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The goal could be efficiently approached by reversing the operations. Instead of trying to reach (tx, ty) from (sx, sy), we attempt to determine whether (sx, sy) can be reached from (tx, ty) using reverse operations. This is achieved by repeatedly applying the inverse operations: (x, x+y) can be reverted to (x, y), and (x+y, y) can be reverted to (x, y). The main idea is to use modulus operations when x != y.
Time Complexity: O(log(max(tx, ty)))
Space Complexity: O(1)
1public class Solution {
2 public bool ReachingPoints(int sx, int sy, int tx, int ty) {
3 while (tx > sx && ty > sy && tx != ty) {
4 if (tx > ty) tx %= ty;
5 else ty %= tx;
6 }
7 if (tx < sx || ty < sy) return false;
8 return tx == sx ? (ty - sy) % sx == 0 : (tx - sx) % sy == 0;
9 }
10}The C# solution uses iterative steps and modulus operations in a similar manner as other solutions. It assures reduction of 'tx' and 'ty' to certain base conditions, verifying if reaching the initial point is feasible.
Another strategy involves recursive backtracking, where the function makes recursive calls to simulate both directions (x + y, y) and (x, x + y) to reach the target point (tx, ty) from the start point (sx, sy). Although less efficient compared to the reverse operation method due to its depth, it's an introductory way to explore the possibilities.
Time Complexity: Exponential in the worst case
Space Complexity: Recursive stack size
1Java's recursion demonstrates similar exploratory steps assessing both logical transitions from (sx, sy) recursively, aiming towards achieving the endpoint. It tends to become impractical beyond smaller values of 'tx', 'ty' due to inherent recursive limitations.