How do you solve Check if Point Is Reachable on LeetCode?

Check if Point Is Reachable (LeetCode #2543) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(log(max(targetX, targetY))) time complexity and O(1) space complexity. Key insight: The ability to reverse operations allows for a more efficient solution.

What is the brute force solution for Check if Point Is Reachable?

The brute force approach for Check if Point Is Reachable is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves exploring all possible paths from (1, 1) to (targetX, targetY) using a recursive depth-first search (DFS). This method checks every possible move until it either finds a valid path or exhausts all options. The optimal Optimal Solution approach improves this to O(log(max(targetX, targetY))).

What algorithm or data structure is used to solve Check if Point Is Reachable?

Check if Point Is Reachable uses the following concepts: Math, Number Theory. The recommended patterns to study are: Greedy Algorithms, Mathematical Induction.

What are the interview tips for Check if Point Is Reachable?

Always think about both forward and backward approaches to a problem. Clarify the constraints and limits of the problem to avoid unnecessary computations. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Check if Point Is Reachable?

Not considering the reverse operations effectively, leading to unnecessary complexity. Failing to handle edge cases where one coordinate becomes zero.

#2543

Check if Point Is Reachable

Hard

MathNumber TheoryGreedy AlgorithmsMathematical Induction

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(log(max(targetX, targetY)))
Space	O(1)	O(1)

💡

Intuition

Time O(log(max(targetX, targetY)))Space O(1)

Instead of exploring all paths, we can work backwards from (targetX, targetY) to (1, 1). By applying reverse operations, we can determine if we can reach the starting point. This significantly reduces the number of checks needed.

⚙️

Algorithm

3 steps

1Step 1: While targetX > 1 and targetY > 1, apply reverse operations: if targetX > targetY, set targetX = targetX - targetY; otherwise, set targetY = targetY - targetX.
2Step 2: If either targetX or targetY is even, divide it by 2 if possible.
3Step 3: If we reach (1, 1), return true; otherwise, return false.

solution.py7 lines

1def canReach(targetX, targetY):
2    while targetX > 1 and targetY > 1:
3        if targetX > targetY:
4            targetX -= targetY
5        else:
6            targetY -= targetX
7    return targetX == 1 or targetY == 1

ℹ

Complexity note: The time complexity is O(log(max(targetX, targetY))) because each operation effectively reduces one of the coordinates, leading to logarithmic steps to reach (1, 1). The space complexity is O(1) since we are using a constant amount of space.

1The ability to reverse operations allows for a more efficient solution.
2Understanding the relationship between the coordinates helps in determining reachability.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.