Sponsored
Sponsored
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)
1#include <stdbool.h>
2bool 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 if (tx == sx) return (ty - sy) % sx == 0;
9 return (tx - sx) % sy == 0;
10}This solution uses a loop to apply the reverse operations until tx or ty is reduced to sx or sy. It checks if further reduction is possible using modulus operations, ensuring the feasibility of reaching (sx, sy) based on the remainder. The logic ensures that one either complements the other by reducing one coordinate using the other. This approach is efficient due to its use of modulus operation, providing a quick reduction of the problem space.
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
1In JavaScript, recursion is deployed to simulate both transformation directions. It becomes impractical on larger scale inputs due to the exponential growth of recursive call depth required to check multiple permutations possible to reach the point.