Watch 10 video solutions for Swim in Rising Water, a hard level problem involving Array, Binary Search, Depth-First Search. This walkthrough by NeetCode has 80,300 views views. Want to try solving it yourself? Practice on FleetCode or read the detailed text solution.
You are given an n x n integer matrix grid where each value grid[i][j] represents the elevation at that point (i, j).
The rain starts to fall. At time t, the depth of the water everywhere is t. 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 stay within the boundaries of the grid during your swim.
Return the least time until you can reach the bottom right square (n - 1, n - 1) if you start at the top left square (0, 0).
Example 1:
Input: grid = [[0,2],[1,3]] Output: 3 Explanation: At time 0, you are in grid location (0, 0). You cannot go anywhere else because 4-directionally adjacent neighbors have a higher elevation than t = 0. You cannot reach point (1, 1) until time 3. When the depth of water is 3, we can swim anywhere inside the grid.
Example 2:
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 Explanation: The final route is shown. We need to wait until time 16 so that (0, 0) and (4, 4) are connected.
Constraints:
n == grid.lengthn == grid[i].length1 <= n <= 500 <= grid[i][j] < n2grid[i][j] is unique.Problem Overview: You are given an array-based n x n grid where each cell represents the elevation of the land. Water rises over time, and at time t you can only enter cells with elevation ≤ t. The goal is to reach the bottom-right cell from the top-left as early as possible.
Approach 1: Dijkstra's Algorithm with Min-Heap (Time: O(n2 log n), Space: O(n2))
This problem behaves like a shortest-path problem where the "cost" of a path is the maximum elevation encountered along the route. Use a min-heap (priority queue) to always expand the cell with the smallest elevation seen so far. Start from (0,0), push it into the heap, and repeatedly pop the lowest elevation cell. From each cell, explore its four neighbors and push them into the heap if they haven't been visited. The key insight: the moment you pop the destination cell, the maximum elevation along that path is the minimum time required. This works because Dijkstra always processes the smallest possible "time" first.
The running time is dominated by heap operations across at most n^2 cells, giving O(n^2 log n). A visited matrix tracks processed cells, requiring O(n^2) space.
Approach 2: Binary Search with DFS (Time: O(n2 log n), Space: O(n2))
Another perspective: instead of constructing the path directly, search for the minimum time t that makes the destination reachable. Use binary search over the time range from grid[0][0] to n^2 - 1. For each candidate time, run a DFS (or BFS) starting at (0,0) and only move to cells whose elevation is ≤ t. If the destination becomes reachable, reduce the search range; otherwise increase it.
The DFS explores at most n^2 cells per check. Binary search runs about log(n^2) iterations, producing overall complexity of O(n^2 log n). This approach is conceptually simple because it separates "reachability" from "time optimization".
Recommended for interviews: Dijkstra with a min-heap is the most commonly expected solution. It demonstrates understanding of priority queues and shortest-path reasoning on grids. Binary search + DFS is also strong because it shows problem transformation skills—turning a path problem into a monotonic feasibility check. Explaining both approaches clearly signals strong algorithmic thinking.
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Dijkstra's Algorithm with Min-Heap | O(n^2 log n) | O(n^2) | Best general solution for grid path problems with weighted constraints |
| Binary Search + DFS/BFS | O(n^2 log n) | O(n^2) | When the answer is monotonic and can be validated via reachability |
| Union Find (Disjoint Set) | O(n^2 log n) | O(n^2) | Useful when processing cells in increasing elevation order and merging components |