LeetCode #3235 — HARD

Check if the Rectangle Corner Is Reachable

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given two positive integers xCorner and yCorner, and a 2D array circles, where circles[i] = [x_i, y_i, r_i] denotes a circle with center at (x_i, y_i) and radius r_i.

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

Return true if such a path exists, and false otherwise.

Example 1:

Input: xCorner = 3, yCorner = 4, circles = [[2,1,1]]

Output: true

Explanation:

The black curve shows a possible path between (0, 0) and (3, 4).

Example 2:

Input: xCorner = 3, yCorner = 3, circles = [[1,1,2]]

Output: false

Explanation:

No path exists from (0, 0) to (3, 3).

Example 3:

Input: xCorner = 3, yCorner = 3, circles = [[2,1,1],[1,2,1]]

Output: false

Explanation:

No path exists from (0, 0) to (3, 3).

Example 4:

Input: xCorner = 4, yCorner = 4, circles = [[5,5,1]]

Output: true

Explanation:

Constraints:

3 <= xCorner, yCorner <= 10⁹
1 <= circles.length <= 1000
circles[i].length == 3
1 <= x_i, y_i, r_i <= 10⁹

Step 01

Brute Force Baseline

Problem summary: You are given two positive integers xCorner and yCorner, and a 2D array circles, where circles[i] = [xi, yi, ri] denotes a circle with center at (xi, yi) and radius ri. There is a rectangle in the coordinate plane with its bottom left corner at the origin and top right corner at the coordinate (xCorner, yCorner). You need to check whether there is a path from the bottom left corner to the top right corner such that the entire path lies inside the rectangle, does not touch or lie inside any circle, and touches the rectangle only at the two corners. Return true if such a path exists, and false otherwise.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Math · Union-Find

Example 1

3
4
[[2,1,1]]

Example 2

3
3
[[1,1,2]]

Example 3

3
3
[[2,1,1],[1,2,1]]

Core Insight

What unlocks the optimal approach

Create a graph with <code>n + 4</code> vertices.
Vertices 0 to <code>n - 1</code> represent the circles, vertex <code>n</code> represents upper edge, vertex <code>n + 1</code> represents right edge, vertex <code>n + 2</code> represents lower edge, and vertex <code>n + 3</code> represents left edge.
Add an edge between these vertices if they intersect or touch.
Answer will be <code>false</code> when any of two sides left-right, left-bottom, right-top or top-bottom are reachable using the edges.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3235: Check if the Rectangle Corner Is Reachable
class Solution {
    private int[][] circles;
    private int xCorner, yCorner;
    private boolean[] vis;

    public boolean canReachCorner(int xCorner, int yCorner, int[][] circles) {
        int n = circles.length;
        this.circles = circles;
        this.xCorner = xCorner;
        this.yCorner = yCorner;
        vis = new boolean[n];
        for (int i = 0; i < n; ++i) {
            var c = circles[i];
            int x = c[0], y = c[1], r = c[2];
            if (inCircle(0, 0, x, y, r) || inCircle(xCorner, yCorner, x, y, r)) {
                return false;
            }
            if (!vis[i] && crossLeftTop(x, y, r) && dfs(i)) {
                return false;
            }
        }
        return true;
    }

    private boolean inCircle(long x, long y, long cx, long cy, long r) {
        return (x - cx) * (x - cx) + (y - cy) * (y - cy) <= r * r;
    }

    private boolean crossLeftTop(long cx, long cy, long r) {
        boolean a = Math.abs(cx) <= r && (cy >= 0 && cy <= yCorner);
        boolean b = Math.abs(cy - yCorner) <= r && (cx >= 0 && cx <= xCorner);
        return a || b;
    }

    private boolean crossRightBottom(long cx, long cy, long r) {
        boolean a = Math.abs(cx - xCorner) <= r && (cy >= 0 && cy <= yCorner);
        boolean b = Math.abs(cy) <= r && (cx >= 0 && cx <= xCorner);
        return a || b;
    }

    private boolean dfs(int i) {
        var c = circles[i];
        long x1 = c[0], y1 = c[1], r1 = c[2];
        if (crossRightBottom(x1, y1, r1)) {
            return true;
        }
        vis[i] = true;
        for (int j = 0; j < circles.length; ++j) {
            var c2 = circles[j];
            long x2 = c2[0], y2 = c2[1], r2 = c2[2];
            if (vis[j]) {
                continue;
            }
            if ((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2) > (r1 + r2) * (r1 + r2)) {
                continue;
            }
            if (x1 * r2 + x2 * r1 < (r1 + r2) * xCorner && y1 * r2 + y2 * r1 < (r1 + r2) * yCorner
                && dfs(j)) {
                return true;
            }
        }
        return false;
    }
}

// Accepted solution for LeetCode #3235: Check if the Rectangle Corner Is Reachable
func canReachCorner(xCorner int, yCorner int, circles [][]int) bool {
	inCircle := func(x, y, cx, cy, r int) bool {
		dx, dy := x-cx, y-cy
		return dx*dx+dy*dy <= r*r
	}

	crossLeftTop := func(cx, cy, r int) bool {
		a := abs(cx) <= r && cy >= 0 && cy <= yCorner
		b := abs(cy-yCorner) <= r && cx >= 0 && cx <= xCorner
		return a || b
	}

	crossRightBottom := func(cx, cy, r int) bool {
		a := abs(cx-xCorner) <= r && cy >= 0 && cy <= yCorner
		b := abs(cy) <= r && cx >= 0 && cx <= xCorner
		return a || b
	}

	vis := make([]bool, len(circles))

	var dfs func(int) bool
	dfs = func(i int) bool {
		c := circles[i]
		x1, y1, r1 := c[0], c[1], c[2]
		if crossRightBottom(x1, y1, r1) {
			return true
		}
		vis[i] = true
		for j, c2 := range circles {
			if vis[j] {
				continue
			}
			x2, y2, r2 := c2[0], c2[1], c2[2]
			if (x1-x2)*(x1-x2)+(y1-y2)*(y1-y2) > (r1+r2)*(r1+r2) {
				continue
			}
			if x1*r2+x2*r1 < (r1+r2)*xCorner && y1*r2+y2*r1 < (r1+r2)*yCorner && dfs(j) {
				return true
			}
		}
		return false
	}

	for i, c := range circles {
		x, y, r := c[0], c[1], c[2]
		if inCircle(0, 0, x, y, r) || inCircle(xCorner, yCorner, x, y, r) {
			return false
		}
		if !vis[i] && crossLeftTop(x, y, r) && dfs(i) {
			return false
		}
	}
	return true
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

# Accepted solution for LeetCode #3235: Check if the Rectangle Corner Is Reachable
class Solution:
    def canReachCorner(
        self, xCorner: int, yCorner: int, circles: List[List[int]]
    ) -> bool:
        def in_circle(x: int, y: int, cx: int, cy: int, r: int) -> int:
            return (x - cx) ** 2 + (y - cy) ** 2 <= r**2

        def cross_left_top(cx: int, cy: int, r: int) -> bool:
            a = abs(cx) <= r and 0 <= cy <= yCorner
            b = abs(cy - yCorner) <= r and 0 <= cx <= xCorner
            return a or b

        def cross_right_bottom(cx: int, cy: int, r: int) -> bool:
            a = abs(cx - xCorner) <= r and 0 <= cy <= yCorner
            b = abs(cy) <= r and 0 <= cx <= xCorner
            return a or b

        def dfs(i: int) -> bool:
            x1, y1, r1 = circles[i]
            if cross_right_bottom(x1, y1, r1):
                return True
            vis[i] = True
            for j, (x2, y2, r2) in enumerate(circles):
                if vis[j] or not ((x1 - x2) ** 2 + (y1 - y2) ** 2 <= (r1 + r2) ** 2):
                    continue
                if (
                    (x1 * r2 + x2 * r1 < (r1 + r2) * xCorner)
                    and (y1 * r2 + y2 * r1 < (r1 + r2) * yCorner)
                    and dfs(j)
                ):
                    return True
            return False

        vis = [False] * len(circles)
        for i, (x, y, r) in enumerate(circles):
            if in_circle(0, 0, x, y, r) or in_circle(xCorner, yCorner, x, y, r):
                return False
            if (not vis[i]) and cross_left_top(x, y, r) and dfs(i):
                return False
        return True

// Accepted solution for LeetCode #3235: Check if the Rectangle Corner Is Reachable
impl Solution {
    pub fn can_reach_corner(x_corner: i32, y_corner: i32, circles: Vec<Vec<i32>>) -> bool {
        let n = circles.len();
        let mut vis = vec![false; n];

        let in_circle = |x: i64, y: i64, cx: i64, cy: i64, r: i64| -> bool {
            (x - cx) * (x - cx) + (y - cy) * (y - cy) <= r * r
        };

        let cross_left_top = |cx: i64, cy: i64, r: i64| -> bool {
            let a = cx.abs() <= r && (cy >= 0 && cy <= y_corner as i64);
            let b = (cy - y_corner as i64).abs() <= r && (cx >= 0 && cx <= x_corner as i64);
            a || b
        };

        let cross_right_bottom = |cx: i64, cy: i64, r: i64| -> bool {
            let a = (cx - x_corner as i64).abs() <= r && (cy >= 0 && cy <= y_corner as i64);
            let b = cy.abs() <= r && (cx >= 0 && cx <= x_corner as i64);
            a || b
        };
        fn dfs(
            circles: &Vec<Vec<i32>>,
            vis: &mut Vec<bool>,
            i: usize,
            x_corner: i32,
            y_corner: i32,
            cross_right_bottom: &dyn Fn(i64, i64, i64) -> bool,
        ) -> bool {
            let c = &circles[i];
            let (x1, y1, r1) = (c[0] as i64, c[1] as i64, c[2] as i64);

            if cross_right_bottom(x1, y1, r1) {
                return true;
            }

            vis[i] = true;

            for j in 0..circles.len() {
                if vis[j] {
                    continue;
                }

                let c2 = &circles[j];
                let (x2, y2, r2) = (c2[0] as i64, c2[1] as i64, c2[2] as i64);

                if (x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2) > (r1 + r2) * (r1 + r2) {
                    continue;
                }

                if x1 * r2 + x2 * r1 < (r1 + r2) * x_corner as i64
                    && y1 * r2 + y2 * r1 < (r1 + r2) * y_corner as i64
                    && dfs(circles, vis, j, x_corner, y_corner, cross_right_bottom)
                {
                    return true;
                }
            }
            false
        }

        for i in 0..n {
            let c = &circles[i];
            let (x, y, r) = (c[0] as i64, c[1] as i64, c[2] as i64);

            if in_circle(0, 0, x, y, r) || in_circle(x_corner as i64, y_corner as i64, x, y, r) {
                return false;
            }

            if !vis[i]
                && cross_left_top(x, y, r)
                && dfs(
                    &circles,
                    &mut vis,
                    i,
                    x_corner,
                    y_corner,
                    &cross_right_bottom,
                )
            {
                return false;
            }
        }

        true
    }
}

// Accepted solution for LeetCode #3235: Check if the Rectangle Corner Is Reachable
function canReachCorner(xCorner: number, yCorner: number, circles: number[][]): boolean {
    const inCircle = (x: bigint, y: bigint, cx: bigint, cy: bigint, r: bigint): boolean => {
        const dx = x - cx;
        const dy = y - cy;
        return dx * dx + dy * dy <= r * r;
    };

    const crossLeftTop = (cx: bigint, cy: bigint, r: bigint): boolean => {
        const a = BigInt(Math.abs(Number(cx))) <= r && cy >= 0n && cy <= BigInt(yCorner);
        const b =
            BigInt(Math.abs(Number(cy - BigInt(yCorner)))) <= r &&
            cx >= 0n &&
            cx <= BigInt(xCorner);
        return a || b;
    };

    const crossRightBottom = (cx: bigint, cy: bigint, r: bigint): boolean => {
        const a =
            BigInt(Math.abs(Number(cx - BigInt(xCorner)))) <= r &&
            cy >= 0n &&
            cy <= BigInt(yCorner);
        const b = BigInt(Math.abs(Number(cy))) <= r && cx >= 0n && cx <= BigInt(xCorner);
        return a || b;
    };

    const n = circles.length;
    const vis: boolean[] = new Array(n).fill(false);

    const dfs = (i: number): boolean => {
        const [x1, y1, r1] = circles[i].map(BigInt);
        if (crossRightBottom(x1, y1, r1)) {
            return true;
        }
        vis[i] = true;
        for (let j = 0; j < n; j++) {
            if (vis[j]) continue;
            const [x2, y2, r2] = circles[j].map(BigInt);
            if ((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2) > (r1 + r2) * (r1 + r2)) {
                continue;
            }
            if (
                x1 * r2 + x2 * r1 < (r1 + r2) * BigInt(xCorner) &&
                y1 * r2 + y2 * r1 < (r1 + r2) * BigInt(yCorner) &&
                dfs(j)
            ) {
                return true;
            }
        }
        return false;
    };

    for (let i = 0; i < n; i++) {
        const [x, y, r] = circles[i].map(BigInt);
        if (inCircle(0n, 0n, x, y, r) || inCircle(BigInt(xCorner), BigInt(yCorner), x, y, r)) {
            return false;
        }
        if (!vis[i] && crossLeftTop(x, y, r) && dfs(i)) {
            return false;
        }
    }

    return true;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n^2)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(n) space

Track components with a list or adjacency matrix. Each union operation may need to update all n elements’ component labels, giving O(n) per union. For n union operations total: O(n²). Find is O(1) with direct lookup, but union dominates.

UNION-FIND

O(α(n)) time

O(n) space

With path compression and union by rank, each find/union operation takes O(α(n)) amortized time, where α is the inverse Ackermann function — effectively constant. Space is O(n) for the parent and rank arrays. For m operations on n elements: O(m × α(n)) total.

Shortcut: Union-Find with path compression + rank → O(α(n)) per operation ≈ O(1). Just say “nearly constant.”

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

Check if the Rectangle Corner Is Reachable

Problem Statement

Roadmap

Brute Force Baseline

Baseline thinking

Example 1

Example 2

Example 3

Related Problems

Core Insight

What unlocks the optimal approach

Algorithm Walkthrough

Iteration Checklist

Edge Cases

Full Annotated Code

Interactive Study Demo

Complexity Analysis

Approach Breakdown

Common Mistakes

Off-by-one on range boundaries

Overflow in intermediate arithmetic