1851. Minimum Interval to Include Each Query

Description

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

 

Solutions

Solution: Priority Queue

• Time complexity: O(nlogn+mlogm+nlogm)
• Space complexity: O(m+n)

 

JavaScript

/**
 * @param {number[][]} intervals
 * @param {number[]} queries
 * @return {number[]}
 */
const minInterval = function (intervals, queries) {
  const m = intervals.length;
  const n = queries.length;
  const minHeap = new MinPriorityQueue(({ left, right }) => right - left + 1);
  const indexedQueries = queries.map((query, index) => ({ query, index }));
  const result = Array.from({ length: n }, () => -1);
  let intervalIndex = 0;

  intervals.sort((a, b) => a[0] - b[0]);
  indexedQueries.sort((a, b) => a.query - b.query);

  for (const { query, index } of indexedQueries) {
    while (intervalIndex < m && intervals[intervalIndex][0] <= query) {
      const [left, right] = intervals[intervalIndex];

      minHeap.enqueue({ left, right });
      intervalIndex += 1;
    }

    while (minHeap.size() && minHeap.front().right < query) {
      minHeap.dequeue();
    }

    if (!minHeap.size()) continue;
    const { left, right } = minHeap.front();

    result[index] = right - left + 1;
  }

  return result;
};