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

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

Level

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;
        }
    }
    
  • // 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;
        }
    };
    
  • # 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
    
    

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