LeetCode #2543 — HARD

Check if Point Is Reachable

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

The Problem

Problem Statement

There exists an infinitely large grid. You are currently at point (1, 1), and you need to reach the point (targetX, targetY) using a finite number of steps.

In one step, you can move from point (x, y) to any one of the following points:

(x, y - x)
(x - y, y)
(2 * x, y)
(x, 2 * y)

Given two integers targetX and targetY representing the X-coordinate and Y-coordinate of your final position, return true if you can reach the point from (1, 1) using some number of steps, and false otherwise.

Example 1:

Input: targetX = 6, targetY = 9
Output: false
Explanation: It is impossible to reach (6,9) from (1,1) using any sequence of moves, so false is returned.

Example 2:

Input: targetX = 4, targetY = 7
Output: true
Explanation: You can follow the path (1,1) -> (1,2) -> (1,4) -> (1,8) -> (1,7) -> (2,7) -> (4,7).

Constraints:

1 <= targetX, targetY <= 10⁹

Step 01

Brute Force Baseline

Problem summary: There exists an infinitely large grid. You are currently at point (1, 1), and you need to reach the point (targetX, targetY) using a finite number of steps. In one step, you can move from point (x, y) to any one of the following points: (x, y - x) (x - y, y) (2 * x, y) (x, 2 * y) Given two integers targetX and targetY representing the X-coordinate and Y-coordinate of your final position, return true if you can reach the point from (1, 1) using some number of steps, and false otherwise.

Baseline thinking

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

Pattern signal: Math

Example 1

6
9

Example 2

4
7

Core Insight

What unlocks the optimal approach

Let’s go in reverse order, from (targetX, targetY) to (1, 1). So, now we can move from (x, y) to (x+y, y), (x, y+x), (x/2, y) if x is even, and (x, y/2) if y is even.
When is it optimal to use the third and fourth operations?
Think how GCD of (x, y) is affected if we apply the first two operations.
How can we check if we can reach (1, 1) using the GCD value calculate above?

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 #2543: Check if Point Is Reachable
class Solution {
    public boolean isReachable(int targetX, int targetY) {
        int x = gcd(targetX, targetY);
        return (x & (x - 1)) == 0;
    }

    private int gcd(int a, int b) {
        return b == 0 ? a : gcd(b, a % b);
    }
}

// Accepted solution for LeetCode #2543: Check if Point Is Reachable
func isReachable(targetX int, targetY int) bool {
	x := gcd(targetX, targetY)
	return x&(x-1) == 0
}

func gcd(a, b int) int {
	if b == 0 {
		return a
	}
	return gcd(b, a%b)
}

# Accepted solution for LeetCode #2543: Check if Point Is Reachable
class Solution:
    def isReachable(self, targetX: int, targetY: int) -> bool:
        x = gcd(targetX, targetY)
        return x & (x - 1) == 0

// Accepted solution for LeetCode #2543: Check if Point Is Reachable
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2543: Check if Point Is Reachable
// class Solution {
//     public boolean isReachable(int targetX, int targetY) {
//         int x = gcd(targetX, targetY);
//         return (x & (x - 1)) == 0;
//     }
// 
//     private int gcd(int a, int b) {
//         return b == 0 ? a : gcd(b, a % b);
//     }
// }

// Accepted solution for LeetCode #2543: Check if Point Is Reachable
function isReachable(targetX: number, targetY: number): boolean {
    const x = gcd(targetX, targetY);
    return (x & (x - 1)) === 0;
}

function gcd(a: number, b: number): number {
    return b == 0 ? a : gcd(b, a % b);
}

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(log(\min(targetX, targetY)

Space

O(1)

Approach Breakdown

ITERATIVE

O(n) time

O(1) space

Simulate the process step by step — multiply n times, check each number up to n, or iterate through all possibilities. Each step is O(1), but doing it n times gives O(n). No extra space needed since we just track running state.

MATH INSIGHT

O(log n) time

O(1) space

Math problems often have a closed-form or O(log n) solution hidden behind an O(n) simulation. Modular arithmetic, fast exponentiation (repeated squaring), GCD (Euclidean algorithm), and number theory properties can dramatically reduce complexity.

Shortcut: Look for mathematical properties that eliminate iteration. Repeated squaring → O(log n). Modular arithmetic avoids overflow.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

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.