<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 Solution:
    def canReachCorner(self, X: int, Y: int, circles: List[List[int]]) -> bool:
        N = len(circles)
        bottom_right = N
        top_left = N + 1

        # construct graph representing connections between each circle and the bottom_right and top_left corners
        neighbours = [[] for _ in range(N + 2)]
        for i in range(N):
            x, y, r = circles[i]
            if x + r >= X or y - r <= 0:
                neighbours[i].append(bottom_right)
                neighbours[bottom_right].append(i)
            if x - r <= 0 or y + r >= Y:
                neighbours[i].append(top_left)
                neighbours[top_left].append(i)
            for j in range(i + 1, N):
                xx, yy, rr = circles[j]
                if sqrt((xx - x)**2 + (yy - y)**2) <= r + rr:
                    neighbours[i].append(j)
                    neighbours[j].append(i)
        
        def search(i, dest, visited = [False] * (N + 2)):
            visited[i] = True
            if i == dest:
                return True
            for j in neighbours[i]:
                if not visited[j] and search(j, dest, visited):
                    return True
            return False
        
        # DFS from bottom_right to top_left: if succesful -> no path from bottom left to top right
        return not search(bottom_right, top_left)
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