<p>You are given two positive integers <code>xCorner</code> and <code>yCorner</code>, and a 2D array <code>circles</code>, where <code>circles[i] = [x<sub>i</sub>, y<sub>i</sub>, r<sub>i</sub>]</code> denotes a circle with center at <code>(x<sub>i</sub>, y<sub>i</sub>)</code> and radius <code>r<sub>i</sub></code>.</p>

<p>There is a rectangle in the coordinate plane with its bottom left corner at the origin and top right corner at the coordinate <code>(xCorner, yCorner)</code>. You need to check whether there is a path from the bottom left corner to the top right corner such that the <strong>entire path</strong> lies inside the rectangle, <strong>does not</strong> touch or lie inside <strong>any</strong> circle, and touches the rectangle <strong>only</strong> at the two corners.</p>

<p>Return <code>true</code> if such a path exists, and <code>false</code> otherwise.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">xCorner = 3, yCorner = 4, circles = [[2,1,1]]</span></p>

<p><strong>Output:</strong> <span class="example-io">true</span></p>

<p><strong>Explanation:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2024/05/18/example2circle1.png" style="width: 346px; height: 264px;" /></p>

<p>The black curve shows a possible path between <code>(0, 0)</code> and <code>(3, 4)</code>.</p>
</div>

<p><strong class="example">Example 2:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">xCorner = 3, yCorner = 3, circles = [[1,1,2]]</span></p>

<p><strong>Output:</strong> <span class="example-io">false</span></p>

<p><strong>Explanation:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2024/05/18/example1circle.png" style="width: 346px; height: 264px;" /></p>

<p>No path exists from <code>(0, 0)</code> to <code>(3, 3)</code>.</p>
</div>

<p><strong class="example">Example 3:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">xCorner = 3, yCorner = 3, circles = [[2,1,1],[1,2,1]]</span></p>

<p><strong>Output:</strong> <span class="example-io">false</span></p>

<p><strong>Explanation:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2024/05/18/example0circle.png" style="width: 346px; height: 264px;" /></p>

<p>No path exists from <code>(0, 0)</code> to <code>(3, 3)</code>.</p>
</div>

<p><strong class="example">Example 4:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">xCorner = 4, yCorner = 4, circles = [[5,5,1]]</span></p>

<p><strong>Output:</strong> <span class="example-io">true</span></p>

<p><strong>Explanation:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2024/08/04/rectangles.png" style="width: 346px; height: 264px;" /></p>
</div>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>3 &lt;= xCorner, yCorner &lt;= 10<sup>9</sup></code></li>
	<li><code>1 &lt;= circles.length &lt;= 1000</code></li>
	<li><code>circles[i].length == 3</code></li>
	<li><code>1 &lt;= x<sub>i</sub>, y<sub>i</sub>, r<sub>i</sub> &lt;= 10<sup>9</sup></code></li>
</ul>

class DSU:

    def __init__(self, n):
        self._p = [-1] * n
    
    def find(self, x):
        if self._p[x] < 0:
            return x
        self._p[x] = self.find(self._p[x])
        return self._p[x]
    
    def union(self, x, y):
        px, py = self.find(x), self.find(y)
        if px == py:
            return
        if self._p[px] > self._p[py]:
            px, py = py, px
        self._p[px] += self._p[py]
        self._p[py] = px


class Solution:
    def canReachCorner(self, X: int, Y: int, circles: List[List[int]]) -> bool:
        n = len(circles)
        dsu = DSU(n + 2)
        for i, (x, y, r) in enumerate(circles):
            if (x <= r and y <= Y) or (y + r >= Y and x <= X):
                dsu.union(n, i)
            if (y <= r and x <= X) or (x + r >= X and y <= Y):
                dsu.union(n + 1, i)
            for j in range(i):
                if dsu.find(i) == dsu.find(j):
                    continue
                x2, y2, r2 = circles[j]
                if (x + x2) / 2 <= X and (y + y2) / 2 <= Y and (x - x2) ** 2 + (y - y2) ** 2 <= (r + r2) ** 2:
                    dsu.union(i, j)
        return dsu.find(n) != dsu.find(n + 1)

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