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

All queries are given at one time, so we can solve this problem offline.

1.  Sort input intervals by their start value;

2.  Sort input queries in increasing order;

3.  Create a min heap that keeps an interval's length and end value. The order is determined by interval length so we can answer a query efficiently.

4.  for each query Q, add all intervals whose start values <= Q, then remove all intervals whose end values < Q. The top of the min heap is the answer.

This solution is correct because for two different queries Q1 < Q2,  we process Q1 before Q2. All intervals with end values < Q1 must be also < Q2, so before processing Q2, we have removed all intervals whose end values < Q1. When processing Q2, we remove all intervals with end values in range [Q1, Q2 - 1].

class Solution {
    public int[] minInterval(int[][] intervals, int[] queries) {
        int[][] q = new int[queries.length][2];
        for(int i = 0; i < q.length; i++) {
            q[i][0] = queries[i];
            q[i][1] = i;
        }
        Arrays.sort(q, Comparator.comparingInt(e->e[0]));
        Arrays.sort(intervals, Comparator.comparingInt(e->e[0]));
        PriorityQueue<int[]> pq = new PriorityQueue<>(Comparator.comparingInt(e->e[0]));
        int[] ans = new int[queries.length];     
        
        int i = 0;
        for(int[] v : q) {
            while(i < intervals.length && intervals[i][0] <= v[0]) {
                pq.add(new int[]{intervals[i][1] - intervals[i][0] + 1, intervals[i][1]});
                i++;
            }
            while(pq.size() > 0 && pq.peek()[1] < v[0]) {
                pq.poll();
            }
            ans[v[1]] = pq.size() == 0 ? -1 : pq.peek()[0];
        }
        return ans;
    }
}

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来源： https://www.cnblogs.com/lz87/p/16074650.html