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Final state of the string after modification
  • Last Updated : 25 Nov, 2020

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: 

C++




// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return final
// positions of the boxes
string pushBoxes(string S)
{
    int N = S.length();
    vector<int> force(N, 0);
 
    // Populate forces going
    // from left to right
    int f = 0;
    for(int i = 0; i < N; i++)
    {
        if (S[i] == 'R')
        {
            f = N;
        }
        else if (S[i] == 'L')
        {
            f = 0;
        }
        else
        {
            f = max(f - 1, 0);
        }
        force[i] += f;
    }
 
    // Populate forces going from right to left
    f = 0;
    for(int i = N - 1; i >= 0; i--)
    {
        if (S[i] == 'L')
        {
            f = N;
        }
        else if (S[i] == 'R')
        {
            f = 0;
        }
        else
        {
            f = max(f - 1, 0);
        }
        force[i] -= f;
    }
 
    // return final state of boxes
    string ans;
    for(int f : force)
    {
        ans += f == 0 ? '.' : f > 0 ? 'R' : 'L';
    }
    return ans;
}
 
// Driver code
int main()
{
    string S = ".L.R...LR..L..";
     
    // Function call to print answer
    cout << pushBoxes(S);
}
 
// This code is contributed by sanjeev2552

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))
Output: 
LL.RR.LLRRLL..



 

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