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Problem solution · Python

Maximum Area Rectangle with Point Constraints I

Maximum Area Rectangle with Point Constraints I: a Python solution using segment tree or range structure. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Segment tree or range structure
Source
Kamyu LeetCode Solutions
Length
67 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Segment tree or range structure

For Maximum Area Rectangle with Point Constraints I, the implementation stores interval information in a range-query data structure so updates and queries avoid rescanning the full input.

  1. Choose the aggregate stored for each interval or prefix.
  2. Build or initialize the structure from the input.
  3. Apply updates and combine the affected nodes to answer each query.

Code notes

  • 67 lines of Python from the credited upstream file maximum-area-rectangle-with-point-constraints-i.py.
  • The implementation visibly relies on sequence storage, ordered lookup.
  • No explicit loop blocks detected.

Complexity

Count the build once, then multiply the logarithmic update or query path by the number of operations.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from Kamyu LeetCode Solutions by kamyu104 and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeMaximum Area Rectangle with Point Constraints I · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O(nlogn)# Space: O(n) # sort, fenwick tree, hash tableclass Solution(object):    def maxRectangleArea(self, points):        """        :type points: List[List[int]]        :rtype: int        """        class BIT(object):  # 0-indexed.            def __init__(self, n):                self.__bit = [0]*(n+1)  # Extra one for dummy node.             def add(self, i, val):                i += 1  # Extra one for dummy node.                while i < len(self.__bit):                    self.__bit[i] += val                    i += (i & -i)             def query(self, i):                i += 1  # Extra one for dummy node.                ret = 0                while i > 0:                    ret += self.__bit[i]                    i -= (i & -i)                return ret            points.sort()        y_to_idx = {y:idx for idx, y in enumerate(sorted(set(y for _, y in points)))}        bit = BIT(len(y_to_idx))        lookup = {}        result = -1        for i, (x, y) in enumerate(points):            y_idx = y_to_idx[y]            bit.add(y_idx, +1)            if not (i-1 >= 0 and points[i-1][0] == x):                continue            prev_y_idx = y_to_idx[points[i-1][1]]            curr = bit.query(y_idx)-bit.query(prev_y_idx-1)            if (prev_y_idx, y_idx) in lookup and lookup[prev_y_idx, y_idx][0] == curr-2:                result = max(result, (x-lookup[prev_y_idx, y_idx][1])*(y-points[i-1][1]))            lookup[prev_y_idx, y_idx] = (curr, x)        return result  # Time:  O(n^2)# Space: O(1)# sort, brute forceclass Solution2(object):    def maxRectangleArea(self, points):        """        :type points: List[List[int]]        :rtype: int        """        result = -1        points.sort()        for i in xrange(len(points)-3):            if points[i][0] != points[i+1][0]:                continue            j = next((j for j in xrange(i+2, len(points)-1) if points[i][1] <= points[j][1] <= points[i+1][1]), len(points)-1)            if j == len(points)-1 or not (points[j][0] == points[j+1][0] and points[i][1] == points[j][1] and points[i+1][1] == points[j+1][1]):                continue            result = max(result, (points[i+1][1]-points[i][1])*(points[j][0]-points[i][0]))        return result

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