the problem?
• Two pointer solutions (left and right pointer variables)
• Does Two pointers help us find the kth largest item
• Storing the elements of the array in a HashMap or a Set
• A hashset will keep around items, but we still need to sort to find the kth largest item
• Traversing the array with a sliding window
• Will viewing pieces of the input at a time help us?
• Heap
• Adding to a heap will take O(logk) time.

## 3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: Create a min-heap and limit it's size to k. The first element will be the kth largest element

``````1) Create heap
2) Limit heap to size k
4) Remove k + 1 largest val from heap
5) Return kth largest val in heap``````

⚠️ Common Mistakes

• Remember to use heap, we cannot get O(logk) otherwise

## 4: I-mplement

Implement the code to solve the algorithm.

``````class KthLargest:

def __init__(self, k: int, nums: List[int]):
# Create heap
self.heap = nums
heapq.heapify(self.heap)

# Limit heap to size k
while len(self.heap) > k:
heapq.heappop(self.heap)

def add(self, val: int) -> int:
heapq.heappush(self.heap, val)

# Remove k + 1 largest val from heap
heapq.heappop(self.heap)

# Return kth largest val in heap
return self.heap``````
``````class KthLargest {
// Create heap
private PriorityQueue<Integer> minHeap = new PriorityQueue<>();
private int k;

public KthLargest(int k, int[] nums) {
this.k = k;
for (int i: nums) {

// Limit heap to size k
if (minHeap.size() > k) {
minHeap.poll();
}
}
}

// Remove k + 1 largest val from heap
if (minHeap.size() > k) {
minHeap.poll();
}

// Return kth largest val in heap
return minHeap.peek();
}
}
/**
* Your KthLargest object will be instantiated and called as such:
* KthLargest obj = new KthLargest(k, nums);
*/``````

## 5: R-eview

Review the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.

• Trace through your code with an input to check for the expected output
• Catch possible edge cases and off-by-one errors

## 6: E-valuate

Evaluate the performance of your algorithm and state any strong/weak or future potential work.

Assume `N` represents the number of items in array and `K` represents the number of items in the heap.

• Time Complexity: O(logK), each time we add an item to the heap, it may take O(logK) time.
• Side note the init function take O(NlogN).
• O(N) to create the heap
• O(N-K*logN-K) to remove N-K items from the heap.
• If N and K are equal, then the init function take O(N) time. However, if K is 1, then the init function takes O(NlogN).
• Space Complexity: O(K), we need to store `K` items in the heap 