Problem Description

The problem presents an operation that can transform a point (x, y) in a two-dimensional grid to either (x, x + y) or (x + y, y). Given four integers sx, sy, tx, and ty, which represent the starting point (sx, sy) and the target point (tx, ty), the task is to determine if it is possible to reach the target point from the starting point by repeatedly applying the transformation.

The only allowed operations are:

1. Add the current y value to the x value, changing (x, y) to (x + y, y).
2. Add the current x value to the y value, changing (x, y) to (x, x + y).

The key challenge is to figure out if, by applying these operations starting from (sx, sy), we can eventually obtain the point (tx, ty). If it's possible, return true; otherwise, return false.

Intuition

To solve this problem, we can use the reverse thinking approach. Instead of starting from (sx, sy) and trying to reach (tx, ty) using the allowed operations, we start from (tx, ty) and try to work our way backwards to (sx, sy). This approach is valid because the operations are reversible.

We repeatedly subtract the smaller of tx and ty from the larger until we either pass sx and sy or land exactly on them. During this process, if tx is greater than ty, we know that tx had to be formed by adding ty to a smaller number in the previous step, so we perform tx %= ty. The same reasoning applies if ty is greater than tx.

While doing this, there are three cases:

1. If both tx and ty are greater than sx and sy respectively, we continue reducing tx and ty.
2. If tx becomes equal to sx, we check if ty can be reduced to sy by a multiple of tx. If so, it means we can reach (sx, sy) from (tx, ty).
3. The same check applies if ty becomes equal to sy.

If none of the conditions are met, then it's not possible to transform (sx, sy) to (tx, ty) using the operations, so we return false.

The time complexity of this approach is O(log(max(tx, ty))), which occurs due to the modulo operation that reduces tx and ty exponentially.

Learn more about Math patterns.

Solution Approach

The implementation of the solution follows a reverse approach working backward from the target point (tx, ty) to the start point (sx, sy). Here's how the solution translates into code:

• The solution starts by initializing a while loop that continues as long as both tx and ty are larger than sx and sy, but tx is not equal to ty. This condition allows us to reduce tx and ty until either we can't reduce them anymore without going below sx or sy, or until one of the coordinates matches sx or sy.

• Inside the loop, we use the % (modulo) operator to reverse the operation as follows:

  • If tx > ty, then tx must have been formed by adding some multiple of ty to the previous x value, so we perform tx %= ty. This reverses the last operation that affected tx.

  • If ty > tx, then we apply ty %= tx for the same reason but in the vertical direction.

• After exiting the loop, either tx == sx and/or ty == sy, or both tx and ty have been reduced below sx and sy. We need to handle both cases to complete the algorithm:

  • If tx == sx and ty == sy at the same time, we've found the exact reverse path to (sx, sy), and the function returns true.

  • If tx == sx, then ty must be reducible to sy by repeatedly subtracting tx (the same operation reversed). We check this by ensuring ty > sy and (ty - sy) % tx == 0. If this is the case, we return true.

  • Similarly, if ty == sy, we check if tx > sx and (tx - sx) % ty == 0 to determine if tx can be reduced to sx and return true accordingly.

• If none of the conditions are met, it means that we can't reach (sx, sy) from (tx, ty) by reversing the operations, so the function returns false.

This approach is efficient because it avoids the exponential time complexity of trying all possible paths from (sx, sy) to (tx, ty). Instead, by taking advantage of the mathematical properties of the operation, it quickly navigates to the solution or determines impossibility with a time complexity of O(log(max(tx, ty))).

Example Walkthrough

Let's consider the problem with a small example:

Imagine our starting point is (sx, sy) = (2, 4) and our target point is (tx, ty) = (9, 4). Can we transform the starting point to the target point using the allowed operations?

We apply the reverse approach:

1. We start from (tx, ty) which is (9, 4) and compare it with (sx, sy). Since tx > ty, we perform tx %= ty.
2. After applying the modulo operation, tx is now 9 % 4 = 1.
3. Now, our target point (tx, ty) has changed to (1, 4).
4. We compare tx with sx. They are not equal, so we continue with the loop.
5. Since ty > tx, we perform ty %= tx.
6. After applying the modulo operation, ty is now 4 % 1 = 0. The target point is now (1, 0).
7. Comparing the new ty with sy, we see that ty < sy; we cannot reduce ty anymore without going below sy.

From the example above, we note the following:

• After several steps, we've ended up with ty < sy, which violates the condition that it's only possible to either match the starting point exactly (in which case both tx and ty would match sx and sy) or reduce one of the target coordinates to match the starting point's while checking if the other can achieve the same via multiples.
• Since we could not achieve a situation where tx or ty matches sx or sy respectively without the other target coordinate going below its starting counterpart, it proves that the path from (sx, sy) to (tx, ty) cannot be achieved with the given operations.
• As a result, we would return false for this example.

Key Takeaways:

• The reverse approach significantly simplifies the problem, turning an otherwise computationally expensive search into a series of simple subtraction operations (modulo operation).
• If at any point during the reverse operation one of the target coordinates becomes less than its corresponding start coordinate, we can stop and return false because this means that the target point cannot be reached from the starting point.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the provided code primarily depends on the number of iterations it takes for either tx or ty to be reduced to either sx or sy, or for tx to equal ty. In each iteration, either tx or ty is reduced by a modulo operation, which takes constant time. However, the number of iterations could be large if tx and ty are much larger than sx and sy.

In the worst case, the modulo operation reduces the larger number by an amount proportional to the smaller number. Therefore, the time complexity is logarithmic in relation to the difference between the target and the start coordinates, or more formally, O(log(max(tx - sx, ty - sy))).

Space Complexity

The space complexity of the provided code is O(1). This is because the algorithm only uses a fixed amount of additional space (for variables tx and ty), and there is no additional space usage that grows with the input size.

Learn more about how to find time and space complexity quickly using problem constraints.