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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

Companies:
Bloomberg

Related Topics:
Greedy

Solution 1.

// OJ: https://leetcode.com/problems/check-if-all-the-integers-in-a-range-are-covered/
// Author: github.com/lzl124631x
// 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;
    }
};

Solution 2.

// OJ: https://leetcode.com/problems/check-if-all-the-integers-in-a-range-are-covered/
// Author: github.com/lzl124631x
// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
    bool isCovered(vector<vector<int>>& A, int left, int right) {
        sort(begin(A), end(A));
        if (A[0][0] > left) return false;
        int i = 0, j = A[0][0];
        while (i < A.size()) {
            if (j < A[i][0] - 1) return false;
            j = max(j, A[i][1]);
            ++i;
            if (j >= right) return true;
        }
        return false;
    }
};