Minimum Interval to Include Each Query

// Accepted solution for LeetCode #1851: Minimum Interval to Include Each Query
class Solution {
    public int[] minInterval(int[][] intervals, int[] queries) {
        int n = intervals.length, m = queries.length;
        Arrays.sort(intervals, (a, b) -> a[0] - b[0]);
        int[][] qs = new int[m][0];
        for (int i = 0; i < m; ++i) {
            qs[i] = new int[] {queries[i], i};
        }
        Arrays.sort(qs, (a, b) -> a[0] - b[0]);
        int[] ans = new int[m];
        Arrays.fill(ans, -1);
        PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> a[0] - b[0]);
        int i = 0;
        for (int[] q : qs) {
            while (i < n && intervals[i][0] <= q[0]) {
                int a = intervals[i][0], b = intervals[i][1];
                pq.offer(new int[] {b - a + 1, b});
                ++i;
            }
            while (!pq.isEmpty() && pq.peek()[1] < q[0]) {
                pq.poll();
            }
            if (!pq.isEmpty()) {
                ans[q[1]] = pq.peek()[0];
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1851: Minimum Interval to Include Each Query
func minInterval(intervals [][]int, queries []int) []int {
	n, m := len(intervals), len(queries)
	sort.Slice(intervals, func(i, j int) bool { return intervals[i][0] < intervals[j][0] })
	qs := make([][2]int, m)
	ans := make([]int, m)
	for i := range qs {
		qs[i] = [2]int{queries[i], i}
		ans[i] = -1
	}
	sort.Slice(qs, func(i, j int) bool { return qs[i][0] < qs[j][0] })
	pq := hp{}
	i := 0
	for _, q := range qs {
		x, j := q[0], q[1]
		for i < n && intervals[i][0] <= x {
			a, b := intervals[i][0], intervals[i][1]
			heap.Push(&pq, pair{b - a + 1, b})
			i++
		}
		for len(pq) > 0 && pq[0].r < x {
			heap.Pop(&pq)
		}
		if len(pq) > 0 {
			ans[j] = pq[0].v
		}
	}
	return ans
}

type pair struct{ v, r int }
type hp []pair

func (h hp) Len() int           { return len(h) }
func (h hp) Less(i, j int) bool { return h[i].v < h[j].v }
func (h hp) Swap(i, j int)      { h[i], h[j] = h[j], h[i] }
func (h *hp) Push(v any)        { *h = append(*h, v.(pair)) }
func (h *hp) Pop() any          { a := *h; v := a[len(a)-1]; *h = a[:len(a)-1]; return v }

# Accepted solution for LeetCode #1851: Minimum Interval to Include Each Query
class Solution:
    def minInterval(self, intervals: List[List[int]], queries: List[int]) -> List[int]:
        n, m = len(intervals), len(queries)
        intervals.sort()
        queries = sorted((x, i) for i, x in enumerate(queries))
        ans = [-1] * m
        pq = []
        i = 0
        for x, j in queries:
            while i < n and intervals[i][0] <= x:
                a, b = intervals[i]
                heappush(pq, (b - a + 1, b))
                i += 1
            while pq and pq[0][1] < x:
                heappop(pq)
            if pq:
                ans[j] = pq[0][0]
        return ans

// Accepted solution for LeetCode #1851: Minimum Interval to Include Each Query
/**
 * [1851] Minimum Interval to Include Each Query
 *
 * 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:
 *
 * 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
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/minimum-interval-to-include-each-query/
// discuss: https://leetcode.com/problems/minimum-interval-to-include-each-query/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn min_interval(intervals: Vec<Vec<i32>>, queries: Vec<i32>) -> Vec<i32> {
        vec![]
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    #[ignore]
    fn test_1851_example_1() {
        let intervals = vec![vec![1, 4], vec![2, 4], vec![3, 6], vec![4, 4]];
        let queries = vec![2, 3, 4, 5];

        let result = vec![3, 3, 1, 4];

        assert_eq!(Solution::min_interval(intervals, queries), result);
    }

    #[test]
    #[ignore]
    fn test_1851_example_2() {
        let intervals = vec![vec![2, 3], vec![2, 5], vec![1, 8], vec![20, 25]];
        let queries = vec![2, 19, 5, 22];

        let result = vec![2, -1, 4, 6];

        assert_eq!(Solution::min_interval(intervals, queries), result);
    }
}

// Accepted solution for LeetCode #1851: Minimum Interval to Include Each Query
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1851: Minimum Interval to Include Each Query
// class Solution {
//     public int[] minInterval(int[][] intervals, int[] queries) {
//         int n = intervals.length, m = queries.length;
//         Arrays.sort(intervals, (a, b) -> a[0] - b[0]);
//         int[][] qs = new int[m][0];
//         for (int i = 0; i < m; ++i) {
//             qs[i] = new int[] {queries[i], i};
//         }
//         Arrays.sort(qs, (a, b) -> a[0] - b[0]);
//         int[] ans = new int[m];
//         Arrays.fill(ans, -1);
//         PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> a[0] - b[0]);
//         int i = 0;
//         for (int[] q : qs) {
//             while (i < n && intervals[i][0] <= q[0]) {
//                 int a = intervals[i][0], b = intervals[i][1];
//                 pq.offer(new int[] {b - a + 1, b});
//                 ++i;
//             }
//             while (!pq.isEmpty() && pq.peek()[1] < q[0]) {
//                 pq.poll();
//             }
//             if (!pq.isEmpty()) {
//                 ans[q[1]] = pq.peek()[0];
//             }
//         }
//         return ans;
//     }
// }