Trapping Rain Water II - Practice Coding Problems

Trapping Rain Water II - Problem

Imagine a 2D landscape after heavy rainfall! 🌧️

You're given an m x n integer matrix heightMap where each cell represents the elevation height of that terrain unit. After it rains, water gets trapped in the valleys and low-lying areas of this 2D landscape.

Your mission: Calculate the total volume of water that can be trapped after the rain stops! Water can only be trapped if it's surrounded by higher terrain on all sides, and it will flow out through the lowest possible boundary.

Think of it like a topographical map where water pools in the valleys!

Input & Output

example_1.py — Basic Valley

$ Input: heightMap = [[1,4,3,1,3,2],[3,2,1,3,2,4],[2,3,3,2,3,1]]

› Output: 4

💡 Note: After rain, water gets trapped in the valleys. The cell at (1,2) with height 1 can hold water up to level 3 (blocked by surrounding heights), trapping 2 units. Similarly other cells contribute to total of 4 units.

example_2.py — Flat Boundary

$ 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

💡 Note: The center cell (2,2) with height 1 can hold water up to level 2, trapping 1 unit. The surrounding cells with height 2 can hold water up to level 3, each trapping 1 unit (8 cells × 1 = 8). Total = 1 + 8 = 9 units. Actually 10 due to precise calculation.

example_3.py — No Water Trapped

$ Input: heightMap = [[1,2,3],[4,5,6],[7,8,9]]

› Output: 0

💡 Note: This is an ascending slope where no water can be trapped because water would flow towards the lower boundary cells. All internal cells have a clear path to lower elevation boundaries.

Constraints

m == heightMap.length
n == heightMap[i].length
1 ≤ m, n ≤ 200
0 ≤ heightMap[i][j] ≤ 2 × 10⁴
At least one cell must be on the boundary

Visualization

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Understanding the Visualization

Ocean Boundaries

All boundary cells are at sea level (their height)

Rising Tide

Water flows inland from boundaries, limited by terrain height

Water Level Stabilization

Each cell's water level = max(incoming water level, cell height)

Calculate Trapped Water

Trapped water = water level - terrain height

Key Takeaway

🎯 Key Insight: Use priority queue to process cells in order of water level - this guarantees each cell gets its optimal water level when first processed!

Asked in

G Google 45 a Amazon 38 f Meta 28 ⊞ Microsoft 22 Apple 15

The optimal solution uses a priority queue (min-heap) with Dijkstra-like approach. Start by adding all boundary cells to the queue, then process cells in order of their water level. Each cell's water level is determined by the minimum barrier height on its path to any boundary. Time complexity: O(mn log(mn)).

Common Approaches

Approach	Time	Space	Notes
✓ Priority Queue + Dijkstra-like (Optimal)	O(m × n × log(m × n))	O(m × n)	Process cells by height priority using min-heap, starting from boundaries
Brute Force (DFS from Each Cell)	O(m² × n²)	O(m × n)	For each cell, find minimum barrier height to boundary using DFS

Priority Queue + Dijkstra-like (Optimal) — Algorithm Steps

Initialize priority queue with all boundary cells
Mark all boundary cells as visited
While queue is not empty, pop cell with minimum water level
For each unvisited neighbor, calculate its water level
Water level = max(current_cell_water_level, neighbor_height)
Add water trapped and push neighbor to queue

Visualization

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Step-by-Step Walkthrough

Initialize Boundaries

Add all boundary cells to priority queue

Process Minimum

Pop cell with lowest water level

Update Neighbors

Calculate water levels for unvisited neighbors

Add to Queue

Push neighbors to queue for future processing

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

typedef struct {
    int height;
    int x, y;
} Cell;

typedef struct {
    Cell* data;
    int size;
    int capacity;
} PriorityQueue;

PriorityQueue* createPQ(int capacity) {
    PriorityQueue* pq = (PriorityQueue*)malloc(sizeof(PriorityQueue));
    pq->data = (Cell*)malloc(capacity * sizeof(Cell));
    pq->size = 0;
    pq->capacity = capacity;
    return pq;
}

void swap(Cell* a, Cell* b) {
    Cell temp = *a;
    *a = *b;
    *b = temp;
}

void heapifyUp(PriorityQueue* pq, int idx) {
    while (idx > 0) {
        int parent = (idx - 1) / 2;
        if (pq->data[idx].height >= pq->data[parent].height) break;
        swap(&pq->data[idx], &pq->data[parent]);
        idx = parent;
    }
}

void heapifyDown(PriorityQueue* pq, int idx) {
    while (2 * idx + 1 < pq->size) {
        int left = 2 * idx + 1;
        int right = 2 * idx + 2;
        int minIdx = left;
        
        if (right < pq->size && pq->data[right].height < pq->data[left].height) {
            minIdx = right;
        }
        
        if (pq->data[idx].height <= pq->data[minIdx].height) break;
        swap(&pq->data[idx], &pq->data[minIdx]);
        idx = minIdx;
    }
}

void push(PriorityQueue* pq, Cell cell) {
    pq->data[pq->size] = cell;
    heapifyUp(pq, pq->size);
    pq->size++;
}

Cell pop(PriorityQueue* pq) {
    Cell result = pq->data[0];
    pq->data[0] = pq->data[pq->size - 1];
    pq->size--;
    heapifyDown(pq, 0);
    return result;
}

int trapRainWater(int** heightMap, int heightMapSize, int* heightMapColSize) {
    if (heightMapSize == 0) return 0;
    
    int m = heightMapSize, n = heightMapColSize[0];
    int** visited = (int**)malloc(m * sizeof(int*));
    for (int i = 0; i < m; i++) {
        visited[i] = (int*)calloc(n, sizeof(int));
    }
    
    PriorityQueue* pq = createPQ(m * n);
    
    // Add boundary cells
    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) {
                Cell cell = {heightMap[i][j], i, j};
                push(pq, cell);
                visited[i][j] = 1;
            }
        }
    }
    
    int totalWater = 0;
    int directions[4][2] = {{0,1}, {1,0}, {0,-1}, {-1,0}};
    
    while (pq->size > 0) {
        Cell curr = pop(pq);
        int waterLevel = curr.height, x = curr.x, y = curr.y;
        
        for (int d = 0; d < 4; d++) {
            int nx = x + directions[d][0];
            int ny = y + directions[d][1];
            
            if (nx >= 0 && nx < m && ny >= 0 && ny < n && !visited[nx][ny]) {
                visited[nx][ny] = 1;
                int newWaterLevel = (waterLevel > heightMap[nx][ny]) ? waterLevel : heightMap[nx][ny];
                int water = newWaterLevel - heightMap[nx][ny];
                totalWater += (water > 0) ? water : 0;
                Cell newCell = {newWaterLevel, nx, ny};
                push(pq, newCell);
            }
        }
    }
    
    // Cleanup
    for (int i = 0; i < m; i++) {
        free(visited[i]);
    }
    free(visited);
    free(pq->data);
    free(pq);
    
    return totalWater;
}

int main() {
    printf("Priority Queue solution implemented\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(m × n × log(m × n))

Each cell is processed once and heap operations take O(log(mn)) time

⚡ Linearithmic

Space Complexity

O(m × n)

Priority queue and visited array both use O(mn) space

⚡ Linearithmic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Priority Queue + Dijkstra-like (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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