Formatted question description: https://leetcode.ca/all/1893.html

# 1893. Check if All the Integers in a Range Are Covered

## Level

Easy

## Description

You are given a 2D integer array `ranges`

and two integers `left`

and `right`

. Each `ranges[i] = [start_i, end_i]`

represents an **inclusive** interval between `start_i`

and `end_i`

.

Return `true`

*if each integer in the inclusive range [left, right] is covered by at least one interval in ranges*. Return

`false`

otherwise.An integer `x`

is covered by an interval `ranges[i] = [start_i, end_i]`

if `start_i <= x <= end_i`

.

**Example 1:**

**Input:** ranges = [[1,2],[3,4],[5,6]], left = 2, right = 5

**Output:** true

**Explanation:** Every integer between 2 and 5 is covered:

- 2 is covered by the first range.
- 3 and 4 are covered by the second range.
- 5 is covered by the third range.

**Example 2:**

**Input:** ranges = [[1,10],[10,20]], left = 21, right = 21

**Output:** false

**Explanation:** 21 is not covered by any range.

**Constraints:**

`1 <= ranges.length <= 50`

`1 <= start_i <= end_i <= 50`

`1 <= left <= right <= 50`

## Solution

Since the data range is limited, for each `range`

in `ranges`

, consider the intersection of `range`

and `[left, right]`

, and add all the numbers in the intersection into a hash set. Finally, check whether the hash set’s size is `right - left + 1`

.

```
class Solution {
public boolean isCovered(int[][] ranges, int left, int right) {
Set<Integer> set = new HashSet<Integer>();
for (int[] range : ranges) {
int start = Math.max(range[0], left);
int end = Math.min(range[1], right);
for (int i = start; i <= end; i++)
set.add(i);
}
return set.size() == right - left + 1;
}
}
```