README.md

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 <= 105
• 1 <= queries.length <= 105
• intervals[i].length == 2
• 1 <= lefti <= righti <= 107
• 1 <= queries[j] <= 107

Solutions (Rust)

1. Solution

use std::cmp::Reverse;
use std::collections::BinaryHeap;

impl Solution {
    pub fn min_interval(mut intervals: Vec<Vec<i32>>, queries: Vec<i32>) -> Vec<i32> {
        let mut queries = (0..queries.len())
            .map(|i| (queries[i], i))
            .collect::<Vec<_>>();
        let mut i = 0;
        let mut heap = BinaryHeap::new();
        let mut ret = vec![-1; queries.len()];

        intervals.sort_unstable();
        queries.sort_unstable();

        for &(query, j) in &queries {
            while i < intervals.len() && intervals[i][0] <= query {
                heap.push(Reverse((
                    intervals[i][1] - intervals[i][0] + 1,
                    intervals[i][1],
                )));
                i += 1;
            }

            while let Some(&Reverse((size, right))) = heap.peek() {
                if right < query {
                    heap.pop();
                } else {
                    ret[j] = size;
                    break;
                }
            }
        }

        ret
    }
}