Problem Description

You have an n x n grid where the origin is at the bottom-left corner. You're given a list of non-overlapping rectangles on this grid, where each rectangle is defined by its bottom-left corner (start_x, start_y) and top-right corner (end_x, end_y).

Your task is to determine if you can make exactly two cuts on the grid (either both horizontal or both vertical) that will divide the grid into three sections such that:

1. Each section contains at least one rectangle
2. Each rectangle belongs to exactly one section (no rectangle is split by the cuts)

The solution uses an interval merging approach. For each rectangle, it tracks the start and end coordinates along both x and y axes. By sorting these coordinates and using a sweep line technique with an overlap counter, it determines how many distinct "segments" or groups of rectangles exist along each axis.

The key insight is that if rectangles can be separated into at least 3 groups along either the x-axis (vertical cuts) or y-axis (horizontal cuts), then two cuts can successfully divide them into three sections. The countLineIntersections function counts these groups by tracking when the overlap becomes zero - each time this happens, it indicates a gap between rectangle groups where a cut could be placed.

The algorithm returns true if either the horizontal direction (y-coordinates) or vertical direction (x-coordinates) has at least 3 distinct groups, meaning two cuts in that direction would create the required three sections.

Intuition

The key realization is that we need to find two parallel cuts (either both horizontal or both vertical) that separate all rectangles into three groups without cutting through any rectangle.

Think about what happens when we project all rectangles onto a single axis. If we project onto the x-axis, we get intervals [x1, x2] for each rectangle. Similarly, projecting onto the y-axis gives us intervals [y1, y2]. The question becomes: can we find two points along one of these axes that separate the intervals into three non-overlapping groups?

If rectangles don't overlap in their projections on an axis, there are natural "gaps" between groups of rectangles. These gaps are the only places where we can make cuts. For example, if we have rectangles at x-coordinates [1,3], [2,4], [6,8], and [10,12], the first two rectangles overlap (forming one group), while the third and fourth rectangles are separate. This gives us three groups with two gaps where cuts can be placed.

The clever part is recognizing that we can detect these groups using a sweep line algorithm with an overlap counter. As we sweep across the axis:

• When we encounter a rectangle start, we increment the overlap counter
• When we encounter a rectangle end, we decrement the overlap counter
• When the counter reaches zero, it means we've completely passed through a group of overlapping rectangles

Each time the counter returns to zero represents the end of one connected component of rectangles. If we have at least 3 such components along either axis, we can place our two cuts in the gaps between them.

This is why the solution tracks start/end markers (coordinate, 1) for starts and (coordinate, 0) for ends, sorts them, and counts how many times the overlap drops to zero. Finding 3+ groups on either axis means we can make valid cuts in that direction.

Learn more about Sorting patterns.

Solution Approach

The solution implements a sweep line algorithm with interval merging to detect separable groups of rectangles along each axis.

Step 1: Extract and Mark Coordinates

For each rectangle [x1, y1, x2, y2], we extract:

• Y-coordinates: (y1, 1) for start and (y2, 0) for end
• X-coordinates: (x1, 1) for start and (x2, 0) for end

The marker 1 indicates the beginning of an interval, and 0 indicates the end. This allows us to track when we enter and leave rectangle boundaries.

Step 2: Sort Coordinates

Both coordinate lists are sorted using sort(key=lambda x: (x[0], x[1])). This ensures:

• Primary sorting by coordinate value
• When coordinates are equal, ends (0) come before starts (1)

The second criterion is crucial: when one rectangle ends exactly where another begins (touching but not overlapping), we process the end first, allowing the overlap counter to reach zero and correctly identify them as separate groups.

Step 3: Count Line Intersections (Groups)

The countLineIntersections method processes the sorted coordinates:

lines = 0
overlap = 0
for value, marker in coordinates:
    if marker == 0:  # End of rectangle
        overlap -= 1
    else:  # Start of rectangle (marker == 1)
        overlap += 1
  
    if overlap == 0:
        lines += 1

The overlap counter tracks how many rectangles we're currently "inside":

• When we hit a start marker, we increment overlap
• When we hit an end marker, we decrement overlap
• When overlap reaches 0, we've completely exited a group of connected rectangles, so we increment lines

Step 4: Check Both Directions

The solution returns true if either direction has at least 3 groups:

return self.countLineIntersections(y_coordinates) or self.countLineIntersections(x_coordinates)

If we find 3+ groups along the y-axis, we can make two horizontal cuts between them. If we find 3+ groups along the x-axis, we can make two vertical cuts between them. Either condition satisfies the problem requirements.

Time Complexity: O(n log n) where n is the number of rectangles, due to sorting. Space Complexity: O(n) for storing the coordinate lists.

Example Walkthrough

Let's walk through a concrete example with 4 rectangles on a 10×10 grid:

• Rectangle A: [1, 1, 3, 4] (bottom-left at (1,1), top-right at (3,4))
• Rectangle B: [2, 2, 4, 5] (overlaps with A)
• Rectangle C: [6, 1, 8, 3] (separated from A and B)
• Rectangle D: [7, 6, 9, 8] (separated from all others)

Step 1: Extract Coordinates

For Y-coordinates (horizontal cuts):

• Rectangle A: (1, 1) start, (4, 0) end
• Rectangle B: (2, 1) start, (5, 0) end
• Rectangle C: (1, 1) start, (3, 0) end
• Rectangle D: (6, 1) start, (8, 0) end

Y-coordinates list: [(1,1), (4,0), (2,1), (5,0), (1,1), (3,0), (6,1), (8,0)]

For X-coordinates (vertical cuts):

• Rectangle A: (1, 1) start, (3, 0) end
• Rectangle B: (2, 1) start, (4, 0) end
• Rectangle C: (6, 1) start, (8, 0) end
• Rectangle D: (7, 1) start, (9, 0) end

X-coordinates list: [(1,1), (3,0), (2,1), (4,0), (6,1), (8,0), (7,1), (9,0)]

Step 2: Sort Coordinates

Y-coordinates sorted: [(1,1), (1,1), (2,1), (3,0), (4,0), (5,0), (6,1), (8,0)]

X-coordinates sorted: [(1,1), (2,1), (3,0), (4,0), (6,1), (7,1), (8,0), (9,0)]

Step 3: Count Groups for Y-coordinates

Processing sorted Y-coordinates:

• (1,1): overlap = 1, lines = 0
• (1,1): overlap = 2, lines = 0 (A and C both start at y=1)
• (2,1): overlap = 3, lines = 0 (B starts)
• (3,0): overlap = 2, lines = 0 (C ends)
• (4,0): overlap = 1, lines = 0 (A ends)
• (5,0): overlap = 0, lines = 1 ✓ (B ends - first group complete!)
• (6,1): overlap = 1, lines = 1 (D starts)
• (8,0): overlap = 0, lines = 2 ✓ (D ends - second group complete!)

Result: 2 groups along Y-axis (not enough for 3 sections)

Step 4: Count Groups for X-coordinates

Processing sorted X-coordinates:

• (1,1): overlap = 1, lines = 0 (A starts)
• (2,1): overlap = 2, lines = 0 (B starts, overlaps with A)
• (3,0): overlap = 1, lines = 0 (A ends)
• (4,0): overlap = 0, lines = 1 ✓ (B ends - first group complete!)
• (6,1): overlap = 1, lines = 1 (C starts)
• (7,1): overlap = 2, lines = 1 (D starts, overlaps with C)
• (8,0): overlap = 1, lines = 1 (C ends)
• (9,0): overlap = 0, lines = 2 ✓ (D ends - second group complete!)

Result: 2 groups along X-axis (not enough for 3 sections)

Final Result: Since neither axis has 3+ groups, return false. We cannot make two cuts to create three sections.

Alternative Example for Success: If Rectangle D was at [1, 6, 3, 8] instead, the Y-coordinates would yield 3 groups:

• Group 1: Rectangles A, B, C (y: 1-5)
• Group 2: Rectangle D (y: 6-8)
• Wait, that's still only 2 groups...

Let me correct with a better example. If we had rectangles at:

• A: [1, 1, 2, 2]
• B: [1, 4, 2, 5]
• C: [1, 7, 2, 8]

The Y-coordinate analysis would show 3 distinct groups with gaps at y=3 and y=6, allowing two horizontal cuts to create three sections. The function would return true.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n log n) where n is the number of rectangles.

• Creating coordinate lists: We iterate through all rectangles once to create y_coordinates and x_coordinates lists. Each rectangle contributes 2 points to each list, resulting in O(n) time.
• Sorting: Both y_coordinates and x_coordinates lists contain 2n elements each. Sorting these lists takes O(2n log(2n)) = O(n log n) time for each list.
• Counting line intersections: The countLineIntersections method iterates through 2n coordinates once, taking O(n) time. This method is called twice (once for each dimension).
• Overall: O(n) + O(n log n) + O(n log n) + O(n) + O(n) = O(n log n)

Space Complexity: O(n) where n is the number of rectangles.

• The y_coordinates list stores 2n tuples (start and end points for each rectangle's y-coordinates): O(n) space.
• The x_coordinates list stores 2n tuples (start and end points for each rectangle's x-coordinates): O(n) space.
• The sorting operations may use additional O(n) space depending on the implementation (Python's Timsort).
• Other variables (lines, overlap, loop variables) use O(1) space.
• Overall: O(n) + O(n) + O(n) = O(n)

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Sorting Order for Equal Coordinates

The Pitfall: A critical mistake is sorting the events incorrectly when coordinates are equal. If you sort so that starts (1) come before ends (0) at the same coordinate, the algorithm will fail to detect separate groups when rectangles are touching but not overlapping.

Example of the bug:

# WRONG: This will fail for touching rectangles
coordinates.sort(key=lambda x: x[0])  # Only sorts by position
# or
coordinates.sort(key=lambda x: (x[0], -x[1]))  # Starts before ends

Consider rectangles [0,0,2,2] and [2,0,4,2] that touch at x=2:

• With incorrect sorting: [(0,1), (2,1), (2,0), (4,0)]
• The overlap counter goes: 0→1→2→1→0 (never reaches 0 in the middle)
• Result: Counted as 1 group instead of 2

The Solution:

# CORRECT: Ends must be processed before starts at the same position
coordinates.sort(key=lambda x: (x[0], x[1]))

This ensures:

• Correct sorting: [(0,1), (2,0), (2,1), (4,0)]
• The overlap counter goes: 0→1→0→1→0 (reaches 0 at x=2)
• Result: Correctly counted as 2 groups

2. Counting Segments vs. Counting Gaps

The Pitfall: Confusing whether to count the number of segments (groups of rectangles) or the number of gaps between segments. The algorithm counts segments when overlap == 0, but developers might mistakenly try to count gaps.

Example of the bug:

# WRONG: Trying to count gaps instead of segments
def countLineIntersections(self, coordinates):
    gaps = 0
    overlap = 0
    was_zero = True
  
    for position, marker in coordinates:
        if marker == 0:
            overlap -= 1
        else:
            if was_zero:  # Thinking this is a new gap
                gaps += 1
            overlap += 1
            was_zero = False
      
        if overlap == 0:
            was_zero = True
  
    return gaps >= 2  # Wrong logic

The Solution: Count segments (groups) directly when overlap reaches zero. Remember: 3 segments means 2 possible cut positions between them.

3. Off-by-One Error in Segment Counting

The Pitfall: Incorrectly checking for >= 2 segments instead of >= 3, forgetting that two cuts create three sections.

Example of the bug:

# WRONG: Checking for 2 segments instead of 3
return segment_count >= 2  # This would only allow for 1 cut!

The Solution:

# CORRECT: Need at least 3 segments for 2 cuts
return segment_count >= 3

Remember the relationship: n cuts create n+1 sections, so 2 cuts require 3 segments.