1851. Minimum Interval to Include Each Query

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1851. Minimum Interval to Include Each Query

Description

You are given a 2D integer array intervals, where intervals[i] = [lefti, righti] describes the i^th 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 j^th 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:

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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:

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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

Hints/Notes

Solution

Language: C++

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class Solution {
public:
    vector<int> minInterval(vector<vector<int>>& intervals, vector<int>& queries) {
        vector<int> qindex(queries.size());
        iota(qindex.begin(), qindex.end(), 0);
        auto cmp = [&](int i, int j) {
            return queries[i] < queries[j];
        };
        ranges::sort(qindex, cmp);
        ranges::sort(intervals);
        int i = 0, n = intervals.size();
        vector<int> res(queries.size());
        priority_queue<vector<int>, vector<vector<int>>, greater<vector<int>>> pq;
        for (int idx : qindex) {
            while (i < n && intervals[i][0] <= queries[idx]) {
                pq.push({intervals[i][1] - intervals[i][0] + 1, i});
                i++;
            }
            while (!pq.empty() && intervals[pq.top()[1]][1] < queries[idx]) {
                pq.pop();
            }
            res[idx] = pq.empty() ? -1 : pq.top()[0];
        }
        return res;
    }
};