Merge pull request #117 from BrianLusina/feat/rectangle-aread

feat(puzzles, math): rectangle area and max consecutive ones

Author: BrianLusina

DIRECTORY.md

@@ -704,6 +704,8 @@
     * [Test Pascals Triangle](https://github.com/BrianLusina/PythonSnips/blob/master/pymath/pascals_triangle/test_pascals_triangle.py)
   * Perfect Square
     * [Test Perfect Squares](https://github.com/BrianLusina/PythonSnips/blob/master/pymath/perfect_square/test_perfect_squares.py)
+  * Rectangle Area
+    * [Test Compute Area](https://github.com/BrianLusina/PythonSnips/blob/master/pymath/rectangle_area/test_compute_area.py)
   * Super Size
     * [Test Super Size](https://github.com/BrianLusina/PythonSnips/blob/master/pymath/super_size/test_super_size.py)
   * Triangles

puzzles/arrays/max_consecutive_ones/README.md

@@ -19,6 +19,16 @@ Explanation: [0,0,1,1,*1*,*1*,1,1,1,*1*,1,1,0,0,0,1,1,1,1]
 starred numbers were flipped from 0 to 1. The longest subarray is [1,1,1,1,1,1,1,1,1,1].
 ```
 
+## Max consecutive ones two
+
+You are given a binary array nums (an array that contains only 0s and 1s). Your task is to find the maximum number of 
+consecutive 1s in the array and return it.
+
+### Constraints
+
+- 1 ≤ `nums.length` ≤ 10^3
+- `nums[i` is either 0 or 1
+
 ## Related Topics
 
 - Array

puzzles/arrays/max_consecutive_ones/__init__.py

@@ -20,3 +20,36 @@ def longest_ones(nums: List[int], k: int) -> int:
             left += 1
 
     return right - left + 1
+
+def find_max_consecutive_ones(nums: List[int]) -> int:
+    """
+    Finds the maximum consecutive ones in a binary array and returns it.
+
+    The most straightforward way to solve this is to use a single pass through the array, keeping track of two values
+    as we go. First, we need a counter for the current streak of consecutive 1s we're seeing. Second, we need to
+    remember the maximum streak we've encountered so far.
+
+    As we examine each element, if we see a 1, we increment our current streak counter. If we see a 0, that breaks our
+    streak, so we reset the counter to 0. Importantly, every time we update our current streak, we check whether it's
+    larger than our maximum and update the maximum if needed.
+
+    Time complexity is O(n) where n is the length of the input array
+    Space complexity is O(1) as no extra space is required
+
+    Args:
+        nums(list): a list of 1s and 0s.
+    Returns:
+        int: maximum number of consecutive 1s in the nums binary array
+    """
+    if len(nums) == 0:
+        return 0
+    max_ones = 0
+    current_consecutive = 0
+
+    for num in nums:
+        if num == 1:
+            current_consecutive += 1
+            max_ones = max(max_ones, current_consecutive)
+        else:
+            current_consecutive = 0
+    return max_ones

puzzles/arrays/max_consecutive_ones/test_max_consecutive_ones.py

@@ -1,23 +1,29 @@
 import unittest
-
-from . import longest_ones
+from typing import List
+from parameterized import parameterized
+from puzzles.arrays.max_consecutive_ones import longest_ones, find_max_consecutive_ones
 
 
 class MaxConsecutiveOnesTestCase(unittest.TestCase):
-    def test_one(self):
-        """should return 6 from nums = [1,1,1,0,0,0,1,1,1,1,0], k = 2"""
-        nums = [1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0]
-        k = 2
-        expected = 6
+
+    @parameterized.expand([
+        ([1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0], 2, 6),
+        ([0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 1, 1], 3, 10),
+    ])
+    def test_longest_ones(self, nums: List[int], k: int, expected: int):
         actual = longest_ones(nums, k)
         self.assertEqual(expected, actual)
 
-    def test_two(self):
-        """should return 10 from nums = [0,0,1,1,0,0,1,1,1,0,1,1,0,0,0,1,1,1,1], k = 3"""
-        nums = [0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 1, 1]
-        k = 3
-        expected = 10
-        actual = longest_ones(nums, k)
+    @parameterized.expand([
+        ([1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0], 4),
+        ([0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 1, 1], 4),
+        ([1,1,0,0,1,1,1,0,0,1,0], 3),
+        ([1,1,0,0,1,1], 2),
+        ([1,1,1,0,1,1,1,1], 4),
+        ([0,0,0,0], 0),
+    ])
+    def test_find_max_consecutive_ones(self, nums: List[int], expected: int):
+        actual = find_max_consecutive_ones(nums)
         self.assertEqual(expected, actual)
 
 

pymath/rectangle_area/README.md

@@ -0,0 +1,24 @@
+# Rectangle Area
+
+You are given the coordinates of two axis-aligned rectangles in a 2D plane. Your task is to calculate the total area 
+covered by both rectangles.
+
+The first rectangle is specified by the coordinates of its bottom-left corner (ax1, ay1) and top-right corner (ay1, ay2).
+
+Similarly, the second rectangle is defined by its bottom-left corner (bx1, by1) and top-right corner (bx2, by2).
+
+> Note: The rectangles may overlap.
+
+## Constraints
+
+1. -10^4 ≤ ax1 ≤ ax2 ≤ 10^4
+2. −10^4 ≤ ay1 ≤ ay2 ≤ 10^4
+3. −10^4 ≤ bx1 ≤ bx2 ≤10^4
+4. −10^4 ≤ by1 ≤ by2 ≤ 10^4
+
+## Examples
+
+![Example 1](./images/examples/rectangle_area_example_1.png)
+![Example 2](./images/examples/rectangle_area_example_2.png)
+![Example 3](./images/examples/rectangle_area_example_3.png)
+![Example 4](./images/examples/rectangle_area_example_4.png)

pymath/rectangle_area/__init__.py

@@ -0,0 +1,33 @@
+def compute_area(
+    ax1: int, ay1: int, ax2: int, ay2: int, bx1: int, by1: int, bx2: int, by2: int
+):
+    # 1. Calculate the area of each individual rectangle
+    area_a = (ax2 - ax1) * (ay2 - ay1)
+    area_b = (bx2 - bx1) * (by2 - by1)
+
+    # 2. Calculate the area of the overlap (A intersect B)
+
+    # Determine the coordinates of the overlap rectangle: (ix1, iy1) to (ix2, iy2)
+
+    # The left edge of the overlap is the max of the two left edges
+    ix1 = max(ax1, bx1)
+    # The right edge of the overlap is the min of the two right edges
+    ix2 = min(ax2, bx2)
+
+    # The bottom edge of the overlap is the max of the two bottom edges
+    iy1 = max(ay1, by1)
+    # The top edge of the overlap is the min of the two top edges
+    iy2 = min(ay2, by2)
+
+    # Calculate the width and height of the overlap
+    overlap_width = max(0, ix2 - ix1)
+    overlap_height = max(0, iy2 - iy1)
+
+    # The max(0, ...) ensures that if the rectangles do not overlap
+    # (e.g., ix2 < ix1), the width/height is 0, and the overlap_area is 0.
+    overlap_area = overlap_width * overlap_height
+
+    # 3. Apply the Inclusion-Exclusion Principle
+    total_area = area_a + area_b - overlap_area
+
+    return total_area

pymath/rectangle_area/test_compute_area.py

@@ -0,0 +1,33 @@
+import unittest
+from parameterized import parameterized
+from pymath.rectangle_area import compute_area
+
+
+class RectangleAreaTestCase(unittest.TestCase):
+    @parameterized.expand(
+        [
+            (0, 0, 1, 1, 2, 2, 3, 3, 2),
+            (0, 0, 2, 2, 1, 1, 3, 3, 7),
+            (-1, -1, 2, 2, 0, 0, 1, 1, 9),
+            (0, 0, 0, 0, 1, 1, 2, 2, 1),
+            (-8918, -419, -7715, 577, -8918, -419, -7715, 577, 1198188),
+        ]
+    )
+    def test_compute_area(
+        self,
+        ax1: int,
+        ay1: int,
+        ax2: int,
+        ay2: int,
+        bx1: int,
+        by1: int,
+        bx2: int,
+        by2: int,
+        expected,
+    ):
+        actual = compute_area(ax1, ay1, ax2, ay2, bx1, by1, bx2, by2)
+        self.assertEqual(expected, actual)
+
+
+if __name__ == "__main__":
+    unittest.main()