class Solution:
    '''
    Time Complexity: O(n^2)
    Space Complexity: O(1)
    Where, n is number of rectangles.
    Brute-force approach to check all pairs of rectangles.
    For each pair, find the overlapping width and height.
    NOTE: Just trying to calculate overlapping width and height directly solves to check if there is any overlap or not, 
    you don't need to check that separately.
    PIT FALLS:
    You first mis-understood the question statement, thinking that if one rectangle had multiple overlaps with other rectangles, 
    then all those overlaps should be considered.
    But actually, only overlaps between any two given rectangles is to be considered. 
    Once you start solving you try to find if there is any intersection between 2 rectangles, 
    as it's process you first wrote down this condition `if (xbj < xti <= xtj and ybj < yti <= ytj) or (xbj <= xbi < xtj and ybj <= ybi < ytj):`.
    But when you started writing code for finding overlapping width and height, 
    you realized that you didn't need that condition at all, as the overlapping width and height calculations 
    would automatically take care of non-overlapping rectangles by resulting in zero width or height.
    So you removed that condition from the final code.
    '''
    def largestSquareArea(self, bottomLeft: List[List[int]], topRight: List[List[int]]) -> int:
        area = 0
        n = len(bottomLeft)
        for i in range(n):
            xbi, ybi = bottomLeft[i]
            xti, yti = topRight[i]
            for j in range(i+1, n):
                xbj, ybj = bottomLeft[j]
                xtj, ytj = topRight[j]

                x1 = max(xbi, xbj)
                x2 = min(xti, xtj)
                x = max(x2-x1, 0)
                y1 = max(ybi, ybj)
                y2 = min(yti, ytj)
                y = max(y2-y1, 0)
                s = min(x,y)

                area = max(area, s*s)                    
        return area