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# 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

[English Version in Youtube](https://youtu.be/Z7qiynScOPA)

[中文版解答Youtube Link](https://youtu.be/q5jHwTw139s)

[中文版解答Bilibili Link](https://www.bilibili.com/video/BV14h411m7GA/)

```
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;
    }
};
```


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