# Check if All the Integers in a Range Are Covered

Published: Dec 12, 2022

Easy Array

## Problem 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]`.

Constraints:

• `1 <= ranges.length <= 50`
• `1 <= start[i] <= end[i] <= 50`
• `1 <= left <= right <= 50`

https://leetcode.com/problems/check-if-all-the-integers-in-a-range-are-covered/

## Examples

``````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.
``````

## How to Solve

The brute force solution works in this problem considering the constraints. Check every value between left and right inclusive whether the value is included in the range. If one of the values between left and right is not found in any range, return false. If two loops come to the end, all values are found. Return true.

## Solution

``````class CheckIfAllTheIntegersInARangeAreCovered {
public:
bool isCovered(vector<vector<int>>& ranges, int left, int right) {
for (auto i = left; i <= right; ++i) {
bool found = false;
for (auto &range : ranges) {
int s = range, e = range;
if (s <= i && i <= e) {
found = true;
break;
}
}
if (!found) { return false; }
}
return true;
}
};
``````
``````class CheckIfAllTheIntegersInARangeAreCovered {
public boolean isCovered(int[][] ranges, int left, int right) {
for (int i = left; i <= right; ++i) {
boolean found = false;
for (int[] range : ranges) {
int s = range, e = range;
if (s <= i && i <= e) {
found = true;
break;
}
}
if (!found) { return false; }
}
return true;
}
}
``````
``````/**
* @param {number[][]} ranges
* @param {number} left
* @param {number} right
* @return {boolean}
*/
var isCovered = function(ranges, left, right) {
for (let i = left; i <= right; i++) {
let found = false;
for (const [s, e] of ranges) {
if (s <= i && i <= e) {
found = true;
break;
}
}
if (!found) {
return false;
}
}
return true;
};
``````
``````class CheckIfAllTheIntegersInARangeAreCovered:
def isCovered(self, ranges: List[List[int]], left: int, right: int) -> bool:
for i in range(left, right + 1):
found = False
for s, e in ranges:
if s <= i <= e:
found = True
break
return False
return True
``````
``````# @param {Integer[][]} ranges
# @param {Integer} left
# @param {Integer} right
# @return {Boolean}
def is_covered(ranges, left, right)
(left..right).each do |i|
found = false
ranges.each do |s, e|
if s <= i && i <= e
found = true
break
end
end
if (!found)
return false
end
end
return true
end
``````

## Complexities

• Time: `O(n * m)` – n: right - left, m: number of ranges
• Space: `O(1)`