Minimum Number of Visited Cells in a Grid

Problem

You are given a 0-indexed m x n integer matrix grid. Your initial position is at the top-left cell (0, 0).

Starting from the cell (i, j), you can move to one of the following cells:

• Cells (i, k) with j < k <= grid[i][j] + j (rightward movement), or
• Cells (k, j) with i < k <= grid[i][j] + i (downward movement).

Return the minimum number of cells you need to visit to reach thebottom-right cell (m - 1, n - 1). If there is no valid path, return -1.

Examples

Example 1

![](https://assets.leetcode.com/uploads/2023/01/25/ex1.png)

    
    
    Input: grid = [[3,4,2,1],[4,2,3,1],[2,1,0,0],[2,4,0,0]]
    Output: 4
    Explanation: The image above shows one of the paths that visits exactly 4 cells.
    

Example 2

![](https://assets.leetcode.com/uploads/2023/01/25/ex2.png)

    
    
    Input: grid = [[3,4,2,1],[4,2,1,1],[2,1,1,0],[3,4,1,0]]
    Output: 3
    Explanation: The image above shows one of the paths that visits exactly 3 cells.
    

Example 3

![](https://assets.leetcode.com/uploads/2023/01/26/ex3.png)

    
    
    Input: grid = [[2,1,0],[1,0,0]]
    Output: -1
    Explanation: It can be proven that no path exists.
    

Constraints

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 10^5
• 1 <= m * n <= 10^5
• 0 <= grid[i][j] < m * n
• grid[m - 1][n - 1] == 0

Solution

Method 1 – BFS with Monotonic Queue

Intuition

We want the shortest path from (0,0) to (m-1,n-1) with allowed moves. BFS finds the minimum steps, but to avoid revisiting cells, we use monotonic queues to efficiently track reachable cells in each row and column.

Approach

1. Use BFS starting from (0,0), tracking steps for each cell.
2. For each cell, use two arrays of monotonic queues to track the next unvisited cell in the same row and column.
3. For each move, push reachable cells into the queue and mark as visited.
4. Stop when reaching (m-1,n-1) or queue is empty.
5. Return the minimum steps or -1 if unreachable.

Code

C++

#include <vector>
#include <queue>
#include <set>
using namespace std;
class Solution {
public:
    int minVisitedCells(vector<vector<int>>& grid) {
        int m = grid.size(), n = grid[0].size();
        vector<set<int>> row(m), col(n);
        for (int i = 0; i < m; ++i)
            for (int j = 0; j < n; ++j) {
                row[i].insert(j);
                col[j].insert(i);
            }
        queue<pair<int, int>> q;
        q.push({0, 0});
        vector<vector<int>> dist(m, vector<int>(n, -1));
        dist[0][0] = 1;
        while (!q.empty()) {
            auto [i, j] = q.front(); q.pop();
            if (i == m - 1 && j == n - 1) return dist[i][j];
            auto it = row[i].upper_bound(j);
            while (it != row[i].end() && *it <= j + grid[i][j]) {
                int nj = *it;
                q.push({i, nj});
                dist[i][nj] = dist[i][j] + 1;
                row[i].erase(it++);
                col[nj].erase(i);
            }
            auto it2 = col[j].upper_bound(i);
            while (it2 != col[j].end() && *it2 <= i + grid[i][j]) {
                int ni = *it2;
                q.push({ni, j});
                dist[ni][j] = dist[i][j] + 1;
                col[j].erase(it2++);
                row[ni].erase(j);
            }
        }
        return -1;
    }
};

Go

func minVisitedCells(grid [][]int) int {
    m, n := len(grid), len(grid[0])
    row := make([]map[int]struct{}, m)
    col := make([]map[int]struct{}, n)
    for i := range row {
        row[i] = make(map[int]struct{})
        for j := 0; j < n; j++ {
            row[i][j] = struct{}{}
        }
    }
    for j := range col {
        col[j] = make(map[int]struct{})
        for i := 0; i < m; i++ {
            col[j][i] = struct{}{}
        }
    }
    type pair struct{ i, j int }
    q := []pair{{0, 0}}
    dist := make([][]int, m)
    for i := range dist {
        dist[i] = make([]int, n)
        for j := range dist[i] {
            dist[i][j] = -1
        }
    }
    dist[0][0] = 1
    for len(q) > 0 {
        p := q[0]
        q = q[1:]
        i, j := p.i, p.j
        if i == m-1 && j == n-1 {
            return dist[i][j]
        }
        for nj := j + 1; nj <= j+grid[i][j] && nj < n; nj++ {
            if _, ok := row[i][nj]; ok {
                q = append(q, pair{i, nj})
                dist[i][nj] = dist[i][j] + 1
                delete(row[i], nj)
                delete(col[nj], i)
            }
        }
        for ni := i + 1; ni <= i+grid[i][j] && ni < m; ni++ {
            if _, ok := col[j][ni]; ok {
                q = append(q, pair{ni, j})
                dist[ni][j] = dist[i][j] + 1
                delete(col[j], ni)
                delete(row[ni], j)
            }
        }
    }
    return -1
}

Java

import java.util.*;
class Solution {
    public int minVisitedCells(int[][] grid) {
        int m = grid.length, n = grid[0].length;
        Set<Integer>[] row = new HashSet[m];
        Set<Integer>[] col = new HashSet[n];
        for (int i = 0; i < m; ++i) {
            row[i] = new HashSet<>();
            for (int j = 0; j < n; ++j) row[i].add(j);
        }
        for (int j = 0; j < n; ++j) {
            col[j] = new HashSet<>();
            for (int i = 0; i < m; ++i) col[j].add(i);
        }
        Queue<int[]> q = new LinkedList<>();
        q.add(new int[]{0, 0});
        int[][] dist = new int[m][n];
        for (int[] d : dist) Arrays.fill(d, -1);
        dist[0][0] = 1;
        while (!q.isEmpty()) {
            int[] p = q.poll();
            int i = p[0], j = p[1];
            if (i == m - 1 && j == n - 1) return dist[i][j];
            for (int nj = j + 1; nj <= j + grid[i][j] && nj < n; nj++) {
                if (row[i].contains(nj)) {
                    q.add(new int[]{i, nj});
                    dist[i][nj] = dist[i][j] + 1;
                    row[i].remove(nj);
                    col[nj].remove(i);
                }
            }
            for (int ni = i + 1; ni <= i + grid[i][j] && ni < m; ni++) {
                if (col[j].contains(ni)) {
                    q.add(new int[]{ni, j});
                    dist[ni][j] = dist[i][j] + 1;
                    col[j].remove(ni);
                    row[ni].remove(j);
                }
            }
        }
        return -1;
    }
}

Kotlin

import java.util.LinkedList
class Solution {
    fun minVisitedCells(grid: Array<IntArray>): Int {
        val m = grid.size
        val n = grid[0].size
        val row = Array(m) { mutableSetOf<Int>() }
        val col = Array(n) { mutableSetOf<Int>() }
        for (i in 0 until m) for (j in 0 until n) row[i].add(j)
        for (j in 0 until n) for (i in 0 until m) col[j].add(i)
        val q = LinkedList<Pair<Int, Int>>()
        q.add(0 to 0)
        val dist = Array(m) { IntArray(n) { -1 } }
        dist[0][0] = 1
        while (q.isNotEmpty()) {
            val (i, j) = q.poll()
            if (i == m - 1 && j == n - 1) return dist[i][j]
            for (nj in j + 1..(j + grid[i][j]).coerceAtMost(n - 1)) {
                if (row[i].remove(nj)) {
                    q.add(i to nj)
                    dist[i][nj] = dist[i][j] + 1
                    col[nj].remove(i)
                }
            }
            for (ni in i + 1..(i + grid[i][j]).coerceAtMost(m - 1)) {
                if (col[j].remove(ni)) {
                    q.add(ni to j)
                    dist[ni][j] = dist[i][j] + 1
                    row[ni].remove(j)
                }
            }
        }
        return -1
    }
}

Python

from collections import deque
class Solution:
    def minVisitedCells(self, grid: list[list[int]]) -> int:
        m, n = len(grid), len(grid[0])
        row = [set(range(n)) for _ in range(m)]
        col = [set(range(m)) for _ in range(n)]
        q = deque([(0, 0)])
        dist = [[-1] * n for _ in range(m)]
        dist[0][0] = 1
        while q:
            i, j = q.popleft()
            if i == m - 1 and j == n - 1:
                return dist[i][j]
            for nj in range(j + 1, min(j + grid[i][j] + 1, n)):
                if nj in row[i]:
                    q.append((i, nj))
                    dist[i][nj] = dist[i][j] + 1
                    row[i].remove(nj)
                    col[nj].remove(i)
            for ni in range(i + 1, min(i + grid[i][j] + 1, m)):
                if ni in col[j]:
                    q.append((ni, j))
                    dist[ni][j] = dist[i][j] + 1
                    col[j].remove(ni)
                    row[ni].remove(j)
        return -1

Rust

use std::collections::{VecDeque, HashSet};
impl Solution {
    pub fn min_visited_cells(grid: Vec<Vec<i32>>) -> i32 {
        let m = grid.len();
        let n = grid[0].len();
        let mut row: Vec<HashSet<usize>> = vec![HashSet::new(); m];
        let mut col: Vec<HashSet<usize>> = vec![HashSet::new(); n];
        for i in 0..m {
            for j in 0..n {
                row[i].insert(j);
                col[j].insert(i);
            }
        }
        let mut q = VecDeque::new();
        q.push_back((0, 0));
        let mut dist = vec![vec![-1; n]; m];
        dist[0][0] = 1;
        while let Some((i, j)) = q.pop_front() {
            if i == m - 1 && j == n - 1 {
                return dist[i][j];
            }
            for nj in (j + 1)..=(j + grid[i][j] as usize).min(n - 1) {
                if row[i].remove(&nj) {
                    q.push_back((i, nj));
                    dist[i][nj] = dist[i][j] + 1;
                    col[nj].remove(&i);
                }
            }
            for ni in (i + 1)..=(i + grid[i][j] as usize).min(m - 1) {
                if col[j].remove(&ni) {
                    q.push_back((ni, j));
                    dist[ni][j] = dist[i][j] + 1;
                    row[ni].remove(&j);
                }
            }
        }
        -1
    }
}

TypeScript

class Solution {
    minVisitedCells(grid: number[][]): number {
        const m = grid.length, n = grid[0].length;
        const row: Set<number>[] = Array.from({length: m}, () => new Set<number>());
        const col: Set<number>[] = Array.from({length: n}, () => new Set<number>());
        for (let i = 0; i < m; ++i) for (let j = 0; j < n; ++j) row[i].add(j);
        for (let j = 0; j < n; ++j) for (let i = 0; i < m; ++i) col[j].add(i);
        const dist: number[][] = Array.from({length: m}, () => Array(n).fill(-1));
        dist[0][0] = 1;
        let q: [number, number][] = [[0, 0]];
        while (q.length) {
            const [i, j] = q.shift()!;
            if (i === m - 1 && j === n - 1) return dist[i][j];
            for (let nj = j + 1; nj <= Math.min(j + grid[i][j], n - 1); ++nj) {
                if (row[i].has(nj)) {
                    q.push([i, nj]);
                    dist[i][nj] = dist[i][j] + 1;
                    row[i].delete(nj);
                    col[nj].delete(i);
                }
            }
            for (let ni = i + 1; ni <= Math.min(i + grid[i][j], m - 1); ++ni) {
                if (col[j].has(ni)) {
                    q.push([ni, j]);
                    dist[ni][j] = dist[i][j] + 1;
                    col[j].delete(ni);
                    row[ni].delete(j);
                }
            }
        }
        return -1;
    }
}

Complexity

• ⏰ Time complexity: O(m * n) — Each cell is visited at most once.
• 🧺 Space complexity: O(m * n) — For visited sets and distance array.