838. Push Dominoes

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.

 

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 '.'.

Solutions

Solution 1: Multi-Source BFS

Treat all initially pushed dominoes (L or R) as sources, which simultaneously propagate their forces outward. Use a queue to perform BFS layer by layer (0, 1, 2, ...):

We define \(\text{time[i]}\) to record the first moment when the i-th domino is affected by a force, with -1 indicating it has not been affected yet. We also define \(\text{force[i]}\) as a variable-length list that stores the directions ('L', 'R') of forces acting on the domino at the same moment. Initially, push all indices of L/R dominoes into the queue and set their time to 0.

When dequeuing index i, if \(\text{force[i]}\) contains only one direction, the domino will fall in that direction \(f\). Let the index of the next domino be:

\[ j = \begin{cases} i - 1, & f = L,\\ i + 1, & f = R. \end{cases} \]

If \(0 \leq j < n\):

• If \(\text{time[j]} = -1\), it means j has not been affected yet. Record \(\text{time[j]} = \text{time[i]} + 1\), enqueue it, and append \(f\) to \(\text{force[j]}\).
• If \(\text{time[j]} = \text{time[i]} + 1\), it means j has already been affected by another force at the same "next moment." In this case, append \(f\) to \(\text{force[j]}\), causing a standoff. Subsequently, since \(\text{len(force[j])} = 2\), it will remain upright.

After the queue is emptied, all positions where \(\text{force[i]}\) has a length of 1 will fall in the corresponding direction, while positions with a length of 2 will remain as .. Finally, concatenate the character array to form the answer.

The complexity is \(O(n)\), and the space complexity is \(O(n)\), where \(n\) is the number of dominoes.

Python3JavaC++GoTypeScript

class Solution:
    def pushDominoes(self, dominoes: str) -> str:
        n = len(dominoes)
        q = deque()
        time = [-1] * n
        force = defaultdict(list)
        for i, f in enumerate(dominoes):
            if f != '.':
                q.append(i)
                time[i] = 0
                force[i].append(f)
        ans = ['.'] * n
        while q:
            i = q.popleft()
            if len(force[i]) == 1:
                ans[i] = f = force[i][0]
                j = i - 1 if f == 'L' else i + 1
                if 0 <= j < n:
                    t = time[i]
                    if time[j] == -1:
                        q.append(j)
                        time[j] = t + 1
                        force[j].append(f)
                    elif time[j] == t + 1:
                        force[j].append(f)
        return ''.join(ans)

class Solution {
    public String pushDominoes(String dominoes) {
        int n = dominoes.length();
        Deque<Integer> q = new ArrayDeque<>();
        int[] time = new int[n];
        Arrays.fill(time, -1);
        List<Character>[] force = new List[n];
        for (int i = 0; i < n; ++i) {
            force[i] = new ArrayList<>();
        }
        for (int i = 0; i < n; ++i) {
            char f = dominoes.charAt(i);
            if (f != '.') {
                q.offer(i);
                time[i] = 0;
                force[i].add(f);
            }
        }
        char[] ans = new char[n];
        Arrays.fill(ans, '.');
        while (!q.isEmpty()) {
            int i = q.poll();
            if (force[i].size() == 1) {
                ans[i] = force[i].get(0);
                char f = ans[i];
                int j = f == 'L' ? i - 1 : i + 1;
                if (j >= 0 && j < n) {
                    int t = time[i];
                    if (time[j] == -1) {
                        q.offer(j);
                        time[j] = t + 1;
                        force[j].add(f);
                    } else if (time[j] == t + 1) {
                        force[j].add(f);
                    }
                }
            }
        }
        return new String(ans);
    }
}

class Solution {
public:
    string pushDominoes(string dominoes) {
        int n = dominoes.size();
        queue<int> q;
        vector<int> time(n, -1);
        vector<string> force(n);
        for (int i = 0; i < n; i++) {
            if (dominoes[i] == '.') continue;
            q.emplace(i);
            time[i] = 0;
            force[i].push_back(dominoes[i]);
        }

        string ans(n, '.');
        while (!q.empty()) {
            int i = q.front();
            q.pop();
            if (force[i].size() == 1) {
                char f = force[i][0];
                ans[i] = f;
                int j = (f == 'L') ? (i - 1) : (i + 1);
                if (j >= 0 && j < n) {
                    int t = time[i];
                    if (time[j] == -1) {
                        q.emplace(j);
                        time[j] = t + 1;
                        force[j].push_back(f);
                    } else if (time[j] == t + 1)
                        force[j].push_back(f);
                }
            }
        }
        return ans;
    }
};

func pushDominoes(dominoes string) string {
    n := len(dominoes)
    q := []int{}
    time := make([]int, n)
    for i := range time {
        time[i] = -1
    }
    force := make([][]byte, n)
    for i, c := range dominoes {
        if c != '.' {
            q = append(q, i)
            time[i] = 0
            force[i] = append(force[i], byte(c))
        }
    }

    ans := bytes.Repeat([]byte{'.'}, n)
    for len(q) > 0 {
        i := q[0]
        q = q[1:]
        if len(force[i]) > 1 {
            continue
        }
        f := force[i][0]
        ans[i] = f
        j := i - 1
        if f == 'R' {
            j = i + 1
        }
        if 0 <= j && j < n {
            t := time[i]
            if time[j] == -1 {
                q = append(q, j)
                time[j] = t + 1
                force[j] = append(force[j], f)
            } else if time[j] == t+1 {
                force[j] = append(force[j], f)
            }
        }
    }
    return string(ans)
}

function pushDominoes(dominoes: string): string {
    const n = dominoes.length;
    const q: number[] = [];
    const time: number[] = Array(n).fill(-1);
    const force: string[][] = Array.from({ length: n }, () => []);

    for (let i = 0; i < n; i++) {
        const f = dominoes[i];
        if (f !== '.') {
            q.push(i);
            time[i] = 0;
            force[i].push(f);
        }
    }

    const ans: string[] = Array(n).fill('.');
    let head = 0;
    while (head < q.length) {
        const i = q[head++];
        if (force[i].length === 1) {
            const f = force[i][0];
            ans[i] = f;
            const j = f === 'L' ? i - 1 : i + 1;
            if (j >= 0 && j < n) {
                const t = time[i];
                if (time[j] === -1) {
                    q.push(j);
                    time[j] = t + 1;
                    force[j].push(f);
                } else if (time[j] === t + 1) {
                    force[j].push(f);
                }
            }
        }
    }
    return ans.join('');
}

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