LeetCode #407 — HARD

Trapping Rain Water II

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

Given an m x n integer matrix heightMap representing the height of each unit cell in a 2D elevation map, return the volume of water it can trap after raining.

Example 1:

Input: heightMap = [[1,4,3,1,3,2],[3,2,1,3,2,4],[2,3,3,2,3,1]]
Output: 4
Explanation: After the rain, water is trapped between the blocks.
We have two small ponds 1 and 3 units trapped.
The total volume of water trapped is 4.

Example 2:

Input: heightMap = [[3,3,3,3,3],[3,2,2,2,3],[3,2,1,2,3],[3,2,2,2,3],[3,3,3,3,3]]
Output: 10

Constraints:

m == heightMap.length
n == heightMap[i].length
1 <= m, n <= 200
0 <= heightMap[i][j] <= 2 * 10⁴

Step 01

Brute Force Baseline

Problem summary: Given an m x n integer matrix heightMap representing the height of each unit cell in a 2D elevation map, return the volume of water it can trap after raining.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[[1,4,3,1,3,2],[3,2,1,3,2,4],[2,3,3,2,3,1]]

Example 2

[[3,3,3,3,3],[3,2,2,2,3],[3,2,1,2,3],[3,2,2,2,3],[3,3,3,3,3]]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #407: Trapping Rain Water II
class Solution {
    public int trapRainWater(int[][] heightMap) {
        int m = heightMap.length, n = heightMap[0].length;
        boolean[][] vis = new boolean[m][n];
        PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> a[0] - b[0]);
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (i == 0 || i == m - 1 || j == 0 || j == n - 1) {
                    pq.offer(new int[] {heightMap[i][j], i, j});
                    vis[i][j] = true;
                }
            }
        }
        int ans = 0;
        int[] dirs = {-1, 0, 1, 0, -1};
        while (!pq.isEmpty()) {
            var p = pq.poll();
            for (int k = 0; k < 4; ++k) {
                int x = p[1] + dirs[k], y = p[2] + dirs[k + 1];
                if (x >= 0 && x < m && y >= 0 && y < n && !vis[x][y]) {
                    ans += Math.max(0, p[0] - heightMap[x][y]);
                    vis[x][y] = true;
                    pq.offer(new int[] {Math.max(p[0], heightMap[x][y]), x, y});
                }
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #407: Trapping Rain Water II
func trapRainWater(heightMap [][]int) (ans int) {
	m, n := len(heightMap), len(heightMap[0])
	pq := hp{}
	vis := make([][]bool, m)
	for i, row := range heightMap {
		vis[i] = make([]bool, n)
		for j, v := range row {
			if i == 0 || i == m-1 || j == 0 || j == n-1 {
				heap.Push(&pq, tuple{v, i, j})
				vis[i][j] = true
			}
		}
	}
	dirs := []int{-1, 0, 1, 0, -1}
	for len(pq) > 0 {
		p := heap.Pop(&pq).(tuple)
		for k := 0; k < 4; k++ {
			x, y := p.i+dirs[k], p.j+dirs[k+1]
			if x >= 0 && x < m && y >= 0 && y < n && !vis[x][y] {
				ans += max(0, p.v-heightMap[x][y])
				vis[x][y] = true
				heap.Push(&pq, tuple{max(p.v, heightMap[x][y]), x, y})
			}
		}
	}
	return
}

type tuple struct{ v, i, j int }
type hp []tuple

func (h hp) Len() int           { return len(h) }
func (h hp) Less(i, j int) bool { return h[i].v < h[j].v }
func (h hp) Swap(i, j int)      { h[i], h[j] = h[j], h[i] }
func (h *hp) Push(v any)        { *h = append(*h, v.(tuple)) }
func (h *hp) Pop() any          { a := *h; v := a[len(a)-1]; *h = a[:len(a)-1]; return v }

# Accepted solution for LeetCode #407: Trapping Rain Water II
class Solution:
    def trapRainWater(self, heightMap: List[List[int]]) -> int:
        m, n = len(heightMap), len(heightMap[0])
        vis = [[False] * n for _ in range(m)]
        pq = []
        for i in range(m):
            for j in range(n):
                if i == 0 or i == m - 1 or j == 0 or j == n - 1:
                    heappush(pq, (heightMap[i][j], i, j))
                    vis[i][j] = True
        ans = 0
        dirs = (-1, 0, 1, 0, -1)
        while pq:
            h, i, j = heappop(pq)
            for a, b in pairwise(dirs):
                x, y = i + a, j + b
                if x >= 0 and x < m and y >= 0 and y < n and not vis[x][y]:
                    ans += max(0, h - heightMap[x][y])
                    vis[x][y] = True
                    heappush(pq, (max(h, heightMap[x][y]), x, y))
        return ans

// Accepted solution for LeetCode #407: Trapping Rain Water II
struct Solution;

use std::cmp::Reverse;
use std::collections::BinaryHeap;

impl Solution {
    fn trap_rain_water(height_map: Vec<Vec<i32>>) -> i32 {
        let n = height_map.len();
        let m = height_map[0].len();
        let mut queue: BinaryHeap<(Reverse<i32>, usize, usize)> = BinaryHeap::new();
        let mut visited = vec![vec![false; m]; n];
        for i in 0..n {
            for j in 0..m {
                if i == 0 || j == 0 || i == n - 1 || j == m - 1 {
                    visited[i][j] = true;
                    queue.push((Reverse(height_map[i][j]), i, j));
                }
            }
        }
        let mut res = 0;
        while let Some((Reverse(h), i, j)) = queue.pop() {
            if i > 0 && !visited[i - 1][j] {
                let ii = i - 1;
                visited[ii][j] = true;
                if h > height_map[ii][j] {
                    res += h - height_map[ii][j];
                    queue.push((Reverse(h), ii, j));
                } else {
                    queue.push((Reverse(height_map[ii][j]), ii, j));
                }
            }
            if j > 0 && !visited[i][j - 1] {
                let jj = j - 1;
                visited[i][jj] = true;
                if h > height_map[i][jj] {
                    res += h - height_map[i][jj];
                    queue.push((Reverse(h), i, jj));
                } else {
                    queue.push((Reverse(height_map[i][jj]), i, jj));
                }
            }
            if i + 1 < n && !visited[i + 1][j] {
                let ii = i + 1;
                visited[ii][j] = true;
                if h > height_map[ii][j] {
                    res += h - height_map[ii][j];
                    queue.push((Reverse(h), ii, j));
                } else {
                    queue.push((Reverse(height_map[ii][j]), ii, j));
                }
            }
            if j + 1 < m && !visited[i][j + 1] {
                let jj = j + 1;
                visited[i][jj] = true;
                if h > height_map[i][jj] {
                    res += h - height_map[i][jj];
                    queue.push((Reverse(h), i, jj));
                } else {
                    queue.push((Reverse(height_map[i][jj]), i, jj));
                }
            }
        }
        res
    }
}

#[test]
fn test() {
    let height_map = vec_vec_i32![[1, 4, 3, 1, 3, 2], [3, 2, 1, 3, 2, 4], [2, 3, 3, 2, 3, 1]];
    let res = 4;
    assert_eq!(Solution::trap_rain_water(height_map), res);
}

// Accepted solution for LeetCode #407: Trapping Rain Water II
function trapRainWater(heightMap: number[][]): number {
    const m = heightMap.length;
    const n = heightMap[0].length;
    const vis: boolean[][] = Array.from({ length: m }, () => new Array(n).fill(false));
    const pq = new PriorityQueue<number[]>((a, b) => a[0] - b[0]);
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            if (i === 0 || i === m - 1 || j === 0 || j === n - 1) {
                pq.enqueue([heightMap[i][j], i, j]);
                vis[i][j] = true;
            }
        }
    }

    let ans = 0;
    const dirs = [-1, 0, 1, 0, -1];
    while (!pq.isEmpty()) {
        const [h, i, j] = pq.dequeue();
        for (let k = 0; k < 4; ++k) {
            const x = i + dirs[k];
            const y = j + dirs[k + 1];
            if (x >= 0 && x < m && y >= 0 && y < n && !vis[x][y]) {
                ans += Math.max(0, h - heightMap[x][y]);
                vis[x][y] = true;
                pq.enqueue([Math.max(h, heightMap[x][y]), x, y]);
            }
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(m × n × log (m × n)

Space

O(m × n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

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.

OPTIMIZED

O(n) time

O(1) space

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.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

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.