Move from brute-force thinking to an efficient approach using two pointers strategy.
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, anddominoes[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.length1 <= n <= 105dominoes[i] is either 'L', 'R', or '.'.Problem summary: 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
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Two Pointers · Dynamic Programming
"RR.L"
".L.R...LR..L.."
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #838: Push Dominoes
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);
}
}
// Accepted solution for LeetCode #838: Push Dominoes
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)
}
# Accepted solution for LeetCode #838: Push Dominoes
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)
// Accepted solution for LeetCode #838: Push Dominoes
struct Solution;
impl Solution {
fn push_dominoes(dominoes: String) -> String {
let mut dot = 0;
let mut right = false;
let mut res = "".to_string();
for c in dominoes.chars().chain("R".chars()) {
match c {
'R' => {
if dot > 0 {
if right {
for _ in 0..=dot {
res.push('R');
}
} else {
for _ in 0..dot {
res.push('.');
}
}
dot = 0;
} else {
if right {
res.push('R');
}
}
right = true;
}
'L' => {
if !right {
for _ in 0..1 + dot {
res.push('L');
}
} else {
for _ in 0..=dot / 2 {
res.push('R');
}
if dot % 2 == 1 {
res.push('.');
}
for _ in 0..=dot / 2 {
res.push('L');
}
}
right = false;
dot = 0;
}
_ => {
dot += 1;
}
}
}
res
}
}
#[test]
fn test() {
let dominoes = ".L.R...LR..L..".to_string();
let res = "LL.RR.LLRRLL..".to_string();
assert_eq!(Solution::push_dominoes(dominoes), res);
let dominoes = "RR.L".to_string();
let res = "RR.L".to_string();
assert_eq!(Solution::push_dominoes(dominoes), res);
let dominoes = "R".to_string();
let res = "R".to_string();
assert_eq!(Solution::push_dominoes(dominoes), res);
}
// Accepted solution for LeetCode #838: Push Dominoes
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('');
}
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.
Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.
Review these before coding to avoid predictable interview regressions.
Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.
Usually fails on: A valid pair can be skipped when only one side should move.
Fix: Move exactly one pointer per decision branch based on invariant.
Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.
Usually fails on: Correctness breaks on cases that differ only in hidden state.
Fix: Define state so each unique subproblem maps to one DP cell.