LeetCode 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 <= 10^5

• 1 <= queries.length <= 10^5

• queries[i].length == 2

• 1 <= lefti <= righti <= 10^7

• 1 <= queries[j] <= 10^7

Solution

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class Solution {
public:
    vector<int> minInterval(vector<vector<int>>& intervals, vector<int>& queries) {
        int m = queries.size();

        sort(intervals.begin(), intervals.end(),[](const vector<int>& lhs, const vector<int>& rhs){
            return lhs[1] - lhs[0] < rhs[1] - rhs[0];
        });
        
        set<pair<int, int>> myset;
        for (int i = 0; i < m; i++) {
            myset.insert({queries[i], i});
        }
        
        vector<int> ans(m, -1);
        for (const vector<int>& interval : intervals) {
            auto it = myset.lower_bound({interval[0], 0});
            
            while (it != myset.end() && it->first <= interval[1]) {
                ans[it->second] = interval[1] - interval[0] + 1;
                it = myset.erase(it);
            }
        }
        
        return ans;
    }
};