Formatted question description: https://leetcode.ca/all/780.html

# 780. Reaching Points (Hard)

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) to (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



Note:

• sx, sy, tx, ty will all be integers in the range [1, 10^9].

Related Topics:
Math

## Solution 1.

Starting from (sx, sy) will get TLE since there are many branches we need to try.

Instead, we should start from (tx, ty) since there could be only a single valid route.

If tx > ty, the previous point must be (tx - ty, ty).

If tx < ty, the previous point must be (tx, ty - tx).

tx and ty can’t be the same otherwise the previous point will have 0 value in the coordinates.

If tx > ty, instead of keeping subtracting ty which could be slow when tx is very large and ty is very small, we do tx %= ty. The tx < ty case is symmetrical.

We loop until tx <= sx or ty <= sy, then we have two valid cases:

• sx == tx, ty >= sy and ty - sy is divisible by sx.
• sy == sy, sx >= tx and tx - sx is divisible by sy.
// OJ: https://leetcode.com/problems/reaching-points/

// Time: O(log(max(tx, ty)))
// Space: O(1)
// Ref: https://leetcode.com/problems/reaching-points/discuss/114856/JavaC%2B%2BPython-Modulo-from-the-End
class Solution {
public:
bool reachingPoints(int sx, int sy, int tx, int ty) {
while (sx < tx && sy < ty) {
if (tx < ty) ty %= tx;
else tx %= ty;
}
return (sx == tx && sy <= ty && (ty - sy) % sx == 0)
|| (sy == ty && sx <= tx && (tx - sx) % sy == 0);
}
};


Java

class Solution {
public boolean reachingPoints(int sx, int sy, int tx, int ty) {
while (tx >= sx && ty >= sy) {
if (tx > ty) {
if (ty > sy)
tx %= ty;
else
return (tx - sx) % ty == 0;
} else if (tx < ty) {
if (tx > sx)
ty %= tx;
else
return (ty - sy) % tx == 0;
} else
break;
}
return tx == sx && ty == sy;
}
}