feat(algorithms, intervals): remove covered intervals

Author: BrianLusina

algorithms/intervals/remove_intervals/README.md

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+# Remove Covered Intervals
+
+Given an array of intervals, where each interval is represented as intervals[i]=[li,ri) (indicating the range from 
+li to ri, inclusive of li and exclusive of ri), remove all intervals that are completely covered by another interval in 
+the list. Return the count of intervals that remain after removing the covered ones.
+
+> Note An interval [a, b) is considered covered by another interval [c,d) if and only if c ⇐ a and b ⇐ d.
+
+## Constraints
+
+- 1 <= intervals.length <= 10^4
+- intervals[i].length == 2
+- 0 <= li < ri <= 10^5
+- All the give intervals are unique
+
+## Examples
+
+![Example 1](./images/examples/remove_covered_intervals_example_1.png)
+![Example 2](./images/examples/remove_covered_intervals_example_2.png)
+![Example 3](./images/examples/remove_covered_intervals_example_3.png)
+![Example 4](./images/examples/remove_covered_intervals_example_4.png)
+![Example 5](./images/examples/remove_covered_intervals_example_5.png)
+
+
+## Solution
+
+The first step is to simplify the process by sorting the intervals. Sorting by the start point in ascending order is 
+straightforward and simplifies the iteration process. However, an important edge case arises when two intervals share 
+the same start point. In such scenarios, sorting solely by the start point would fail to correctly identify covered 
+intervals. To handle this, we sort intervals with the same start point by their endpoint in descending order, ensuring 
+that longer intervals come first. This sorting strategy guarantees that if one interval covers another, it will be 
+positioned earlier in the sorted list. Once the intervals are sorted, we iterate through them while keeping track of the 
+maximum endpoint seen so far. If the current interval’s end point exceeds this maximum, it is not covered, so we increment 
+the count and update the maximum end. The interval is covered and skipped if the endpoint is less than or equal to the 
+maximum. After completing the iteration, the final count reflects the remaining non-covered intervals.
+
+Now, let’s look at the solution steps below:
+
+1. If the start points are the same, sort the intervals by the start point in ascending order, otherwise, sort by the 
+   endpoint in descending order to prioritize longer intervals. 
+2. Initialize the count with zero to track the remaining (non-covered) intervals. 
+3. Initialize prev_end with zero to track the maximum end value we’ve seen.. 
+4. Start iterating through intervals for each interval [start, end] in the sorted list:
+   - If end > prev_end, any previous interval does not cover the interval. 
+     - Increment count by 1 
+     - Update prev_end to end. 
+   - Else:
+     - A previous interval covers the interval, so we skip it.
+
+5. After iterating through all the intervals, the return count is the final value, representing the remaining intervals.
+
+Let’s look at the following illustration to get a better understanding of the solution:
+
+![Solution 1](./images/solutions/remove_covered_intervals_solution_1.png)
+![Solution 2](./images/solutions/remove_covered_intervals_solution_2.png)
+![Solution 3](./images/solutions/remove_covered_intervals_solution_3.png)
+![Solution 4](./images/solutions/remove_covered_intervals_solution_4.png)
+![Solution 5](./images/solutions/remove_covered_intervals_solution_5.png)
+![Solution 6](./images/solutions/remove_covered_intervals_solution_6.png)
+![Solution 7](./images/solutions/remove_covered_intervals_solution_7.png)
+
+### Time Complexity
+
+The time complexity of the solution is O(n logn), where n is the number of intervals. This is because sorting the 
+intervals takes O(n logn) time, and the subsequent iteration through the intervals takes O(n) time. Therefore, the 
+overall time complexity is dominated by the sorting step, resulting in O(n logn).
+
+### Space Complexity
+
+The sorting operation has a space complexity of O(n) in the worst case due to additional memory required for temporary 
+arrays during the sorting process. Apart from the space used by the built-in sorting algorithm, the algorithm’s space 
+complexity is constant, O(1). Therefore, the overall space complexity of the solution is O(n).

algorithms/intervals/remove_intervals/__init__.py

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+from typing import List
+
+
+def remove_covered_intervals(intervals: List[List[int]]) -> int:
+    """
+    Finds the number of intervals that are not covered by any other interval in the list.
+
+    This uses a greedy approach to find the number of intervals that are not covered by any other interval in the list.
+    Note An interval [a, b) is considered covered by another interval [c,d) if and only if c ⇐ a and b ⇐ d.
+
+    So, the timeline will look something like this:
+    c---a---b---d
+
+    If b <= d, then [a, b) is covered by [c, d)
+
+    Args:
+        intervals (List[List[int]]): A list of intervals, where each interval is represented as [start, end]
+    Returns:
+        int: The number of intervals that are not covered by any other interval in the list
+    """
+    # early return if there are no intervals to begin with
+    if not intervals:
+        return 0
+
+    # Sort intervals by start time in ascending order and then by end time in descending order. We sort by the end time
+    # to remove a tie-breaker where two intervals have the same start time.
+    # This will incur a time complexity of O(nlogn). We sort in place, so, we do not
+    # incur any additional space complexity by copying over to a new list. The assumption made here is that it is okay
+    # to mutate the input list.
+    intervals.sort(key=lambda x: (x[0], -x[1]))
+
+    # keep track of the last max end seen so far, we use a large negative infinity to cover all possible numbers
+    max_end_seen = float('-inf')
+    count = 0
+
+    # We then iterate through the given intervals
+    for _, current_end in intervals:
+        if current_end > max_end_seen:
+            count += 1
+            max_end_seen = current_end
+
+    # return the count of non-overlapping intervals
+    return count