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0838 - Push Dominoes (Medium)

https://leetcode.com/problems/push-dominoes/

Problem Statement​

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.

Example 1:

Input: dominoes = "RR.L"
Output: "RR.L"
Explanation: The first domino expends no additional force on the second domino.

Example 2:

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

Constraints:

  • n == dominoes.length
  • 1 <= n <= 105
  • dominoes[i] is either 'L', 'R', or '.'.

Approach 1: Two Pointers​

Written by @wingkwong
class Solution {
public:
    string pushDominoes(string dominoes) {
        // l is the left pointer
        int l = 0, n = dominoes.size();
        // r is the right pointer
        for (int r = 0; r < n; r++) {
            if (dominoes[r] == '.') {
                // case 1. meeting `.`, then skip it
                continue;
            } else if ((dominoes[r] == dominoes[l]) || (dominoes[l] == '.' && dominoes[r] == 'L')) {
                // case 2. both end is equal, i.e. dominoes[r] == dominoes[l]
                // then fill all the dots between both end 
                // e.g. L....L -> LLLLLL
                // e.g. R....R -> RRRRRR
                // case 2.1 if the left end is . and the right end is L, 
                // i.e. dominoes[l] == '.' && dominoes[r] == 'L'
                // then we need to fill them from `l` to `r` in this case
                for (int k = l; k < r; k++) dominoes[k] = dominoes[r];
            } else if (dominoes[l] == 'L' && dominoes[r] == 'R') {
                // case 3. left end is L and right end is R
                // e.g. L.....R
                // then do nothing
            }  else if (dominoes[l] == 'R' && dominoes[r] == 'L') {
                // case 4. left end is R and right end is L
                // if we have odd number of dots between them (let's say m dots), 
                // then we can only add (m / 2) Ls and (m / 2) Rs. 
                // p.s `/` here is integer division. e.g. 3 / 2 = 1
                // e.g. R...L -> RR.LL 
                // if we have even number of dots between them (let's say m dots), 
                // then we can only add (m / 2) Ls and (m / 2) Rs. 
                // e.g. R....L -> RRRLLL
                int m = (r - l - 1) / 2;
                for (int k = 1; k <= m; k++) dominoes[r - k] = 'L', dominoes[l + k] = 'R';
            }
            // update left pointer
            l = r;
        }
        // case 5. if the left dominoe is `R`, then fill all 'R' till the end
        // e.g. LL.R. -> LL.RR
        if (dominoes[l] == 'R') for (int k = l; k < n; k++) dominoes[k] = 'R';
        return dominoes;
    }
};