How do you solve Check if All the Integers in a Range Are Covered on LeetCode?

Check if All the Integers in a Range Are Covered (LeetCode #1893) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a boolean array allows for efficient tracking of coverage.

What is the time complexity of Check if All the Integers in a Range Are Covered?

The optimal solution for Check if All the Integers in a Range Are Covered runs in O(n) time and uses O(n) space. The time complexity is O(n) because we iterate through the ranges and then through the boolean array, which is efficient given the constraints.

What is the brute force solution for Check if All the Integers in a Range Are Covered?

The brute force approach for Check if All the Integers in a Range Are Covered is Brute Force, with O(n²) time complexity and O(1) space complexity. We will check each integer in the range [left, right] to see if it is covered by any of the given intervals. This method is straightforward but can be inefficient for larger ranges. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Check if All the Integers in a Range Are Covered?

Check if All the Integers in a Range Are Covered uses the following concepts: Array, Hash Table, Prefix Sum. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Check if All the Integers in a Range Are Covered?

Clarify the range boundaries before starting. Discuss your thought process while coding. Consider edge cases, such as when left equals right.

What are common mistakes when solving Check if All the Integers in a Range Are Covered?

Not checking all integers in the range [left, right]. Assuming a range covers all integers without verifying.

#1893

Check if All the Integers in a Range Are Covered

Easy

ArrayHash TablePrefix SumHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Instead of checking each integer individually, we can create a boolean array to mark covered integers in the range. This allows us to efficiently track coverage in a single pass.

⚙️

Algorithm

4 steps

1Step 1: Create a boolean array of size 51 (to cover integers from 1 to 50).
2Step 2: For each range in ranges, mark the corresponding indices in the boolean array as true.
3Step 3: Check each integer from left to right and see if it is marked as true in the boolean array. If any integer is false, return false.
4Step 4: If all integers are true, return true.

solution.py9 lines

1def isCovered(ranges, left, right):
2    covered = [False] * 51
3    for start, end in ranges:
4        for i in range(start, end + 1):
5            covered[i] = True
6    for x in range(left, right + 1):
7        if not covered[x]:
8            return False
9    return True

ℹ

Complexity note: The time complexity is O(n) because we iterate through the ranges and then through the boolean array, which is efficient given the constraints.

1Using a boolean array allows for efficient tracking of coverage.
2Iterating through ranges can be done in a single pass to mark coverage.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.