Problem Link
Author: Alexandru Valeanu
Tester: Hasan Jaddouh
Editorialist: Bhuvnesh Jain
Difficulty
MEDIUM
Prerequisites
Recursion, Bitmask Dynamic Programming, Hashing
Problem
You are given a grid, R*C, where each location has some number written on it. You are also given N teleportation pairs (dx, dy), meaning that you can go from (a, b) to (e, f) iff |a-e| = dx and |b-f| = dy, where |x| denotes the absolute value of x. You are required to find a sequence of steps so that you use as maximum number of teleportation pairs (atmost N), each one only once, and try to maximise the sum of the number on the locations you visited.
Quick Explanation
Since, we need to use each teleportation pair once, we can store a bitmask where the set bits denotes which what teleportation pairs are still not used up. Then we can simply iterate over the possible next states based on the current location and bitmask and use dynamic programming to solve the above problem. Since, the number of locations that might be visited are limited, try to optimise the dynamic programming solution using hashing to handle the case of excess memory usage.
Explanation
Subtask - 1
No dynmaic programming or any advanced techniques are required to solve this subtask. Only simple recursion which traverses all the possible positions depending on the remaining teleportation pairs is enough. But this solution is quite slow to pass other subtasks.
Subtask - 2
We can simply see that subproblems of many recursive calls of above solution are overlapping. To clearing differentiaite between the different states, we create an dynamic programming array consisiting of 3 states, namely the current x-coordinate, current y-coordinate and the mask denoting the teleportation pairs still to be used. Thus our dynamic programming array looks like dp[x][y][mask].
Building the above the dynamic programming bottom up, would lead to complexity of O(R*C*N*{2}^{N}), which is quite slow for final 2 subtasks. Also, the memory used by the above approach will be of the order of O(R*C*{2}^{N}), which would cause memory limit issues with last 2 subtasks as well.
Subtask - 3
In the above approach it is easy to see that most of the states would be unreachable. For example, there might not exist any sequence of steps (of length <= N), such that we can reach from (a, b) to (c, d). Also, there might be cases we could reach from (a, b) to (c, d) only through particular sequence of moves, meaning that all âmaskâ positions might not be visited also. Thus, we need to avoid these redundant calculations in the above approach and also try to find an more strict upper-bound on the number of visited states in the above dynmaic programming solution.
Let us first define what a valid teleportation is. A teleporation is said to be valid if from a location on a grid, we can go to a location present on the grid. It is invalid, if we go to a location outside the grid.
We try to find a tighter upperbound for the number of states visited. Now, let us say, we are at a particular location, (a, b), then we have 5 possible choices to take from the current location, i.e. either stay in the current location (since we canât use any valid teleporation) or jump in any of the 4 directions (if jump is valid i.e. inside the grid). Also, at each steps, we would have to choose from K of the remaining teleportation pairs to go to next point. Thus, we can say that we have an upperbound of O(min(N*{5}^{N}, R*C*N*2^{N})) on the number of states that are visited. (Why minimum is taken is because, we either visit all the states in which case the bound is simply R*C*N*{2}^{N} or we canât take some valid teleportation pair at some location and visit less states, leading to N*{5}^{N} bound).
Now, let us try to write the above solution in top-down appraoch i.e. using recursion with memoisation. This, will lead us to visiting only those states which are reachable from starting state i.e dp[Sx][Sy][{2}^{N}-1]. Thus, in recursive solution for subtask - 1, we simply add the dynamic programming approach. But there is still a catch here i.e the memory requirement. The memory required by this solution is still O(R*C*{2}^{N}), which limits us from passing this subtask. But our time complexity of the solution has decreased. To get a trade of between the time and space complexity, we can use âmapsâ or âunordered mapsâ to store the list of possible âmasksâ or âlocationsâ as per implementation. This will increase the time complexity by logarithmic factor, but the space complexity would now be order of the number of possible states. Using this optimisation, we can easily pass the given subtask.
Subtask - 4
We are almost close to the full solution now. We just need to lower the trade off we did between memory and time in the last approach. For this, we will use hashing approach, to remove the logarithmic factor from the above last approach. (It may seem that unordered_map would suffice for this subtask too, but since they number of states in the unordered map are large, rehash would be called many times, which is a costly operation and lead to the solution becoming slower).
First let us consider mapping 3 states into a single number without any collision at all. Since, (a, b) are grid points, we can represent each grid point into a single integer as a * C + b. Thus now, the states reduce to 2. We can again apply similar operation to reduce the states to 1. For example mapping (a, b, mask) to single integer (a * C + b) * {2}^{N} + mask. The order of this number is O(R*C*{2}^{N}). We can simply hash this number to another value, i.e. build a hash table for these values. Suppose, we have successfully build our hash table, then we can easily see that we can retrieve the elements in O(1), thus removing the logarithmic factor from above solution. Care must be taken while choosing the size of hash table. Taking it too small would be of no use due to large number of collisions. We should chose the size around the number of states that can be reached, thus reducing the number of collisions while inserting the keys into it. Also, we can either implement hashing using linear probing or separate chaining. The author chose linear-probing to implement the hashing technique.
For more details, you can refer to the authorâs solution below.
Testerâs Solution for full problem
The testerâs solution used the fact that we have 5 choices at each and every point on the grid. Thus similar to bitmask where we have 2 bits to represent a states, he represented each states used base 5 encoding where 0 means we stay at the same location, and 1 to 4 means we chose one of the 4 directions. Simliar to iterative bitmasking solution which are written traversing each mask from 0 to 2^N - 1, the solution traverse each base 5 mask from 0 to 5^N - 1. Then based on the values in the mask, we just perform the teleportation and check whether we land at valid grid point or not. If yes, we update the answer for that mask. The final answer is the maximum over all the mask values calculated. The time complexity of this solution is exactly N * 5^N, and it doesnât use any hashing. Also, the 2 dimensions where we store the current x and y coordinate in above solution is not required here as it is captured in the base case, where we store the cost of starting point.
For more details, you can refer to testerâs solution below.
Time Complexity
O(min(N*{5}^N, R*C*N*{2}^{N}))
Space Complexity
Size of Hash Table ~ O(min({5}^N, R*C*{2}^{N}))
Solution Links
Setterâs solution
Testerâs solution
Editorialist solution - For 70 points, (No implementation of hash table)
