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

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

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;
}
}

############

class Solution {
public boolean isCovered(int[][] ranges, int left, int right) {
int[] diff = new int[52];
for (int[] range : ranges) {
int l = range[0], r = range[1];
++diff[l];
--diff[r + 1];
}
int cur = 0;
for (int i = 0; i < diff.length; ++i) {
cur += diff[i];
if (i >= left && i <= right && cur == 0) {
return false;
}
}
return true;
}
}

• // OJ: https://leetcode.com/problems/check-if-all-the-integers-in-a-range-are-covered/
// Time: O(N)
// Space: O(N)
class Solution {
public:
bool isCovered(vector<vector<int>>& A, int left, int right) {
unordered_set<int> s;
for (auto &v : A) {
for (int i = max(left,v[0]); i <= min(right,v[1]); ++i) {
s.insert(i);
}
}
return s.size() == right - left + 1;
}
};

• class Solution:
def isCovered(self, ranges: List[List[int]], left: int, right: int) -> bool:
diff = [0] * 52
for l, r in ranges:
diff[l] += 1
diff[r + 1] -= 1
cur = 0
for i, df in enumerate(diff):
cur += df
if left <= i <= right and cur == 0:
return False
return True

############

# 1893. Check if All the Integers in a Range Are Covered
# https://leetcode.com/problems/check-if-all-the-integers-in-a-range-are-covered/

class Solution:
def isCovered(self, ranges: List[List[int]], left: int, right: int) -> bool:
A = [0] * 52

for start,end in ranges:
A[start] += 1
A[end + 1] -= 1

for i in range(1, 52):
A[i] += A[i - 1]

for i in range(left, right + 1):
if A[i] <= 0: return False

return True

• func isCovered(ranges [][]int, left int, right int) bool {
diff := [52]int{}
for _, rg := range ranges {
l, r := rg[0], rg[1]
diff[l]++
diff[r+1]--
}
cur := 0
for i, x := range diff {
cur += x
if i >= left && i <= right && cur <= 0 {
return false
}
}
return true
}

• function isCovered(ranges: number[][], left: number, right: number): boolean {
const diff = new Array(52).fill(0);
for (const [l, r] of ranges) {
++diff[l];
--diff[r + 1];
}
let cur = 0;
for (let i = 0; i < 52; ++i) {
cur += diff[i];
if (i >= left && i <= right && cur <= 0) {
return false;
}
}
return true;
}

• /**
* @param {number[][]} ranges
* @param {number} left
* @param {number} right
* @return {boolean}
*/
var isCovered = function (ranges, left, right) {
const diff = new Array(52).fill(0);
for (const [l, r] of ranges) {
++diff[l];
--diff[r + 1];
}
let cur = 0;
for (let i = 0; i < 52; ++i) {
cur += diff[i];
if (i >= left && i <= right && cur <= 0) {
return false;
}
}
return true;
};