Published: Sep 27, 2022

## Introduction

Depending on how to think, this problem has a few approaches to solve. The two pointers, DP would be one of them. Also, a string manipulation approach is there.

## Problem Description

There are n dominoes in a line, and we place each domino vertically upright. In the beginning, we simultaneously push some of the dominoes either to the left or to the right.

After each second, each domino that is falling to the left pushes the adjacent domino on the left. Similarly, the dominoes falling to the right push their adjacent dominoes standing on the right. When a vertical domino has dominoes falling on it from both sides, it stays still due to the balance of the forces. For the purposes of this question, we will consider that a falling domino expends no additional force to a falling or already fallen domino.

You are given a string

`dominoes`

representing the initial state where:

`dominoes[i] = 'L'`

, if the ith domino has been pushed to the left,`dominoes[i] = 'R'`

, if the ith domino has been pushed to the right, and`dominoes[i] = '.'`

, if the ith domino has not been pushed.Return a string representing the final state.

Constraints:

`n == dominoes.length`

`1 <= n <= 10**5`

`dominoes[i]`

is either ‘L’, ‘R’, or ‘.’.

## Examples

```
Example 1
Input: dominoes = "RR.L"
Output: "RR.L"
```

```
Example 2
Input: dominoes = ".L.R...LR..L.."
Output: "LL.RR.LLRRLL.."
```

```
Example 3
Input: dominoes = "..R.."
Output: "..RRR"
```

## Analysis

The solution here took the string manipulation approach.
If the fragment is ‘R.L’, we want to keep it as is. So, replace it by ‘** ’ for now.
While ‘R.’ or ‘.L” patterns should be replaced by ‘RR’ or ‘LL’.
If there’s no more change, the process is over.
In the end, replace ‘**’ to the original ‘R.L’.

## Solution

```
lass PushDominoes:
def pushDominoes(self, dominoes: str) -> str:
nochange = '***'
result = ''
while result != dominoes:
result = dominoes
dominoes = dominoes.replace('R.L', nochange)
dominoes = dominoes.replace('R.', 'RR')
dominoes = dominoes.replace('.L', 'LL')
return result.replace(nochange, 'R.L')
```

## Complexities

- Time:
`O(n)`

- Space:
`O(1)`