LeetCode #2658 — MEDIUM

Maximum Number of Fish in a Grid

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

Problem Statement

You are given a 0-indexed 2D matrix grid of size m x n, where (r, c) represents:

A land cell if grid[r][c] = 0, or
A water cell containing grid[r][c] fish, if grid[r][c] > 0.

A fisher can start at any water cell (r, c) and can do the following operations any number of times:

Catch all the fish at cell (r, c), or
Move to any adjacent water cell.

Return the maximum number of fish the fisher can catch if he chooses his starting cell optimally, or 0 if no water cell exists.

An adjacent cell of the cell (r, c), is one of the cells (r, c + 1), (r, c - 1), (r + 1, c) or (r - 1, c) if it exists.

Example 1:

Input: grid = [[0,2,1,0],[4,0,0,3],[1,0,0,4],[0,3,2,0]]
Output: 7
Explanation: The fisher can start at cell (1,3) and collect 3 fish, then move to cell (2,3) and collect 4 fish.

Example 2:

Input: grid = [[1,0,0,0],[0,0,0,0],[0,0,0,0],[0,0,0,1]]
Output: 1
Explanation: The fisher can start at cells (0,0) or (3,3) and collect a single fish.

Constraints:

m == grid.length
n == grid[i].length
1 <= m, n <= 10
0 <= grid[i][j] <= 10

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed 2D matrix grid of size m x n, where (r, c) represents: A land cell if grid[r][c] = 0, or A water cell containing grid[r][c] fish, if grid[r][c] > 0. A fisher can start at any water cell (r, c) and can do the following operations any number of times: Catch all the fish at cell (r, c), or Move to any adjacent water cell. Return the maximum number of fish the fisher can catch if he chooses his starting cell optimally, or 0 if no water cell exists. An adjacent cell of the cell (r, c), is one of the cells (r, c + 1), (r, c - 1), (r + 1, c) or (r - 1, c) if it exists.

Baseline thinking

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

Pattern signal: Array · Union-Find

Example 1

[[0,2,1,0],[4,0,0,3],[1,0,0,4],[0,3,2,0]]

Example 2

[[1,0,0,0],[0,0,0,0],[0,0,0,0],[0,0,0,1]]

Core Insight

What unlocks the optimal approach

Run DFS from each non-zero cell.
Each time you pick a cell to start from, add up the number of fish contained in the cells you visit.

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

Upper-end input sizes

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 #2658: Maximum Number of Fish in a Grid
class Solution {
    private int[][] grid;
    private int m;
    private int n;

    public int findMaxFish(int[][] grid) {
        m = grid.length;
        n = grid[0].length;
        this.grid = grid;
        int ans = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] > 0) {
                    ans = Math.max(ans, dfs(i, j));
                }
            }
        }
        return ans;
    }

    private int dfs(int i, int j) {
        int cnt = grid[i][j];
        grid[i][j] = 0;
        int[] dirs = {-1, 0, 1, 0, -1};
        for (int k = 0; k < 4; ++k) {
            int x = i + dirs[k], y = j + dirs[k + 1];
            if (x >= 0 && x < m && y >= 0 && y < n && grid[x][y] > 0) {
                cnt += dfs(x, y);
            }
        }
        return cnt;
    }
}

// Accepted solution for LeetCode #2658: Maximum Number of Fish in a Grid
func findMaxFish(grid [][]int) (ans int) {
	m, n := len(grid), len(grid[0])
	dirs := [5]int{-1, 0, 1, 0, -1}
	var dfs func(i, j int) int
	dfs = func(i, j int) int {
		cnt := grid[i][j]
		grid[i][j] = 0
		for k := 0; k < 4; k++ {
			x, y := i+dirs[k], j+dirs[k+1]
			if x >= 0 && x < m && y >= 0 && y < n && grid[x][y] > 0 {
				cnt += dfs(x, y)
			}
		}
		return cnt
	}
	for i := range grid {
		for j := range grid[i] {
			if grid[i][j] > 0 {
				ans = max(ans, dfs(i, j))
			}
		}
	}
	return
}

# Accepted solution for LeetCode #2658: Maximum Number of Fish in a Grid
class Solution:
    def findMaxFish(self, grid: List[List[int]]) -> int:
        def dfs(i: int, j: int) -> int:
            cnt = grid[i][j]
            grid[i][j] = 0
            for a, b in pairwise((-1, 0, 1, 0, -1)):
                x, y = i + a, j + b
                if 0 <= x < m and 0 <= y < n and grid[x][y]:
                    cnt += dfs(x, y)
            return cnt

        m, n = len(grid), len(grid[0])
        ans = 0
        for i in range(m):
            for j in range(n):
                if grid[i][j]:
                    ans = max(ans, dfs(i, j))
        return ans

// Accepted solution for LeetCode #2658: Maximum Number of Fish in a Grid
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2658: Maximum Number of Fish in a Grid
// class Solution {
//     private int[][] grid;
//     private int m;
//     private int n;
// 
//     public int findMaxFish(int[][] grid) {
//         m = grid.length;
//         n = grid[0].length;
//         this.grid = grid;
//         int ans = 0;
//         for (int i = 0; i < m; ++i) {
//             for (int j = 0; j < n; ++j) {
//                 if (grid[i][j] > 0) {
//                     ans = Math.max(ans, dfs(i, j));
//                 }
//             }
//         }
//         return ans;
//     }
// 
//     private int dfs(int i, int j) {
//         int cnt = grid[i][j];
//         grid[i][j] = 0;
//         int[] dirs = {-1, 0, 1, 0, -1};
//         for (int k = 0; k < 4; ++k) {
//             int x = i + dirs[k], y = j + dirs[k + 1];
//             if (x >= 0 && x < m && y >= 0 && y < n && grid[x][y] > 0) {
//                 cnt += dfs(x, y);
//             }
//         }
//         return cnt;
//     }
// }

// Accepted solution for LeetCode #2658: Maximum Number of Fish in a Grid
function findMaxFish(grid: number[][]): number {
    const m = grid.length;
    const n = grid[0].length;

    const dirs = [-1, 0, 1, 0, -1];
    const dfs = (i: number, j: number): number => {
        let cnt = grid[i][j];
        grid[i][j] = 0;
        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 && grid[x][y] > 0) {
                cnt += dfs(x, y);
            }
        }
        return cnt;
    };

    let ans = 0;
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            if (grid[i][j] > 0) {
                ans = Math.max(ans, dfs(i, j));
            }
        }
    }
    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)

Space

O(m × n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(n) space

Track components with a list or adjacency matrix. Each union operation may need to update all n elements’ component labels, giving O(n) per union. For n union operations total: O(n²). Find is O(1) with direct lookup, but union dominates.

UNION-FIND

O(α(n)) time

O(n) space

With path compression and union by rank, each find/union operation takes O(α(n)) amortized time, where α is the inverse Ackermann function — effectively constant. Space is O(n) for the parent and rank arrays. For m operations on n elements: O(m × α(n)) total.

Shortcut: Union-Find with path compression + rank → O(α(n)) per operation ≈ O(1). Just say “nearly constant.”

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.