README_EN.md

1851. Minimum Interval to Include Each Query

中文文档

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

Python3

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

Java

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;
    }
}

C++

class Solution {
public:
    vector<int> minInterval(vector<vector<int>>& intervals, vector<int>& queries) {
        int n = intervals.size(), m = queries.size();
        sort(intervals.begin(), intervals.end());
        using pii = pair<int, int>;
        vector<pii> qs;
        for (int i = 0; i < m; ++i) {
            qs.emplace_back(queries[i], i);
        }
        sort(qs.begin(), qs.end());
        vector<int> ans(m, -1);
        priority_queue<pii, vector<pii>, greater<pii>> pq;
        int i = 0;
        for (auto& [x, j] : qs) {
            while (i < n && intervals[i][0] <= x) {
                int a = intervals[i][0], b = intervals[i][1];
                pq.emplace(b - a + 1, b);
                ++i;
            }
            while (!pq.empty() && pq.top().second < x) {
                pq.pop();
            }
            if (!pq.empty()) {
                ans[j] = pq.top().first;
            }
        }
        return ans;
    }
};

Go

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 interface{}) { *h = append(*h, v.(pair)) }
func (h *hp) Pop() interface{}   { a := *h; v := a[len(a)-1]; *h = a[:len(a)-1]; return v }

...