780. Reaching Points

Problem Description

A move consists of taking a point (x, y) and transforming it to either (x, x+y) or (x+y, y).

Given a starting point (sx, sy) and a target point (tx, ty), return True if and only if a sequence of moves exists to transform the point (sx, sy) into (tx, ty). Otherwise, return False.

Examples:

Input: sx = 1, sy = 1, tx = 3, ty = 5
Output: true
Explanation:
One series of moves that transforms the starting point to the target is:
(1, 1) -> (1, 2)
(1, 2) -> (3, 2)
(3, 2) -> (3, 5)

Input: sx = 1, sy = 1, tx = 2, ty = 2
Output: false

Input: sx = 1, sy = 1, tx = 1, ty = 1
Output: true

Constraints:

• 1 ≤ sx, sy, tx, ty ≤ 10^9

Python Solution

class Solution:
    def reachingPoints(self, sx: int, sy: int, tx: int, ty: int) -> bool:
        while tx >= sx and ty >= sy:
            if tx == sx and ty == sy:
                return True
            if tx > ty:
                if ty == sy:
                    return (tx - sx) % ty == 0
                tx %= ty
            else:
                if tx == sx:
                    return (ty - sy) % tx == 0
                ty %= tx
        return False

Implementation Notes:

• Uses reverse simulation approach
• Key insight: work backwards from target to start
• Time complexity: O(log max(tx, ty))
• Space complexity: O(1)

Java Solution

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) {
                if (ty == sy) {
                    return (tx - sx) % ty == 0;
                }
                tx %= ty;
            } else {
                if (tx == sx) {
                    return (ty - sy) % tx == 0;
                }
                ty %= tx;
            }
        }
        return false;
    }
}

C++ Solution

class Solution {
public:
    bool reachingPoints(int sx, int sy, int tx, int ty) {
        while (tx >= sx && ty >= sy) {
            if (tx == sx && ty == sy) {
                return true;
            }
            if (tx > ty) {
                if (ty == sy) {
                    return (tx - sx) % ty == 0;
                }
                tx %= ty;
            } else {
                if (tx == sx) {
                    return (ty - sy) % tx == 0;
                }
                ty %= tx;
            }
        }
        return false;
    }
};

JavaScript Solution

function reachingPoints(sx, sy, tx, ty) {
    while (tx >= sx && ty >= sy) {
        if (tx === sx && ty === sy) {
            return true;
        }
        if (tx > ty) {
            if (ty === sy) {
                return (tx - sx) % ty === 0;
            }
            tx %= ty;
        } else {
            if (tx === sx) {
                return (ty - sy) % tx === 0;
            }
            ty %= tx;
        }
    }
    return false;
}

C# Solution

public class Solution {
    public bool ReachingPoints(int sx, int sy, int tx, int ty) {
        while (tx >= sx && ty >= sy) {
            if (tx == sx && ty == sy) {
                return true;
            }
            if (tx > ty) {
                if (ty == sy) {
                    return (tx - sx) % ty == 0;
                }
                tx %= ty;
            } else {
                if (tx == sx) {
                    return (ty - sy) % tx == 0;
                }
                ty %= tx;
            }
        }
        return false;
    }
}

Implementation Notes:

• Uses reverse simulation approach
• Key insight: work backwards from target to start
• Time complexity: O(log max(tx, ty))
• Space complexity: O(1)