# 1851 Minimum Interval to Include Each Query

You are given a 2D integer array `intervals`, where `intervals[i] = [lefti, righti]` describes the `ith` interval starting at `lefti` and ending at `righti` (inclusive). The size of an interval is defined as the number of integers it contains, or more formally `righti - lefti + 1`.

You are also given an integer array `queries`. The answer to the `jth` query is the size of the smallest interval `i` such that `lefti <= queries[j] <= righti`. If no such interval exists, the answer is `-1`.

Return an array containing the answers to the queries.

Example 1:

``````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.
``````

Example 2:

``````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.
``````

Constraints:

• `1 <= intervals.length <= 105`
• `1 <= queries.length <= 105`
• `intervals[i].length == 2`
• `1 <= lefti <= righti <= 107`
• `1 <= queries[j] <= 107`
 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 `````` ``````class Solution: def minInterval(self, intervals: List[List[int]], queries: List[int]) -> List[int]: intervals = sorted(intervals)[::-1] queries = sorted([(q, i) for i, q in enumerate(queries)]) heap = [] res = [-1] * len(queries) for q, qi in queries: while intervals and intervals[-1][0] <= q: i, j = intervals.pop() if j >= q: heapq.heappush(heap, (j - i + 1, j)) while heap and heap[0][1] < q: heapq.heappop(heap) if heap: res[qi] = heap[0][0] return res``````