What is the time complexity of Swim in Rising Water?

The optimal solution has a time complexity of O(n^2 log n).

What is the space complexity of Swim in Rising Water?

The space complexity is O(n^2).

What data structures are used to solve Swim in Rising Water?

This problem uses Array, Binary Search, Depth-First Search, Breadth-First Search, Union Find, Heap (Priority Queue), Matrix. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Swim in Rising Water?

Insight 1: The problem can be solved using a priority queue to explore the grid, prioritizing cells with the lowest required water level (time). Insight 2: Binary search can be used to find the minimum time, by checking if a path exists at a given time. However, the provided solution uses a more efficient priority queue approach directly.

Which companies ask LeetCode 778. Swim in Rising Water in interviews?

This problem is frequently asked at DE Shaw, DoorDash, PhonePe, WeRide.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Swim in Rising Water - LeetCode 778 Solution

Swim in Rising Water - Complete Solution Guide

Swim in Rising Water is LeetCode problem 778, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given an n x n integer matrix grid where each value grid[i][j] represents the elevation at that point (i, j) . It starts raining, and water gradually rises over time. At time t , the water level is t , meaning any cell with elevation less than equal to t is submerged or reachable. You can swim from a square to another 4-directionally adjacent square if and only if the elevation of both squares individually are at most t . You can swim infinite distances in zero time. Of course, you must

Detailed Explanation

The problem requires finding the minimum time needed to reach the bottom-right cell (n-1, n-1) of an n x n grid, starting from the top-left cell (0, 0). Each cell in the grid has an elevation value. You can only swim between adjacent cells if the water level, represented by time `t`, is greater than or equal to the elevation of both cells. The goal is to find the smallest `t` that allows a path from (0, 0) to (n-1, n-1). Essentially, we need to find the path where the maximum elevation along that path is minimized.

Solution Approach

The provided solution employs Dijkstra's algorithm using a priority queue. The priority queue stores tuples of (time, row, col), where 'time' represents the minimum water level needed to reach that cell. The algorithm starts at the top-left cell and iteratively explores its neighbors. For each neighbor, the algorithm calculates the new required time to reach that neighbor (the maximum of the current time and the neighbor's elevation). The algorithm keeps track of visited cells to prevent revisiting them. The search continues until the bottom-right cell is reached. The final time is the minimum time required to reach the destination.

Step-by-Step Algorithm

Step 1: Initialize a priority queue `pq` with the starting cell (0, 0) and its elevation (grid[0][0]). Also, initialize a `visited` set to keep track of visited cells.
Step 2: While the priority queue is not empty, extract the cell with the minimum time from the priority queue.
Step 3: If the extracted cell is the destination cell (n-1, n-1), return the time associated with that cell.
Step 4: For each of the four neighbors (up, down, left, right) of the current cell:
Step 5: Check if the neighbor is within the grid boundaries and has not been visited.
Step 6: If the neighbor is valid, calculate the new required time to reach the neighbor as the maximum of the current time and the elevation of the neighbor.
Step 7: Add the neighbor to the `visited` set and push it into the priority queue with the new time and its row and column coordinates.
Step 8: Repeat steps 2-7 until the destination is reached.

Key Insights

Insight 1: The problem can be solved using a priority queue to explore the grid, prioritizing cells with the lowest required water level (time).
Insight 2: Binary search can be used to find the minimum time, by checking if a path exists at a given time. However, the provided solution uses a more efficient priority queue approach directly.
Insight 3: The key is to maintain a set of visited cells to avoid cycles and to only explore unvisited neighbors.

Complexity Analysis

Time Complexity: O(n^2 log n)

Space Complexity: O(n^2)

Topics

This problem involves: Array, Binary Search, Depth-First Search, Breadth-First Search, Union Find, Heap (Priority Queue), Matrix.

Companies

Asked at: DE Shaw, DoorDash, PhonePe, WeRide.

Swim in Rising Water | LeetCode 778

Hard

Array Binary Search Depth-First Search Breadth-First Search Union Find Heap (Priority Queue)Matrix

Time: O(n^2 log n)Space: O(n^2)

Solution Code

Available in:

import java.util.PriorityQueue;
import java.util.HashSet;
import java.util.Set;

class Solution {
    public int swimInWater(int[][] grid) {
        int n = grid.length;
        PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> a[0] - b[0]); // (time, row, col)
        pq.offer(new int[]{grid[0][0], 0, 0});
        Set<String> visited = new HashSet<>();
        visited.add(0 + "," + 0);

        while (!pq.isEmpty()) {
            int[] curr = pq.poll();
            int time = curr[0];
            int r = curr[1];
            int c = curr[2];

            if (r == n - 1 && c == n - 1) {
                return time;
            }

            int[][] directions = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
            for (int[] dir : directions) {
                int nr = r + dir[0];
                int nc = c + dir[1];

                if (0 <= nr && nr < n && 0 <= nc && nc < n && !visited.contains(nr + "," + nc)) {
                    visited.add(nr + "," + nc);
                    int newTime = Math.max(time, grid[nr][nc]);
                    pq.offer(new int[]{newTime, nr, nc});
                }
            }
        }

        return -1; // Should not reach here if the grid is valid
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 778: Swim in Rising Water - Detailed explanation and solution approaches.

Topics

Array Binary Search Depth-First Search Breadth-First Search Union Find Heap (Priority Queue)Matrix

Companies

DE ShawDoorDashPhonePeWeRide

Stats

Difficulty:Hard

Languages:4

Companies:4

Problem #:778

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Initialize a priority queue `pq` with the starting cell (0, 0) and its elevation (grid[0][0]). Also, initialize a `visited` set to keep track of visited cells.
2
Step 2: While the priority queue is not empty, extract the cell with the minimum time from the priority queue.
3
Step 3: If the extracted cell is the destination cell (n-1, n-1), return the time associated with that cell.
4
Step 4: For each of the four neighbors (up, down, left, right) of the current cell:
5
Step 5: Check if the neighbor is within the grid boundaries and has not been visited.
6
Step 6: If the neighbor is valid, calculate the new required time to reach the neighbor as the maximum of the current time and the elevation of the neighbor.
7
Step 7: Add the neighbor to the `visited` set and push it into the priority queue with the new time and its row and column coordinates.
8
Step 8: Repeat steps 2-7 until the destination is reached.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n^2 log n)

💾

Space Complexity

O(n^2)

📚 Quadratic space - Uses lots of memory!

📝

Why?

The time complexity is dominated by the priority queue operations. The `visited` set ensures that each cell is processed at most once. The space complexity is due to the storage of the `visited` set and the priority queue.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Binary Search

Perform a binary search on the possible range of time values (0 to n^2 - 1). For each time value 't', use Depth First Search (DFS) or Breadth First Search (BFS) to check if a path exists from (0, 0) to (n-1, n-1) where all cell elevations along the path are less than or equal to 't'.

Time: O(n^2 log n), where n is the dimension of the grid. The binary search takes O(log n^2) = O(log n) steps, and each step involves a DFS/BFS which takes O(n^2) time.Space: O(n^2), due to the visited array for DFS/BFS.

Trade-offs: The binary search approach may be slightly less efficient in practice than the priority queue approach because the DFS/BFS has to traverse much of the graph for each binary search step. However it can be easier to understand initially.

Union Find

Iterate through all possible water levels (0 to n^2 - 1). At each water level 't', check if any new cells are submerged (elevation <= t). Use the Union-Find data structure to connect adjacent submerged cells. The algorithm terminates when the top-left and bottom-right cells are connected (i.e., they belong to the same connected component).

Time: O(n^2 α(n^2)), where α is the inverse Ackermann function, which grows very slowly. In practice, it's almost constant time. The n^2 part comes from the potential need to check all grid values as 't' increases.Space: O(n^2)

Trade-offs: Union Find can be more efficient if the elevations are processed in sorted order and the 'union' operations are relatively quick compared to the priority queue updates. However, implementation can be more complex.

✨

Key Insights

1Insight 1: The problem can be solved using a priority queue to explore the grid, prioritizing cells with the lowest required water level (time).
2Insight 2: Binary search can be used to find the minimum time, by checking if a path exists at a given time. However, the provided solution uses a more efficient priority queue approach directly.
3Insight 3: The key is to maintain a set of visited cells to avoid cycles and to only explore unvisited neighbors.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Edge Cases

Edge case 1: n = 1 (1x1 grid). The algorithm should correctly return the elevation of the single cell (grid[0][0]). This is handled correctly as the starting node itself is the destination.
Edge case 2: Large elevation differences between adjacent cells. The priority queue ensures that the algorithm explores the path with the minimum maximum elevation first.
Edge case 3: When all grid values are the same, algorithm should return the grid value. This case is correctly handled.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not keeping track of visited cells. This can lead to infinite loops and incorrect results.

✗Common mistake 2: Incorrectly calculating the new time to reach a neighbor. It should be the maximum of the current time and the neighbor's elevation, not the sum or some other operation.

✗Common mistake 3: Not handling the edge case where the grid is only 1x1. The code should return the elevation of the single cell in this case.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Pathfinding in a terrain with varying elevations and water levels. For example, finding the safest path for a robot to navigate a flooded area.

🏦Real-world scenario 2: Network routing with varying link capacities. The 'elevation' could represent the capacity of a network link, and the goal is to find the path with the highest minimum capacity.

🎮Real-world scenario 3: Resource allocation in a system with constraints. The 'elevation' could represent the resource requirement of a task, and the goal is to find a schedule that minimizes the maximum resource requirement at any given time.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 29, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I avoid off-by-one errors in binary search?

Use the pattern: left = 0, right = n-1, while (left <= right), mid = left + (right - left) / 2. Decide if you need to find exact match, first occurrence, or last occurrence, and adjust your conditions accordingly. Test with arrays of size 0, 1, and 2.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Swim in Rising Water"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether binary search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?