1851. 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 <= 105

• 1 <= queries.length <= 105

• intervals[i].length == 2

• 1 <= lefti <= righti <= 107

• 1 <= queries[j] <= 107

Solution

/**
 * @param {number[][]} intervals
 * @param {number[]} queries
 * @return {number[]}
 */
var minInterval = function(intervals, queries) {
    var starts = intervals.sort((a, b) => a[0] - b[0]);
    var nums = queries.map((num, i) => [num, i]).sort((a, b) => a[0] - b[0]);
    var queue = new MinPriorityQueue();
    var res = Array(queries.length);
    var j = 0;
    for (var i = 0; i < nums.length; i++) {
        while (starts[j] && starts[j][0] <= nums[i][0]) queue.enqueue(starts[j], starts[j][1] - starts[j++][0] + 1);
        while (queue.size() && queue.front().element[1] < nums[i][0]) queue.dequeue();
        res[nums[i][1]] = queue.size() ? queue.front().priority : -1;
    }
    return res;
};

Explain:

nope.

Complexity:

• Time complexity : O(nlog(n) + mlog(m) + mlog(n)).
• Space complexity : O(n + m).