LeetCode #2617 — HARD

Minimum Number of Visited Cells in a Grid

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

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 the bottom-right cell (m - 1, n - 1). If there is no valid path, return -1.

Example 1:

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:

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:

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 <= 10⁵
1 <= m * n <= 10⁵
0 <= grid[i][j] < m * n
grid[m - 1][n - 1] == 0

Step 01

Brute Force Baseline

Problem summary: 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 the bottom-right cell (m - 1, n - 1). If there is no valid path, return -1.

Baseline thinking

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

Pattern signal: Array · Dynamic Programming · Stack · Union-Find

Example 1

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

Example 2

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

Example 3

[[2,1,0],[1,0,0]]

Core Insight

What unlocks the optimal approach

For each cell (i,j), it is critical to find out the minimum number of steps to reach (i,j), denoted dis[i][j], quickly, given the tight constraint.
Calculate dis[i][j] going left to right, top to bottom.
Suppose we want to calculate dis[i][j], keep track of a priority queue that stores (dis[i][k], i, k) for all k ≤ j, and another priority queue that stores (dis[k][j], k, j) for all k ≤ i.

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

Largest constraint values

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 #2617: Minimum Number of Visited Cells in a Grid
class Solution {
    public int minimumVisitedCells(int[][] grid) {
        int m = grid.length, n = grid[0].length;
        int[][] dist = new int[m][n];
        PriorityQueue<int[]>[] row = new PriorityQueue[m];
        PriorityQueue<int[]>[] col = new PriorityQueue[n];
        for (int i = 0; i < m; ++i) {
            Arrays.fill(dist[i], -1);
            row[i] = new PriorityQueue<>((a, b) -> a[0] == b[0] ? a[1] - b[1] : a[0] - b[0]);
        }
        for (int i = 0; i < n; ++i) {
            col[i] = new PriorityQueue<>((a, b) -> a[0] == b[0] ? a[1] - b[1] : a[0] - b[0]);
        }
        dist[0][0] = 1;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                while (!row[i].isEmpty() && grid[i][row[i].peek()[1]] + row[i].peek()[1] < j) {
                    row[i].poll();
                }
                if (!row[i].isEmpty() && (dist[i][j] == -1 || row[i].peek()[0] + 1 < dist[i][j])) {
                    dist[i][j] = row[i].peek()[0] + 1;
                }
                while (!col[j].isEmpty() && grid[col[j].peek()[1]][j] + col[j].peek()[1] < i) {
                    col[j].poll();
                }
                if (!col[j].isEmpty() && (dist[i][j] == -1 || col[j].peek()[0] + 1 < dist[i][j])) {
                    dist[i][j] = col[j].peek()[0] + 1;
                }
                if (dist[i][j] != -1) {
                    row[i].offer(new int[] {dist[i][j], j});
                    col[j].offer(new int[] {dist[i][j], i});
                }
            }
        }
        return dist[m - 1][n - 1];
    }
}

// Accepted solution for LeetCode #2617: Minimum Number of Visited Cells in a Grid
func minimumVisitedCells(grid [][]int) int {
	m, n := len(grid), len(grid[0])
	dist := make([][]int, m)
	row := make([]hp, m)
	col := make([]hp, n)
	for i := range dist {
		dist[i] = make([]int, n)
		for j := range dist[i] {
			dist[i][j] = -1
		}
	}
	dist[0][0] = 1
	for i := 0; i < m; i++ {
		for j := 0; j < n; j++ {
			for len(row[i]) > 0 && grid[i][row[i][0].second]+row[i][0].second < j {
				heap.Pop(&row[i])
			}
			if len(row[i]) > 0 && (dist[i][j] == -1 || row[i][0].first+1 < dist[i][j]) {
				dist[i][j] = row[i][0].first + 1
			}
			for len(col[j]) > 0 && grid[col[j][0].second][j]+col[j][0].second < i {
				heap.Pop(&col[j])
			}
			if len(col[j]) > 0 && (dist[i][j] == -1 || col[j][0].first+1 < dist[i][j]) {
				dist[i][j] = col[j][0].first + 1
			}
			if dist[i][j] != -1 {
				heap.Push(&row[i], pair{dist[i][j], j})
				heap.Push(&col[j], pair{dist[i][j], i})
			}
		}
	}
	return dist[m-1][n-1]
}

type pair struct {
	first  int
	second int
}

type hp []pair

func (a hp) Len() int      { return len(a) }
func (a hp) Swap(i, j int) { a[i], a[j] = a[j], a[i] }
func (a hp) Less(i, j int) bool {
	return a[i].first < a[j].first || (a[i].first == a[j].first && a[i].second < a[j].second)
}
func (a *hp) Push(x any) { *a = append(*a, x.(pair)) }
func (a *hp) Pop() any   { l := len(*a); t := (*a)[l-1]; *a = (*a)[:l-1]; return t }

# Accepted solution for LeetCode #2617: Minimum Number of Visited Cells in a Grid
class Solution:
    def minimumVisitedCells(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])
        dist = [[-1] * n for _ in range(m)]
        dist[0][0] = 1
        row = [[] for _ in range(m)]
        col = [[] for _ in range(n)]
        for i in range(m):
            for j in range(n):
                while row[i] and grid[i][row[i][0][1]] + row[i][0][1] < j:
                    heappop(row[i])
                if row[i] and (dist[i][j] == -1 or dist[i][j] > row[i][0][0] + 1):
                    dist[i][j] = row[i][0][0] + 1
                while col[j] and grid[col[j][0][1]][j] + col[j][0][1] < i:
                    heappop(col[j])
                if col[j] and (dist[i][j] == -1 or dist[i][j] > col[j][0][0] + 1):
                    dist[i][j] = col[j][0][0] + 1
                if dist[i][j] != -1:
                    heappush(row[i], (dist[i][j], j))
                    heappush(col[j], (dist[i][j], i))
        return dist[-1][-1]

// Accepted solution for LeetCode #2617: Minimum Number of Visited Cells in a Grid
/**
 * [2617] Minimum Number of Visited Cells in a Grid
 */
pub struct Solution {}

// submission codes start here
use std::{cmp::Reverse, collections::BinaryHeap};

impl Solution {
    pub fn minimum_visited_cells(grid: Vec<Vec<i32>>) -> i32 {
        let height = grid.len();
        let width = grid[0].len();

        let mut column_heaps: Vec<BinaryHeap<(Reverse<i32>, usize)>> =
            vec![BinaryHeap::new(); width];
        let mut row_heap: BinaryHeap<(Reverse<i32>, usize)> = BinaryHeap::new();

        let mut dis = 0;
        for (i, row) in grid.iter().enumerate() {
            row_heap.clear();

            for (j, &node) in row.iter().enumerate() {
                while !row_heap.is_empty() && row_heap.peek().unwrap().1 < j {
                    row_heap.pop();
                }

                let column_heap = &mut column_heaps[j];
                while !column_heap.is_empty() && column_heap.peek().unwrap().1 < i {
                    column_heap.pop();
                }

                dis = if i > 0 || j > 0 { i32::MAX } else { 1 };

                if let Some((d, _)) = row_heap.peek() {
                    dis = d.0 + 1;
                }

                if let Some((d, _)) = column_heap.peek() {
                    dis = dis.min(d.0 + 1);
                }

                if node > 0 && dis < i32::MAX {
                    let node = node as usize;
                    row_heap.push((Reverse(dis), node + j));
                    column_heap.push((Reverse(dis), node + i));
                }
            }
        }

        if dis < i32::MAX {
            dis
        } else {
            -1
        }
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_2617() {}
}

// Accepted solution for LeetCode #2617: Minimum Number of Visited Cells in a Grid
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #2617: Minimum Number of Visited Cells in a Grid
// class Solution {
//     public int minimumVisitedCells(int[][] grid) {
//         int m = grid.length, n = grid[0].length;
//         int[][] dist = new int[m][n];
//         PriorityQueue<int[]>[] row = new PriorityQueue[m];
//         PriorityQueue<int[]>[] col = new PriorityQueue[n];
//         for (int i = 0; i < m; ++i) {
//             Arrays.fill(dist[i], -1);
//             row[i] = new PriorityQueue<>((a, b) -> a[0] == b[0] ? a[1] - b[1] : a[0] - b[0]);
//         }
//         for (int i = 0; i < n; ++i) {
//             col[i] = new PriorityQueue<>((a, b) -> a[0] == b[0] ? a[1] - b[1] : a[0] - b[0]);
//         }
//         dist[0][0] = 1;
//         for (int i = 0; i < m; ++i) {
//             for (int j = 0; j < n; ++j) {
//                 while (!row[i].isEmpty() && grid[i][row[i].peek()[1]] + row[i].peek()[1] < j) {
//                     row[i].poll();
//                 }
//                 if (!row[i].isEmpty() && (dist[i][j] == -1 || row[i].peek()[0] + 1 < dist[i][j])) {
//                     dist[i][j] = row[i].peek()[0] + 1;
//                 }
//                 while (!col[j].isEmpty() && grid[col[j].peek()[1]][j] + col[j].peek()[1] < i) {
//                     col[j].poll();
//                 }
//                 if (!col[j].isEmpty() && (dist[i][j] == -1 || col[j].peek()[0] + 1 < dist[i][j])) {
//                     dist[i][j] = col[j].peek()[0] + 1;
//                 }
//                 if (dist[i][j] != -1) {
//                     row[i].offer(new int[] {dist[i][j], j});
//                     col[j].offer(new int[] {dist[i][j], i});
//                 }
//             }
//         }
//         return dist[m - 1][n - 1];
//     }
// }

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 × log (m × n)

Space

O(m × n)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

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.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Breaking monotonic invariant

Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.

Usually fails on: Indices point to blocked elements and outputs shift.

Fix: Pop while invariant is violated before pushing current element.