Off-by-one on range boundaries
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Move from brute-force thinking to an efficient approach using array strategy.
There is an m x n rectangular island that borders both the Pacific Ocean and Atlantic Ocean. The Pacific Ocean touches the island's left and top edges, and the Atlantic Ocean touches the island's right and bottom edges.
The island is partitioned into a grid of square cells. You are given an m x n integer matrix heights where heights[r][c] represents the height above sea level of the cell at coordinate (r, c).
The island receives a lot of rain, and the rain water can flow to neighboring cells directly north, south, east, and west if the neighboring cell's height is less than or equal to the current cell's height. Water can flow from any cell adjacent to an ocean into the ocean.
Return a 2D list of grid coordinates result where result[i] = [ri, ci] denotes that rain water can flow from cell (ri, ci) to both the Pacific and Atlantic oceans.
Example 1:
Input: heights = [[1,2,2,3,5],[3,2,3,4,4],[2,4,5,3,1],[6,7,1,4,5],[5,1,1,2,4]]
Output: [[0,4],[1,3],[1,4],[2,2],[3,0],[3,1],[4,0]]
Explanation: The following cells can flow to the Pacific and Atlantic oceans, as shown below:
[0,4]: [0,4] -> Pacific Ocean
[0,4] -> Atlantic Ocean
[1,3]: [1,3] -> [0,3] -> Pacific Ocean
[1,3] -> [1,4] -> Atlantic Ocean
[1,4]: [1,4] -> [1,3] -> [0,3] -> Pacific Ocean
[1,4] -> Atlantic Ocean
[2,2]: [2,2] -> [1,2] -> [0,2] -> Pacific Ocean
[2,2] -> [2,3] -> [2,4] -> Atlantic Ocean
[3,0]: [3,0] -> Pacific Ocean
[3,0] -> [4,0] -> Atlantic Ocean
[3,1]: [3,1] -> [3,0] -> Pacific Ocean
[3,1] -> [4,1] -> Atlantic Ocean
[4,0]: [4,0] -> Pacific Ocean
[4,0] -> Atlantic Ocean
Note that there are other possible paths for these cells to flow to the Pacific and Atlantic oceans.
Example 2:
Input: heights = [[1]] Output: [[0,0]] Explanation: The water can flow from the only cell to the Pacific and Atlantic oceans.
Constraints:
m == heights.lengthn == heights[r].length1 <= m, n <= 2000 <= heights[r][c] <= 105Problem summary: There is an m x n rectangular island that borders both the Pacific Ocean and Atlantic Ocean. The Pacific Ocean touches the island's left and top edges, and the Atlantic Ocean touches the island's right and bottom edges. The island is partitioned into a grid of square cells. You are given an m x n integer matrix heights where heights[r][c] represents the height above sea level of the cell at coordinate (r, c). The island receives a lot of rain, and the rain water can flow to neighboring cells directly north, south, east, and west if the neighboring cell's height is less than or equal to the current cell's height. Water can flow from any cell adjacent to an ocean into the ocean. Return a 2D list of grid coordinates result where result[i] = [ri, ci] denotes that rain water can flow from cell (ri, ci) to both the Pacific and Atlantic oceans.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array
[[1,2,2,3,5],[3,2,3,4,4],[2,4,5,3,1],[6,7,1,4,5],[5,1,1,2,4]]
[[1]]
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #417: Pacific Atlantic Water Flow
class Solution {
public List<List<Integer>> pacificAtlantic(int[][] heights) {
int m = heights.length, n = heights[0].length;
boolean[][] vis1 = new boolean[m][n];
boolean[][] vis2 = new boolean[m][n];
Deque<int[]> q1 = new ArrayDeque<>();
Deque<int[]> q2 = new ArrayDeque<>();
int[] dirs = {-1, 0, 1, 0, -1};
for (int i = 0; i < m; ++i) {
q1.offer(new int[] {i, 0});
vis1[i][0] = true;
q2.offer(new int[] {i, n - 1});
vis2[i][n - 1] = true;
}
for (int j = 0; j < n; ++j) {
q1.offer(new int[] {0, j});
vis1[0][j] = true;
q2.offer(new int[] {m - 1, j});
vis2[m - 1][j] = true;
}
BiConsumer<Deque<int[]>, boolean[][]> bfs = (q, vis) -> {
while (!q.isEmpty()) {
var cell = q.poll();
int x = cell[0], y = cell[1];
for (int k = 0; k < 4; ++k) {
int nx = x + dirs[k], ny = y + dirs[k + 1];
if (nx >= 0 && nx < m && ny >= 0 && ny < n && !vis[nx][ny]
&& heights[nx][ny] >= heights[x][y]) {
vis[nx][ny] = true;
q.offer(new int[] {nx, ny});
}
}
}
};
bfs.accept(q1, vis1);
bfs.accept(q2, vis2);
List<List<Integer>> ans = new ArrayList<>();
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (vis1[i][j] && vis2[i][j]) {
ans.add(List.of(i, j));
}
}
}
return ans;
}
}
// Accepted solution for LeetCode #417: Pacific Atlantic Water Flow
func pacificAtlantic(heights [][]int) [][]int {
m, n := len(heights), len(heights[0])
vis1 := make([][]bool, m)
vis2 := make([][]bool, m)
for i := range vis1 {
vis1[i] = make([]bool, n)
vis2[i] = make([]bool, n)
}
q1, q2 := [][2]int{}, [][2]int{}
dirs := [5]int{-1, 0, 1, 0, -1}
for i := 0; i < m; i++ {
q1 = append(q1, [2]int{i, 0})
vis1[i][0] = true
q2 = append(q2, [2]int{i, n - 1})
vis2[i][n-1] = true
}
for j := 0; j < n; j++ {
q1 = append(q1, [2]int{0, j})
vis1[0][j] = true
q2 = append(q2, [2]int{m - 1, j})
vis2[m-1][j] = true
}
bfs := func(q [][2]int, vis [][]bool) {
for len(q) > 0 {
x, y := q[0][0], q[0][1]
q = q[1:]
for k := 0; k < 4; k++ {
nx, ny := x+dirs[k], y+dirs[k+1]
if nx >= 0 && nx < m && ny >= 0 && ny < n &&
!vis[nx][ny] && heights[nx][ny] >= heights[x][y] {
vis[nx][ny] = true
q = append(q, [2]int{nx, ny})
}
}
}
}
bfs(q1, vis1)
bfs(q2, vis2)
var ans [][]int
for i := 0; i < m; i++ {
for j := 0; j < n; j++ {
if vis1[i][j] && vis2[i][j] {
ans = append(ans, []int{i, j})
}
}
}
return ans
}
# Accepted solution for LeetCode #417: Pacific Atlantic Water Flow
class Solution:
def pacificAtlantic(self, heights: List[List[int]]) -> List[List[int]]:
def bfs(q: Deque[Tuple[int, int]], vis: List[List[bool]]) -> None:
while q:
x, y = q.popleft()
for dx, dy in pairwise(dirs):
nx, ny = x + dx, y + dy
if (
0 <= nx < m
and 0 <= ny < n
and not vis[nx][ny]
and heights[nx][ny] >= heights[x][y]
):
vis[nx][ny] = True
q.append((nx, ny))
m, n = len(heights), len(heights[0])
vis1 = [[False] * n for _ in range(m)]
vis2 = [[False] * n for _ in range(m)]
q1: Deque[Tuple[int, int]] = deque()
q2: Deque[Tuple[int, int]] = deque()
dirs = (-1, 0, 1, 0, -1)
for i in range(m):
q1.append((i, 0))
vis1[i][0] = True
q2.append((i, n - 1))
vis2[i][n - 1] = True
for j in range(n):
q1.append((0, j))
vis1[0][j] = True
q2.append((m - 1, j))
vis2[m - 1][j] = True
bfs(q1, vis1)
bfs(q2, vis2)
return [(i, j) for i in range(m) for j in range(n) if vis1[i][j] and vis2[i][j]]
// Accepted solution for LeetCode #417: Pacific Atlantic Water Flow
use std::collections::VecDeque;
impl Solution {
pub fn pacific_atlantic(heights: Vec<Vec<i32>>) -> Vec<Vec<i32>> {
let (m, n) = (heights.len(), heights[0].len());
let mut vis1 = vec![vec![false; n]; m];
let mut vis2 = vec![vec![false; n]; m];
let mut q1 = VecDeque::new();
let mut q2 = VecDeque::new();
let dirs = [-1, 0, 1, 0, -1];
for i in 0..m {
q1.push_back((i, 0));
vis1[i][0] = true;
q2.push_back((i, n - 1));
vis2[i][n - 1] = true;
}
for j in 0..n {
q1.push_back((0, j));
vis1[0][j] = true;
q2.push_back((m - 1, j));
vis2[m - 1][j] = true;
}
let bfs = |q: &mut VecDeque<(usize, usize)>, vis: &mut Vec<Vec<bool>>| {
while let Some((x, y)) = q.pop_front() {
for k in 0..4 {
let nx = x as i32 + dirs[k];
let ny = y as i32 + dirs[k + 1];
if nx >= 0
&& nx < m as i32
&& ny >= 0
&& ny < n as i32
&& !vis[nx as usize][ny as usize]
&& heights[nx as usize][ny as usize] >= heights[x][y]
{
vis[nx as usize][ny as usize] = true;
q.push_back((nx as usize, ny as usize));
}
}
}
};
bfs(&mut q1, &mut vis1);
bfs(&mut q2, &mut vis2);
let mut ans = vec![];
for i in 0..m {
for j in 0..n {
if vis1[i][j] && vis2[i][j] {
ans.push(vec![i as i32, j as i32]);
}
}
}
ans
}
}
// Accepted solution for LeetCode #417: Pacific Atlantic Water Flow
function pacificAtlantic(heights: number[][]): number[][] {
const m = heights.length,
n = heights[0].length;
const vis1: boolean[][] = Array.from({ length: m }, () => Array(n).fill(false));
const vis2: boolean[][] = Array.from({ length: m }, () => Array(n).fill(false));
const q1: [number, number][] = [];
const q2: [number, number][] = [];
const dirs = [-1, 0, 1, 0, -1];
for (let i = 0; i < m; ++i) {
q1.push([i, 0]);
vis1[i][0] = true;
q2.push([i, n - 1]);
vis2[i][n - 1] = true;
}
for (let j = 0; j < n; ++j) {
q1.push([0, j]);
vis1[0][j] = true;
q2.push([m - 1, j]);
vis2[m - 1][j] = true;
}
const bfs = (q: [number, number][], vis: boolean[][]) => {
while (q.length) {
const [x, y] = q.shift()!;
for (let k = 0; k < 4; ++k) {
const nx = x + dirs[k],
ny = y + dirs[k + 1];
if (
nx >= 0 &&
nx < m &&
ny >= 0 &&
ny < n &&
!vis[nx][ny] &&
heights[nx][ny] >= heights[x][y]
) {
vis[nx][ny] = true;
q.push([nx, ny]);
}
}
}
};
bfs(q1, vis1);
bfs(q2, vis2);
const ans: number[][] = [];
for (let i = 0; i < m; ++i) {
for (let j = 0; j < n; ++j) {
if (vis1[i][j] && vis2[i][j]) {
ans.push([i, j]);
}
}
}
return ans;
}
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.
Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.
Review these before coding to avoid predictable interview regressions.
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.