Find kth largest element in interval list

Question

Given a list of intervals, each defined by a start and end point, write a function to find the kth largest element among the merged intervals. The function should take an array of intervals and an integer k as input and return the kth largest element in the merged list of intervals, where the merged list contains all unique points from the intervals. If k is greater than the number of unique points, return -1.

Codemia · Accepted Answer

To solve the problem of finding the kth largest element among merged intervals, we can recognize that the underlying pattern involves sorting and unique point extraction. The primary challenge is merging overlapping intervals and then identifying the unique points in the merged list. The key steps involve:
1. Merging overlapping intervals to avoid duplicates.
2. Extracting unique points from the merged intervals.
3. Sorting the unique points to efficiently retrieve the kth largest element.
This approach leverages sorting and list traversal, making it suitable given the constraints.

Here’s a step-by-step breakdown of the algorithm for this specific problem:
1. **Merge Intervals**: First, sort the list of intervals based on their start points. Then, iterate through the sorted intervals and merge any overlapping intervals. For example, given intervals `[[1, 3], [2, 4], [5, 7]]`, after sorting and merging, we would get `[[1, 4], [5, 7]]`.
2. **Extract Unique Points**: From the merged intervals, extract the unique points. The merged intervals `[[1, 4], [5, 7]]` yield unique points: `[1, 2, 3, 4, 5, 6, 7]`.
3. **Sort Unique Points**: Sort the unique points to facilitate the retrieval of the kth largest element. In this case, the sorted unique points remain `[1, 2, 3, 4, 5, 6, 7]`.
4. **Retrieve kth Largest**: Check if k is within the bounds of the list length. If valid, return the element at the index `-k`. If k exceeds the number of unique points, return -1.

Here’s the Python solution for this specific problem:
```python
def kth_largest_in_intervals(intervals, k):
    if k <= 0:
        return -1

# Step 1: Merge Intervals
    intervals.sort(key=lambda x: x[0])
    merged = []
    for interval in intervals:
        if not merged or merged[-1][1] < interval[0]:
            merged.append(interval)
        else:
            merged[-1][1] = max(merged[-1][1], interval[1])

# Step 2: Extract Unique Points
    unique_points = set()
    for start, end in merged:
        for point in range(start, end + 1):
            unique_points.add(point)

# Step 3: Sort Unique Points
    sorted_points = sorted(unique_points)

# Step 4: Retrieve kth Largest
    if k > len(sorted_points):
        return -1
    return sorted_points[-k]
```
This code accurately implements the outlined approach.

The time and space complexity for this solution is as follows:
- **Time Complexity**: O(n log n) due to the sorting step of the intervals, where n is the number of intervals. Merging intervals and collecting unique points takes O(n), but the sorting dominates.
- **Space Complexity**: O(m) where m is the number of unique points generated from the merged intervals. This is due to the storage of unique points in a set.

Specific edge cases to consider for this problem include:
1. **Empty Input**: If the intervals list is empty, the function should return -1 regardless of k.
2. **Non-overlapping Intervals**: If all intervals are separate, ensure they are merged correctly and that unique points are extracted accurately.
3. **k Greater Than Points**: If k exceeds the number of unique points after merging, the function should return -1.
4. **Single Interval**: If there's only one interval, the unique points should be correctly extracted based on that interval.

Find kth largest element in interval list

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