LeetCode 780: Reaching Points Solution

Problem Explanation

Explanation

To solve this problem, we can work backwards from the target point (tx, ty) to the starting point (sx, sy). We can keep subtracting the smaller of the two coordinates from the larger one until we reach the starting point or go out of bounds. If we reach the starting point, we return true; otherwise, we return false.

Algorithm

While tx >= sx and ty >= sy:
- If tx == sx and ty == sy, return true.
- If tx > ty, subtract ty from tx.
- If ty > tx, subtract tx from ty.
If we go out of bounds, return false.

Time Complexity

The time complexity of this algorithm is O(log(max(tx, ty))) because at each step, we are reducing one of the coordinates by the other one.

Space Complexity

The space complexity of this algorithm is O(1) as we are using constant extra space.

Solution Code

class Solution {
    public boolean reachingPoints(int sx, int sy, int tx, int ty) {
        while (tx >= sx && ty >= sy) {
            if (tx == sx && ty == sy) {
                return true;
            }
            if (tx > ty) {
                tx -= ty;
            } else {
                ty -= tx;
            }
        }
        return false;
    }
}

Frequently Asked Questions

How to solve LeetCode 780 (Reaching Points)?

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What is the time complexity of LeetCode 780 (Reaching Points)?

The time complexity for LeetCode 780 (Reaching Points) varies by solution. Check the detailed explanation section for specific complexities in Java, C++, and Python implementations.

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780. Reaching Points