3781 Maximize The Distance Between Points On A Square
Maximize the Distance Between Points on a Square 
You are given an integer side, representing the edge length of a square with corners at (0, 0), (0, side), (side, 0), and (side, side) on a Cartesian plane.
You are also given a positive integer k and a 2D integer array points, where points[i] = [xi, yi] represents the coordinate of a point lying on the boundary of the square.
You need to select k elements among points such that the minimum Manhattan distance between any two points is maximized.
Return the maximum possible minimum Manhattan distance between the selected k points.
| The Manhattan Distance between two cells (xi, yi) and (xj, yj) is | xi - xj | + | yi - yj | . |
Example 1:
Input: side = 2, points = [[0,2],[2,0],[2,2],[0,0]], k = 4
Output: 2
Explanation:
Select all four points.
Example 2:
Input: side = 2, points = [[0,0],[1,2],[2,0],[2,2],[2,1]], k = 4
Output: 1
Explanation:
Select the points (0, 0), (2, 0), (2, 2), and (2, 1).
Example 3:
Input: side = 2, points = [[0,0],[0,1],[0,2],[1,2],[2,0],[2,2],[2,1]], k = 5
Output: 1
Explanation:
Select the points (0, 0), (0, 1), (0, 2), (1, 2), and (2, 2).
Constraints:
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1 <= side <= 109
4 <= points.length <= min(4 * side, 15 * 103)
points[i] == [xi, yi]
The input is generated such that:
points[i] lies on the boundary of the square.
All points[i] are **unique**.
4 <= k <= min(25, points.length)
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class Solution:
def maxDistance(self, side: int, points: List[List[int]], k: int) -> int:
arr = []
for x, y in points:
if x == 0:
arr.append(y)
elif y == side:
arr.append(side + x)
elif x == side:
arr.append(side * 3 - y)
else:
arr.append(side * 4 - x)
arr.sort()
def check(limit: int) -> bool:
perimeter = side * 4
for start in arr:
end = start + perimeter - limit
cur = start
for _ in range(k - 1):
idx = bisect_left(arr, cur + limit)
if idx == len(arr) or arr[idx] > end:
cur = -1
break
cur = arr[idx]
if cur >= 0:
return True
return False
lo, hi = 1, side
ans = 0
while lo <= hi:
mid = (lo + hi) // 2
if check(mid):
lo = mid + 1
ans = mid
else:
hi = mid - 1
return ans