How do you solve Minimum Rectangles to Cover Points on LeetCode?

Minimum Rectangles to Cover Points (LeetCode #3111) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(1) space complexity. Key insight: The y-values of the points do not affect the rectangle count, only the x-values matter.

What is the time complexity of Minimum Rectangles to Cover Points?

The optimal solution for Minimum Rectangles to Cover Points runs in O(n log n) time and uses O(1) space. The complexity is O(n log n) due to the sorting step, while the greedy selection of rectangles runs in O(n).

What is the brute force solution for Minimum Rectangles to Cover Points?

The brute force approach for Minimum Rectangles to Cover Points is Brute Force, with O(n²) time complexity and O(1) space complexity. We can try to cover each point with rectangles by checking every possible rectangle placement. This approach is straightforward but inefficient as it considers all combinations of points. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Minimum Rectangles to Cover Points?

Minimum Rectangles to Cover Points uses the following concepts: Array, Greedy, Sorting. The recommended patterns to study are: Greedy, Sorting.

What are the interview tips for Minimum Rectangles to Cover Points?

Always clarify the constraints and requirements of the problem before diving into coding. Think about edge cases, such as all points being on the same x-coordinate or having very small widths. Practice explaining your thought process as you code, as this demonstrates your understanding to the interviewer.

What are common mistakes when solving Minimum Rectangles to Cover Points?

Not sorting the points before processing, which can lead to incorrect rectangle placements. Failing to account for all points when determining the width of the rectangles.

#3111

Minimum Rectangles to Cover Points

Medium

ArrayGreedySortingGreedySorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n log n)Space O(1)

By sorting the points and using a greedy approach, we can minimize the number of rectangles needed by always extending the rectangle to cover as many points as possible within the width constraint.

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Algorithm

4 steps

1Step 1: Sort the points based on their x-coordinates.
2Step 2: Initialize a count for rectangles and start from the first point.
3Step 3: For each point, create a rectangle that extends to the right until the width exceeds w, covering all points in that range.
4Step 4: Move to the next uncovered point and repeat until all points are covered.

solution.py11 lines

1def minRectangles(points, w):
2    points.sort()
3    count = 0
4    i = 0
5    n = len(points)
6    while i < n:
7        count += 1
8        start = points[i][0]
9        while i < n and points[i][0] <= start + w:
10            i += 1
11    return count

ℹ

Complexity note: The complexity is O(n log n) due to the sorting step, while the greedy selection of rectangles runs in O(n).

1The y-values of the points do not affect the rectangle count, only the x-values matter.
2Sorting the points allows for a more systematic approach to covering them efficiently.

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