## Minimum Interval to Include Each Query solution leetcode

Table of Contents

# SOLUTION

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You are given a 2D integer array `intervals`

, where `intervals[i] = [left`

describes the _{i}, right_{i}]`i`

interval starting at ^{th}`left`

and ending at _{i}`right`

_{i}**(inclusive)**. The **size** of an interval is defined as the number of integers it contains, or more formally `right`

._{i} - left_{i} + 1

You are also given an integer array `queries`

. The answer to the `j`

query is the ^{th}**size of the smallest interval** `i`

such that `left`

. If no such interval exists, the answer is _{i} <= queries[j] <= right_{i}`-1`

.

Return *an array containing the answers to the queries*.

**Example 1: Minimum Interval to Include Each Query solution leetcode**

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: Minimum Interval to Include Each Query solution leetcode**

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 <= left`

_{i}<= right_{i}<= 10^{7}`1 <= queries[j] <= 10`

^{7}

# SOLUTION

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