# 28.01.2025 [2658. Maximum Number of Fish in a Grid]

Largest region in 2D grid #medium #bfs #dfs

28.01.2025

2658. Maximum Number of Fish in a Grid medium blog post substack youtube

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Problem TLDR

Largest region in 2D grid #medium #bfs #dfs

Intuition

Do Depth/Breadth-First Search from any cell with fish.

Approach

• we can reuse grid to mark visited cells

Complexity

• Time complexity: $$O(nm)$$

• Space complexity: $$O(nm)$$

Code

    fun findMaxFish(g: Array<IntArray>) = 
        (0..<g.size * g[0].size).maxOf { yx ->
            fun dfs(y: Int, x: Int): Int = 
                if (min(y, x) < 0 || y == g.size || x == g[0].size || g[y][x] < 1) 0
                else g[y][x].also { g[y][x] = 0 } + 
                    dfs(y - 1, x) + dfs(y + 1, x) + dfs(y, x - 1) + dfs(y, x + 1)
            dfs(yx / g[0].size, yx % g[0].size)
        }

    pub fn find_max_fish(mut g: Vec<Vec<i32>>) -> i32 {
        (0..g.len() * g[0].len()).map(|i| { let mut s = 0; 
            let mut q = VecDeque::from([(i / g[0].len(), i % g[0].len())]);
            while let Some((y, x)) = q.pop_front() { if g[y][x] > 0 {
                s += g[y][x]; g[y][x] = 0; 
                if y > 0 { q.push_back((y - 1, x))}
                if x > 0 { q.push_back((y, x - 1))}
                if y < g.len() - 1 { q.push_back((y + 1, x))}
                if x < g[0].len() - 1 { q.push_back((y, x + 1))}
            }}; s
        }).max().unwrap()
    }

    int findMaxFish(vector<vector<int>>& g) {
        int n = size(g), m = size(g[0]), r = 0;
        auto d = [&](this auto const& d, int y, int x) -> int {
            return min(y, x) < 0 || y == n || x == m || !g[y][x] ? 0 :
            exchange(g[y][x], 0) + d(y - 1, x) + d(y + 1, x) + d(y, x - 1) + d(y, x + 1);
        };
        for (int i = 0; i < m * n; ++i) r = max(r, d(i / m, i % m));
        return r;
    }