Problem Statement

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.

Problem Link

Minimum Interval to Include Each Query - LeetCode

Code

class Solution {
public:
    vector<int> minInterval(vector<vector<int>>& intervals, vector<int>& queries) {
        vector<int> Q = queries; //creating temporary vector for queries
        
        sort(Q.begin(),Q.end()); //sorting in ascending order
        sort(intervals.begin(),intervals.end()); //sorting in ascending order
        
        unordered_map<int,int> m; //stores query q and its corresponding minimum interval
        priority_queue<pair<int,int>,vector<pair<int,int>>,greater<>> pq; ///minheap based on the length of the interval, alse stores its corresponding right value
        int i = 0;
        int n = intervals.size();
        
        for(int q:Q)
        {
            for(;i<n&&intervals[i][0]<=q;i++)
            {
                int len = intervals[i][1] - intervals[i][0] + 1;
                pq.push({len,intervals[i][1]}); //only push when left value of interval is <= q
            }
            while(pq.size()&&pq.top().second<q)
                pq.pop(); //pop when right value of interval is less than q
            
            //after this if our pq is not empty then it will contain the required min interval at its root(top)
            m[q] = pq.size() ? pq.top().first : -1; 
        }
        vector<int> res;
        for(int q:queries)
            res.push_back(m[q]);
        
        return res;
            
    }
};

Complexity

Time O(nlogn + qlogq) Space O(n+q)