Swim in Rising Water

Problem Highlights

• 🔗 Leetcode Link: https://leetcode.com/problems/swim-in-rising-water/
• 💡 Problem Difficulty: Hard
• ⏰ Time to complete: __ mins
• 🛠️ Topics: Graphs, Depth-First Search, Breadth-First Search
• 🗒️ Similar Questions: Path With Minimum Effort

1: U-nderstand

Understand what the interviewer is asking for by using test cases and questions about the problem.

• Established a set (2-3) of test cases to verify their own solution later.
• Established a set (1-2) of edge cases to verify their solution handles complexities.
• Have fully understood the problem and have no clarifying questions.
• Have you verified any Time/Space Constraints for this problem?

• Can we think of the grid as a weighted graph?

  • Yes, to solve this problem, we have to think about the grid as a weighted graph. Each cell is a node. An edge is a connector between two of such nodes. The weight of an edge is calculated as the maximum value of two nodes connected by that edge. That way we reduced the problem to graph traversal problem with some constraints. We no longer want to search for the shortest path, but a path with the minimum-maximum value.

• Where do I start on the grid?

  • Start with (0,0)corner. On each moment of time we choose node with smallest value to visit. In this way when we reached (N-1, N-1)corner, the answer will be the maximum of visited cells so far.

    HAPPY CASE
    Input: grid = [[0,2],[1,3]]
    Output: 3
    
    Input: grid = [[0,1,2,3,4],[24,23,22,21,5],[12,13,14,15,16],[11,17,18,19,20],[10,9,8,7,6]]
    Output: 16

2: M-atch

Match what this problem looks like to known categories of problems, e.g. Linked List or Dynamic Programming, and strategies or patterns in those categories.

For graph problems, some things we want to consider are:

• BFS: We can utilize BFS and Djikstra’s algorithm. This problem is to find shortest path, the only thing we need to pay attention to is the weight of edge from grid[0][0] to grid[n-1][n-1]. We can take grid 0, 1], [2, 3 as an example: We can treat this as a graph with 4 vertices 0,1,2,3 and there are 4 edges. {0, 1} with weight 1, {0, 2} with weight 2, {1, 3} with weight 3 and {2, 3} with weight 3 respectively. Each time we choose the smallest edge {m, n} to simulate at time max(m, n) and rain falls. So we first take edge {0, 1} then {0, 2}. When we are processing edge {1, 3}, the original result is 2(getting from {0, 2}), but since original time is less than we currently need, we need time 3 to reach vertex 3.

For another case like grid 3, 2], [1, 0, the first part is the same. To reach from vertex 3 to vertex 1, we need time 3. Then to reach from vertex 1 to vertex 0, we also only need time 3, since we can move infinite distance in one move. We can see this like at time 3, water can move from vertex 3 to vertex 1 then vertex 0 directly.

From 2 cases above, we can get the most important thing in this problem that the time needs to reach grid[i][j] is max(grid[i][j], grid[i'][j']), where grid[i'][j'] is the cell to get to grid[i][j].

• DFS: Another option is to a DFS. How do we know that a path exists? Here, we will travel in all the four directions (up, down, left, right) by not visiting the node we have visited earlier. And in this traversal we start from [0,0] and if we happen to touch [n-1,m-1], then we can say that a path exists. This is general Depth First Traversal, but we have another constraint that we cannot move to a height > t, so we include a constraint that we can move in any of the four directions, if and only if, the height of the building in that direction is less than t.
• Adjacency List: We can use an adjacency list to store graph, but equation variables cannot be directly translated to an index.
• Adjacency Matrix: We can use an adjacency matrix to store graph, but will cause runtime slowdowns of O(N^2) for a sparse graph and run into similar issues as with adjacency list.
• Map: We can use a map to store the edges in the graph to lookup equations by name, and store all neighbors by name.
• Topological Sort: We can use topological sort to traverse the graph, similar to BFS and DFS, and find the result.
• Union Find: Are there find and union operations here? Can you perform a find operation where you can determine which subset a particular element is in? This can be used for determining if two elements are in the same subset. Can you perform a union operation where you join two subsets into a single subset? Can you check if the two subsets belong to same set? If no, then we cannot perform union.

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

- We start from 0, 0, and push this onto priority_queue
- Then push all possible moves from this position onto queue
- Queue is designed such that the first move to be processed will be the one with lowest value in grid[][]
- When moving to new position, if the new position value is lower then previous, retain previous

Continue this until bottom right element is filled. This will be minimum.

⚠️ Common Mistakes

• The brute force solution may not garner the best time T. We can use either Dijkstra's, or binary search for the best time T where you can reach the end if you only step on squares at most T.

4: I-mplement

Implement the code to solve the algorithm.

    class Solution {
        int[][] directions = new int[][]{{0,1}, {0,-1}, {1,0}, {-1,0}};
        public int swimInWater(int[][] grid) {
            int rowLen = grid.length;
            int colLen = grid[0].length;
            Queue<int[]> pq = new PriorityQueue<>((a,b) -> Integer.compare(a[2],b[2]));
            
            pq.add(new int[]{0, 0, grid[0][0]});
            
            int result = grid[rowLen-1][colLen-1];
            
            boolean[][] visited = new boolean[rowLen][colLen];
            
            while(pq.isEmpty()==false) {
                int[] currentCell = pq.poll();
                
                result = Math.max(result, currentCell[2]);
                
                visited[currentCell[0]][currentCell[1]] = true;
                
                if(currentCell[0]==rowLen-1 && currentCell[1]==colLen-1) {
                    break;
                }
                
                for(int[] dir : directions) {
                    int newRow = currentCell[0] + dir[0];
                    int newCol = currentCell[1] + dir[1];
                    
                    if(newRow < 0 || newRow >= rowLen || newCol < 0 || newCol >= colLen) {
                        continue;
                    }
                    
                    if(visited[newRow][newCol]) {
                        continue;
                    }
                    
                    pq.add(new int[]{newRow, newCol, grid[newRow][newCol]});
                }
            }
            return result;
        }
    }

    class Solution:
        def swimInWater(self, grid: List[List[int]]) -> int:
            directions = [(0, 1), (0, -1), (-1, 0), (1, 0)]
            N = len(grid)
            visited = [[False] * N for _ in range(N)]
    
            pq = []
            ans = 0
            heapq.heappush(pq, (grid[0][0], 0, 0))
    
            while pq:
                time, r, c = heapq.heappop(pq)
                if visited[r][c]:
                    continue
                visited[r][c] = True
                ans = max(ans, time)
    
                if r == N - 1 and c == N - 1:
                    break
    
                for dir_r, dir_c in directions:
                    rr, cc = r + dir_r, c + dir_c
                    if not self._isValid(rr, cc, N, visited):
                        continue
                    heapq.heappush(pq, (grid[rr][cc], rr, cc))
    
            return ans
    
        def _isValid(self, r, c, N, visited):
            return r >= 0 and r < N and c >= 0 and c < N and not visited[r][c]

5: R-eview

Review the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.

• Trace through your code with an input to check for the expected output
• Catch possible edge cases and off-by-one errors and verify the code works for the happy and edge cases you created in the “Understand” section

6: E-valuate

Evaluate the performance of your algorithm and state any strong/weak or future potential work.

Time Complexity: O(r*c* logn), where r, c = rows and columns in grid and n is the max cell elevation found in the grid
Space Complexity: O(r * c), where r, c = rows and columns in grid and n is the max cell elevation found in the grid