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 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 <= 10^5
1 <= queries.length <= 10^5
intervals[i].length == 2
1 <= lefti <= righti <= 10^7
1 <= queries[j] <= 10^7

Solution

class Solution {
    public int[] minInterval(int[][] intervals, int[] queries) {
        int numQuery = queries.length;

        int[][] queriesWithIndex = new int[numQuery][2];
        for (int i = 0; i < numQuery; i++) {
            queriesWithIndex[i] = new int[] {queries[i], i};
        }

        Arrays.sort(intervals, Comparator.comparingInt(a -> a[0]));
        Arrays.sort(queriesWithIndex, Comparator.comparingInt(a -> a[0]));

        PriorityQueue<int[]> minHeap =
                new PriorityQueue<>(Comparator.comparingInt(a -> (a[1] - a[0])));

        int[] result = new int[numQuery];

        int j = 0;
        for (int i = 0; i < queries.length; i++) {
            int queryVal = queriesWithIndex[i][0];
            int queryIndex = queriesWithIndex[i][1];

            while (j < intervals.length && intervals[j][0] <= queryVal) {
                minHeap.add(intervals[j]);
                j++;
            }

            while (!minHeap.isEmpty() && minHeap.peek()[1] < queryVal) {
                minHeap.remove();
            }
            result[queryIndex] =
                    minHeap.isEmpty() ? -1 : (minHeap.peek()[1] - minHeap.peek()[0] + 1);
        }

        return result;
    }
}

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).