1893. Check if All the Integers in a Range Are Covered
Description
You are given a 2D integer array ranges and two integers left and right. Each ranges[i] = [starti, endi] represents an inclusive interval between starti and endi.
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] = [starti, endi] if starti <= x <= endi.
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 <= 501 <= starti <= endi <= 501 <= left <= right <= 50
Solutions
Solution 1: Difference Array
We can use the idea of a difference array to create a difference array $\textit{diff}$ of length $52$.
Next, we iterate through the array $\textit{ranges}$. For each interval $[l, r]$, we increment $\textit{diff}[l]$ by $1$ and decrement $\textit{diff}[r + 1]$ by $1$.
Then, we iterate through the difference array $\textit{diff}$, maintaining a prefix sum $s$. For each position $i$, we increment $s$ by $\textit{diff}[i]$. If $s \le 0$ and $left \le i \le right$, it indicates that an integer $i$ within the interval $[left, right]$ is not covered, and we return $\textit{false}$.
If we finish iterating through the difference array $\textit{diff}$ without returning $\textit{false}$, it means that every integer within the interval $[left, right]$ is covered by at least one interval in $\textit{ranges}$, and we return $\textit{true}$.
The time complexity is $O(n + M)$, and the space complexity is $O(M)$. Here, $n$ is the length of the array $\textit{ranges}$, and $M$ is the maximum value of the interval, which in this case is $M \le 50$.
Python3
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
s = 0
for i, x in enumerate(diff):
s += x
if s <= 0 and left <= i <= right:
return False
return True
Java
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 s = 0;
for (int i = 0; i < diff.length; ++i) {
s += diff[i];
if (s <= 0 && left <= i && i <= right) {
return false;
}
}
return true;
}
}
C++
class Solution {
public:
bool isCovered(vector<vector<int>>& ranges, int left, int right) {
vector<int> diff(52);
for (auto& range : ranges) {
int l = range[0], r = range[1];
++diff[l];
--diff[r + 1];
}
int s = 0;
for (int i = 0; i < diff.size(); ++i) {
s += diff[i];
if (s <= 0 && left <= i && i <= right) {
return false;
}
}
return true;
}
};
Go
func isCovered(ranges [][]int, left int, right int) bool {
diff := [52]int{}
for _, e := range ranges {
l, r := e[0], e[1]
diff[l]++
diff[r+1]--
}
s := 0
for i, x := range diff {
s += x
if s <= 0 && left <= i && i <= right {
return false
}
}
return true
}
TypeScript
function isCovered(ranges: number[][], left: number, right: number): boolean {
const diff: number[] = Array(52).fill(0);
for (const [l, r] of ranges) {
++diff[l];
--diff[r + 1];
}
let s = 0;
for (let i = 0; i < diff.length; ++i) {
s += diff[i];
if (s <= 0 && left <= i && i <= right) {
return false;
}
}
return true;
}
JavaScript
/**
* @param {number[][]} ranges
* @param {number} left
* @param {number} right
* @return {boolean}
*/
var isCovered = function (ranges, left, right) {
const diff = Array(52).fill(0);
for (const [l, r] of ranges) {
++diff[l];
--diff[r + 1];
}
let s = 0;
for (let i = 0; i < diff.length; ++i) {
s += diff[i];
if (s <= 0 && left <= i && i <= right) {
return false;
}
}
return true;
};