Minimum Interval to Include Each Query

lcpy gen -n 1851   # by problem number
lcpy gen -s minimum_interval_to_include_each_query   # by problem name

Input: intervals = [[1,4],[2,4],[3,6],[4,4]], queries = [2,3,4,5]
Output: [3,3,1,4]
Explanation: The queries are processed as follows:
- Query = 2: The interval [2,4] is the smallest interval containing 2. The answer is 4 - 2 + 1 = 3.
- Query = 3: The interval [2,4] is the smallest interval containing 3. The answer is 4 - 2 + 1 = 3.
- Query = 4: The interval [4,4] is the smallest interval containing 4. The answer is 4 - 4 + 1 = 1.
- Query = 5: The interval [3,6] is the smallest interval containing 5. The answer is 6 - 3 + 1 = 4.

Input: intervals = [[2,3],[2,5],[1,8],[20,25]], queries = [2,19,5,22]
Output: [2,-1,4,6]
Explanation: The queries are processed as follows:
- Query = 2: The interval [2,3] is the smallest interval containing 2. The answer is 3 - 2 + 1 = 2.
- Query = 19: None of the intervals contain 19. The answer is -1.
- Query = 5: The interval [2,5] is the smallest interval containing 5. The answer is 5 - 2 + 1 = 4.
- Query = 22: The interval [20,25] is the smallest interval containing 22. The answer is 25 - 20 + 1 = 6.

import heapq


class Solution:
    # Time: O((n + q) log n) - n intervals, q queries
    # Space: O(n)
    def min_interval(self, intervals: list[list[int]], queries: list[int]) -> list[int]:
        intervals.sort()
        sorted_queries = sorted(range(len(queries)), key=lambda i: queries[i])

        result = [-1] * len(queries)
        min_heap: list[tuple[int, int]] = []  # (size, right)
        interval_idx = 0

        for query_idx in sorted_queries:
            query_val = queries[query_idx]

            # Add all intervals that start at or before this query value.
            while interval_idx < len(intervals) and intervals[interval_idx][0] <= query_val:
                left, right = intervals[interval_idx]
                heapq.heappush(min_heap, (right - left + 1, right))
                interval_idx += 1

            # Remove intervals that ended before this query value.
            while min_heap and min_heap[0][1] < query_val:
                heapq.heappop(min_heap)

            if min_heap:
                result[query_idx] = min_heap[0][0]

        return result