Given a string S of length n representing n boxes adjacent to each other.
- A character R at index i represents that i-th box is being pushed towards the right.
- On the other hand, L at index i represents that i-th box is being pushed towards the left.
- And a . represents an empty space.
We start with initial configuration, at every time unit, a box being pushed toward right pushes next box toward right and same is true for a box being pushed toward left. The task is to print the final positions of all boxes when no more movements are possible. If there is a substring “R.L”, then middle box is being pushed from both directions and therefore not moved any more.
Examples:
Input: S = “RR.L”
Output: RR.L
The first and second boxes are being toward right. The middle dot is being pushed from both directions, so does not move.Input: S = “R..R…L.”
Output: RRRRR.LL.
At time unit 1 :
The first box pushes second box.
Third box pushes fourth box.
Second last box pushes third last box, configuration becomes
“RR.RR.LL.”
At time unit 2 :
Second box pushes third box, string becomes
“RRRRR.LL.”
Approach: We can calculate the net force applied on every box. The forces we care about are how close a box is to a leftward R or to a rightward L i.e. the more closer we are, the stronger will be the force.
- Scanning from left to right, force decays by 1 in every iteration, and resets to N if we meet an R again.
- Similarly, from right to left direction, we can find the forces from the right (closeness to L).
- For some box result[i], if the forces are equal then the result is . as both side apply same force to it. Otherwise, our result is implied by whichever force is stronger.
For string S = “R..R…L.”
The forces going from left to right are [7, 6, 5, 7, 6, 5, 4, 0, 0].
The forces going from right to left are [0, 0, 0, 0, -4, -5, -6, -7, 0].
Combining them (taking their arithmetic addition at each index) gives result = [7, 6, 5, 7, 2, 0, -2, -7, 0].
Thus, the final answer is RRRRR.LL.
Below is the implementation of above approach:
Python3
# Python3 implementation of above approach # Function to return final # positions of the boxes def pushBoxes(S): N = len(S) force = [0] * N # Populate forces going from left to right f = 0 for i in range(0, N): if S[i] == 'R': f = N elif S[i] == 'L': f = 0 else: f = max(f - 1, 0) force[i] += f # Populate forces going from right to left f = 0 for i in range(N - 1, -1, -1): if S[i] == 'L': f = N elif S[i] == 'R': f = 0 else: f = max(f - 1, 0) force[i] -= f # return final state of boxes return "".join('.' if f == 0 else 'R' if f > 0 else 'L' for f in force) # Driver code S = ".L.R...LR..L.." # Function call to print answer print(pushBoxes(S)) |
LL.RR.LLRRLL..
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