leetcode - 780. Reaching Points

KpLn_HJL2023-10-21 8:28

Description

Given four integers sx, sy, tx, and ty, return true if it is possible to convert the point (sx, sy) to the point (tx, ty) through some operations, or false otherwise.

The allowed operation on some point (x, y) is to convert it to either (x, x + y) or (x + y, y).

Example 1:

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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)

Example 2:

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

Example 3:

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

Constraints:

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1 <= sx, sy, tx, ty <= 10^9

Solution

Shrink 1by1

The possibilities are like a binary tree, use example 1:

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			   1,1
			/		\
		1,2			2,1
		/	\		/ \
	1,3		3,2	  2,3 	3,1
	/ \		/ \		/\
  1,4  4,3 3,5 5,2 ...

So instead of searching from the sx, sy, which is the top of the tree, we could start from the leaf, which is the tx, ty

Note that:
t x , t y = { s x , s x + s y s x + s y , s y tx, ty = \begin{cases} sx, sx+sy \\ sx + sy, sy \end{cases} tx,ty={sx,sx+sysx+sy,sy

So every time shrink the smaller one from tx, ty, which means find the parent of the node, until we find the source node.

Time complexity: o ( log ⁡ max ⁡ ( t x , t y ) ) o(\log \max(tx, ty)) o(logmax(tx,ty))

Space complexity: o ( 1 ) o(1) o(1)

Shrink by potential maximum

It's too slow to shrink one node at a time, we could shrink to the number that is larger than sx or sy

Code

Shrink 1by1 (TLE)

python3 复制代码

class Solution:
    def reachingPoints(self, sx: int, sy: int, tx: int, ty: int) -> bool:
        while (tx != sx or ty != sy) and tx >= 1 and ty >= 1:
            if tx > ty:
                tx, ty = tx % ty, ty
            else:
                tx, ty = tx, ty % tx
        return tx == sx and ty == sy

Shrink by potential maximum

python3 复制代码

class Solution:
    def reachingPoints(self, sx: int, sy: int, tx: int, ty: int) -> bool:
        while (tx != sx or ty != sy) and tx >= 1 and ty >= 1:
            if tx > ty:
                multi_factor = max(1, (tx - sx) // ty)
                tx, ty = tx - multi_factor * ty, ty
            else:
                multi_factor = max(1, (ty - sy) // tx)
                tx, ty = tx, ty - tx * multi_factor
        return tx == sx and ty == sy