LC1851 - Minimum Interval to Include Each Query

Problem

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

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.

Solution

Time complexity of this solution is $O(n\log n)$.

def minInterval(self, intervals: List[List[int]], queries: List[int]) -> List[int]:
	intervals.sort()
	minHeap = []
	res = {}
	ret = []
	i = 0
	
	for query in sorted(queries):	
		while i < len(intervals) and intervals[i][0] <= query:
			left, right = intervals[i]
			heapq.heappush(minHeap, (right - left + 1, right))
			i += 1

		while minHeap and minHeap[0][1] < query:
			heapq.heappop(minHeap)
		
		if minHeap:
			res[query] = minHeap[0][0]
		else:
			res[query] = -1
	
	for query in queries:	
		ret.append(res[query])
	
	return ret