Final state of the string after modification

Given a string S of length n representing n boxes adjacent to each other.

  1. A character R at index i represents that i-th box is being pushed towards the right.
  2. On the other hand, L at index i represents that i-th box is being pushed towards the left.
  3. 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

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# 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))

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Output:

LL.RR.LLRRLL..


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