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:
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)) |
LL.RR.LLRRLL..
Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.