LeetCode #576 — MEDIUM

Out of Boundary Paths

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

The Problem

Problem Statement

There is an m x n grid with a ball. The ball is initially at the position [startRow, startColumn]. You are allowed to move the ball to one of the four adjacent cells in the grid (possibly out of the grid crossing the grid boundary). You can apply at most maxMove moves to the ball.

Given the five integers m, n, maxMove, startRow, startColumn, return the number of paths to move the ball out of the grid boundary. Since the answer can be very large, return it modulo 10⁹ + 7.

Example 1:

Input: m = 2, n = 2, maxMove = 2, startRow = 0, startColumn = 0
Output: 6

Example 2:

Input: m = 1, n = 3, maxMove = 3, startRow = 0, startColumn = 1
Output: 12

Constraints:

1 <= m, n <= 50
0 <= maxMove <= 50
0 <= startRow < m
0 <= startColumn < n

Step 01

Brute Force Baseline

Problem summary: There is an m x n grid with a ball. The ball is initially at the position [startRow, startColumn]. You are allowed to move the ball to one of the four adjacent cells in the grid (possibly out of the grid crossing the grid boundary). You can apply at most maxMove moves to the ball. Given the five integers m, n, maxMove, startRow, startColumn, return the number of paths to move the ball out of the grid boundary. Since the answer can be very large, return it modulo 109 + 7.

Baseline thinking

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

Pattern signal: Dynamic Programming

Example 1

Example 2

Core Insight

What unlocks the optimal approach

Is traversing every path feasible? There are many possible paths for a small matrix. Try to optimize it.
Can we use some space to store the number of paths and update them after every move?
One obvious thing: the ball will go out of the boundary only by crossing it. Also, there is only one possible way the ball can go out of the boundary from the boundary cell except for corner cells. From the corner cell, the ball can go out in two different ways. Can you use this thing to solve the problem?

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 #576: Out of Boundary Paths
class Solution {
    private int m, n;
    private Integer[][][] f;
    private final int mod = (int) 1e9 + 7;

    public int findPaths(int m, int n, int maxMove, int startRow, int startColumn) {
        this.m = m;
        this.n = n;
        f = new Integer[m][n][maxMove + 1];
        return dfs(startRow, startColumn, maxMove);
    }

    private int dfs(int i, int j, int k) {
        if (i < 0 || i >= m || j < 0 || j >= n) {
            return k >= 0 ? 1 : 0;
        }
        if (k <= 0) {
            return 0;
        }
        if (f[i][j][k] != null) {
            return f[i][j][k];
        }
        int ans = 0;
        final int[] dirs = {-1, 0, 1, 0, -1};
        for (int d = 0; d < 4; ++d) {
            int x = i + dirs[d], y = j + dirs[d + 1];
            ans = (ans + dfs(x, y, k - 1)) % mod;
        }
        return f[i][j][k] = ans;
    }
}

// Accepted solution for LeetCode #576: Out of Boundary Paths
func findPaths(m int, n int, maxMove int, startRow int, startColumn int) int {
	f := make([][][]int, m)
	for i := range f {
		f[i] = make([][]int, n)
		for j := range f[i] {
			f[i][j] = make([]int, maxMove+1)
			for k := range f[i][j] {
				f[i][j][k] = -1
			}
		}
	}
	const mod int = 1e9 + 7
	var dfs func(int, int, int) int
	dirs := [5]int{-1, 0, 1, 0, -1}
	dfs = func(i, j, k int) int {
		if i < 0 || i >= m || j < 0 || j >= n {
			if k >= 0 {
				return 1
			}
			return 0
		}
		if k <= 0 {
			return 0
		}
		if f[i][j][k] != -1 {
			return f[i][j][k]
		}
		ans := 0
		for d := 0; d < 4; d++ {
			x, y := i+dirs[d], j+dirs[d+1]
			ans = (ans + dfs(x, y, k-1)) % mod
		}
		f[i][j][k] = ans
		return ans
	}
	return dfs(startRow, startColumn, maxMove)
}

# Accepted solution for LeetCode #576: Out of Boundary Paths
class Solution:
    def findPaths(
        self, m: int, n: int, maxMove: int, startRow: int, startColumn: int
    ) -> int:
        @cache
        def dfs(i: int, j: int, k: int) -> int:
            if not 0 <= i < m or not 0 <= j < n:
                return int(k >= 0)
            if k <= 0:
                return 0
            ans = 0
            for a, b in pairwise(dirs):
                x, y = i + a, j + b
                ans = (ans + dfs(x, y, k - 1)) % mod
            return ans

        mod = 10**9 + 7
        dirs = (-1, 0, 1, 0, -1)
        return dfs(startRow, startColumn, maxMove)

// Accepted solution for LeetCode #576: Out of Boundary Paths
struct Solution;
use std::collections::HashMap;

const MOD: i32 = 1_000_000_007;

impl Solution {
    fn find_paths(m: i32, n: i32, k: i32, i: i32, j: i32) -> i32 {
        let mut memo: HashMap<(usize, usize, usize), i32> = HashMap::new();
        let m = m as usize;
        let n = n as usize;
        let k = k as usize;
        let i = i as usize;
        let j = j as usize;
        Self::dp(i, j, k, &mut memo, m, n)
    }
    fn dp(
        i: usize,
        j: usize,
        k: usize,
        memo: &mut HashMap<(usize, usize, usize), i32>,
        n: usize,
        m: usize,
    ) -> i32 {
        if k == 0 {
            0
        } else {
            if let Some(&res) = memo.get(&(i, j, k)) {
                return res;
            }
            let top = if i > 0 {
                Self::dp(i - 1, j, k - 1, memo, n, m)
            } else {
                1
            };
            let left = if j > 0 {
                Self::dp(i, j - 1, k - 1, memo, n, m)
            } else {
                1
            };
            let bottom = if i + 1 < n {
                Self::dp(i + 1, j, k - 1, memo, n, m)
            } else {
                1
            };
            let right = if j + 1 < m {
                Self::dp(i, j + 1, k - 1, memo, n, m)
            } else {
                1
            };
            let mut res = 0;
            res += top;
            res %= MOD;
            res += left;
            res %= MOD;
            res += right;
            res %= MOD;
            res += bottom;
            res %= MOD;
            memo.insert((i, j, k), res);
            res
        }
    }
}

#[test]
fn test() {
    let m = 2;
    let n = 2;
    let k = 2;
    let i = 0;
    let j = 0;
    let res = 6;
    assert_eq!(Solution::find_paths(m, n, k, i, j), res);
    let m = 1;
    let n = 3;
    let k = 3;
    let i = 0;
    let j = 1;
    let res = 12;
    assert_eq!(Solution::find_paths(m, n, k, i, j), res);
    let m = 36;
    let n = 5;
    let k = 50;
    let i = 15;
    let j = 3;
    let res = 390153306;
    assert_eq!(Solution::find_paths(m, n, k, i, j), res);
}

// Accepted solution for LeetCode #576: Out of Boundary Paths
function findPaths(
    m: number,
    n: number,
    maxMove: number,
    startRow: number,
    startColumn: number,
): number {
    const f = Array.from({ length: m }, () =>
        Array.from({ length: n }, () => Array(maxMove + 1).fill(-1)),
    );
    const mod = 1000000007;
    const dirs = [-1, 0, 1, 0, -1];
    const dfs = (i: number, j: number, k: number): number => {
        if (i < 0 || i >= m || j < 0 || j >= n) {
            return k >= 0 ? 1 : 0;
        }
        if (k <= 0) {
            return 0;
        }
        if (f[i][j][k] !== -1) {
            return f[i][j][k];
        }
        let ans = 0;
        for (let d = 0; d < 4; ++d) {
            const [x, y] = [i + dirs[d], j + dirs[d + 1]];
            ans = (ans + dfs(x, y, k - 1)) % mod;
        }
        return (f[i][j][k] = ans);
    };
    return dfs(startRow, startColumn, maxMove);
}

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 × k)

Space

O(m × n × k)

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.

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.