Interview question: How many rectangles overlap each other in the most places python implementation

# encoding=utf-8
 '''
 Input description :
 The input consists of five lines.
The first line includes an integer n (2 <= n <= 50), which represents the number of rectangles.
The second line includes n integers x1[i] (-10^9 <= x1[i] <= 10^9), which represent the abscissa of the lower left corner.
The third line includes n integers y1[i] (-10^9 <= y1[i] <= 10^9), which represent the ordinate of the lower left corner.
The fourth line includes n integers x2[i] (-10^9 <= x2[i] <= 10^9), representing the abscissa of the upper right corner.
The fifth line includes n integers y2[i] (-10^9 <= y2[i] <= 10^9), which represent the ordinate of the upper right corner.
Output description :
 output a positive integer , indicating how many rectangles overlap each other in the most places , if the rectangles do not overlap each other , output 1 .
Example 1
input 2
 0 90
 0 90
 100 200
 100 200
 output 2
 '''
 import sys



lines = sys.stdin.readlines() #read   multiple lines at a time
 n = int (lines[ 0 ])
x1 = list ( map ( int , lines[ 1 ].split()))   # map batch processing
 y1 = list ( map ( int , lines[ 2 ].split()))
x2 = list(map(int, lines[3].split()))
y2 = list ( map ( int , lines[ 4 ].split()))
 print (x1 + x2)   # [0, 90, 100, 200]   list addition
 print (y1 + y2)   # [0, 90, 100, 200]
 print x1   # [0, 90]
 res = 1
 for x in x1+x2:   # Determine the maximum number of intersections between the lower left or upper right corner of each rectangle and every other rectangle
 for y in y1+y2:    
        cnt = 0   #For the lower left corner and upper right corner, judge independently, max takes the largest value
 for i in range (n):
             if x > x1[i] and y > y1[i] and x <= x2[i] and y <= y2[i]:   #If there is a common area, the abscissa must be greater than x1 , less than or equal to x2 , and the ordinate is greater than y1 , less than or equal to y2
 cnt += 1
 res = max (res , cnt)
 print (res)                                


# ctrl + d   to end the input
 # run->stop