LeetCode #2684 — MEDIUM

Maximum Number of Moves 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 m x n matrix grid consisting of positive integers.

You can start at any cell in the first column of the matrix, and traverse the grid in the following way:

From a cell (row, col), you can move to any of the cells: (row - 1, col + 1), (row, col + 1) and (row + 1, col + 1) such that the value of the cell you move to, should be strictly bigger than the value of the current cell.

Return the maximum number of moves that you can perform.

Example 1:

Input: grid = [[2,4,3,5],[5,4,9,3],[3,4,2,11],[10,9,13,15]]
Output: 3
Explanation: We can start at the cell (0, 0) and make the following moves:
- (0, 0) -> (0, 1).
- (0, 1) -> (1, 2).
- (1, 2) -> (2, 3).
It can be shown that it is the maximum number of moves that can be made.

Example 2:


Input: grid = [[3,2,4],[2,1,9],[1,1,7]]
Output: 0
Explanation: Starting from any cell in the first column we cannot perform any moves.

Constraints:

m == grid.length
n == grid[i].length
2 <= m, n <= 1000
4 <= m * n <= 10⁵
1 <= grid[i][j] <= 10⁶

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed m x n matrix grid consisting of positive integers. You can start at any cell in the first column of the matrix, and traverse the grid in the following way: From a cell (row, col), you can move to any of the cells: (row - 1, col + 1), (row, col + 1) and (row + 1, col + 1) such that the value of the cell you move to, should be strictly bigger than the value of the current cell. Return the maximum number of moves that you can perform.

Baseline thinking

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

Pattern signal: Array · Dynamic Programming

Example 1

[[2,4,3,5],[5,4,9,3],[3,4,2,11],[10,9,13,15]]

Example 2

[[3,2,4],[2,1,9],[1,1,7]]

Step 02

Core Insight

What unlocks the optimal approach

Consider using dynamic programming to find the maximum number of moves that can be made from each cell.
The final answer will be the maximum value in cells of the first column.

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 #2684: Maximum Number of Moves in a Grid
class Solution {
    public int maxMoves(int[][] grid) {
        int m = grid.length, n = grid[0].length;
        Set<Integer> q = IntStream.range(0, m).boxed().collect(Collectors.toSet());
        for (int j = 0; j < n - 1; ++j) {
            Set<Integer> t = new HashSet<>();
            for (int i : q) {
                for (int k = i - 1; k <= i + 1; ++k) {
                    if (k >= 0 && k < m && grid[i][j] < grid[k][j + 1]) {
                        t.add(k);
                    }
                }
            }
            if (t.isEmpty()) {
                return j;
            }
            q = t;
        }
        return n - 1;
    }
}

// Accepted solution for LeetCode #2684: Maximum Number of Moves in a Grid
func maxMoves(grid [][]int) (ans int) {
	m, n := len(grid), len(grid[0])
	q := map[int]bool{}
	for i := range grid {
		q[i] = true
	}
	for j := 0; j < n-1; j++ {
		t := map[int]bool{}
		for i := range q {
			for k := i - 1; k <= i+1; k++ {
				if k >= 0 && k < m && grid[i][j] < grid[k][j+1] {
					t[k] = true
				}
			}
		}
		if len(t) == 0 {
			return j
		}
		q = t
	}
	return n - 1
}

# Accepted solution for LeetCode #2684: Maximum Number of Moves in a Grid
class Solution:
    def maxMoves(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])
        q = set(range(m))
        for j in range(n - 1):
            t = set()
            for i in q:
                for k in range(i - 1, i + 2):
                    if 0 <= k < m and grid[i][j] < grid[k][j + 1]:
                        t.add(k)
            if not t:
                return j
            q = t
        return n - 1

// Accepted solution for LeetCode #2684: Maximum Number of Moves in a Grid
/**
 * [2684] Maximum Number of Moves in a Grid
 */
pub struct Solution {}

// submission codes start here

impl Solution {
    pub fn max_moves(grid: Vec<Vec<i32>>) -> i32 {
        let m = grid.len();
        let n = grid[0].len();

        let mut dp = vec![vec![0; n]; m];

        for j in (0..n - 1).rev() {
            for i in 0..m {
                // [row - 1, col + 1]
                if i != 0 {
                    if grid[i - 1][j + 1] > grid[i][j] {
                        dp[i][j] = dp[i][j].max(dp[i - 1][j + 1] + 1);
                    }
                }

                // [row, col + 1]
                if grid[i][j + 1] > grid[i][j] {
                    dp[i][j] = dp[i][j].max(dp[i][j + 1] + 1);
                }

                // [row + 1, col + 1]
                if i != m - 1 {
                    if grid[i + 1][j + 1] > grid[i][j] {
                        dp[i][j] = dp[i][j].max(dp[i + 1][j + 1] + 1);
                    }
                }
            }
        }

        let mut result = 0;

        for i in 0..m {
            result = result.max(dp[i][0]);
        }

        result
    }
}

// submission codes end

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

    #[test]
    fn test_2684() {
        assert_eq!(
            3,
            Solution::max_moves(vec![
                vec![2, 4, 3, 5],
                vec![5, 4, 9, 3],
                vec![3, 4, 2, 11],
                vec![10, 9, 13, 15]
            ])
        );
        assert_eq!(
            0,
            Solution::max_moves(vec![vec![3, 2, 4], vec![2, 1, 9], vec![1, 1, 7]])
        );
    }
}

// Accepted solution for LeetCode #2684: Maximum Number of Moves in a Grid
function maxMoves(grid: number[][]): number {
    const m = grid.length;
    const n = grid[0].length;
    let q = new Set<number>(Array.from({ length: m }, (_, i) => i));
    for (let j = 0; j < n - 1; ++j) {
        const t = new Set<number>();
        for (const i of q) {
            for (let k = i - 1; k <= i + 1; ++k) {
                if (k >= 0 && k < m && grid[i][j] < grid[k][j + 1]) {
                    t.add(k);
                }
            }
        }
        if (t.size === 0) {
            return j;
        }
        q = t;
    }
    return 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)

Space

O(m)

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.