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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 <= 50
  • 1 <= starti <= endi <= 50
  • 1 <= 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 $M \le 50$.

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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
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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;
    }
}
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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;
    }
};
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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
}
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