Find kth largest element in interval list

Question

Given a linked list, find the maximum path efficiently using BFS.

Codemia · Accepted Answer

To solve the problem of finding the kth largest element in an interval list represented as a linked list, we need to recognize that this is essentially a variant of the selection problem. The optimal approach here can be viewed through the lens of 'selection algorithms' which allow us to efficiently find the kth largest element by utilizing a max-heap (priority queue) data structure. This approach is particularly suited because it allows us to dynamically maintain the largest elements as we traverse the linked list. Given that the problem specifies using BFS, we will treat the linked list as a graph where each node points to the next, making a level-wise traversal possible.

1. **Initialization**: Create a max-heap (using Python's `heapq` which is a min-heap, we will invert values to simulate a max-heap). 
2. **Traversal**: Use a BFS approach to traverse the linked list. Since the linked list is a single path, we will iterate through it sequentially. 
3. **Heap Management**: For each element, we will push its value onto the max-heap. If the size of the heap exceeds k, we will pop the smallest element (which is effectively the k+1th largest element). 
4. **Final Extraction**: Once the entire linked list has been traversed, the root of the max-heap will be the kth largest element.

**Example**: For a linked list with values [3, 1, 4, 1, 5, 9], if k = 3, we would push 3, 1, 4, 1, 5, 9 into the heap, and after processing all elements, the max-heap would have the three largest elements, from which we can extract the 3rd largest.

```python
import heapq

class ListNode:
    def __init__(self, value=0, next=None):
        self.value = value
        self.next = next

def find_kth_largest(head: ListNode, k: int) -> int:
    max_heap = []
    current = head
    
    # Traverse the linked list
    while current:
        # Push the current node's value to the max-heap
        heapq.heappush(max_heap, current.value)
        # Maintain the size of the heap to be at most k
        if len(max_heap) > k:
            heapq.heappop(max_heap)
        current = current.next
    
    # The root of the heap is the kth largest element
    return max_heap[0]
```

The time complexity of this algorithm is O(n log k), where n is the number of elements in the linked list. This is because we process each element once (O(n)) and each insertion and deletion operation in the max-heap takes O(log k). The space complexity is O(k) due to the storage used by the heap to keep track of the k largest elements.

1. **k is larger than the number of elements**: If k is greater than the length of the linked list, we should return an error or a message indicating that k is invalid. 
2. **Empty linked list**: If the linked list is empty, we should also handle this case gracefully, potentially returning None or raising an exception. 
3. **All identical elements**: If all elements in the linked list are the same, the kth largest element should still be that same value regardless of k. 
4. **k is 1**: This should return the maximum element of the list.

Find kth largest element in interval list

Waymo

What the Interviewer Expects

Key Topics to Cover

How to Approach This

Possible Follow-up Questions

Sharpen Your Skills on Codemia

Sample Answer

Problem Analysis

Approach

Submit Your Answer

Waymo Software Engineer Interview Guide

Related Questions

Binary Search on graph

Count kth largest element in array

Find connected components in string

Merge Three Sorted strings

Binary Search on string