LeetCode #980 — HARD

Unique Paths III

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

The Problem

Problem Statement

You are given an m x n integer array grid where grid[i][j] could be:

1 representing the starting square. There is exactly one starting square.
2 representing the ending square. There is exactly one ending square.
0 representing empty squares we can walk over.
-1 representing obstacles that we cannot walk over.

Return the number of 4-directional walks from the starting square to the ending square, that walk over every non-obstacle square exactly once.

Example 1:

Input: grid = [[1,0,0,0],[0,0,0,0],[0,0,2,-1]]
Output: 2
Explanation: We have the following two paths: 
1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2)
2. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2)

Example 2:

Input: grid = [[1,0,0,0],[0,0,0,0],[0,0,0,2]]
Output: 4
Explanation: We have the following four paths: 
1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2),(2,3)
2. (0,0),(0,1),(1,1),(1,0),(2,0),(2,1),(2,2),(1,2),(0,2),(0,3),(1,3),(2,3)
3. (0,0),(1,0),(2,0),(2,1),(2,2),(1,2),(1,1),(0,1),(0,2),(0,3),(1,3),(2,3)
4. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2),(2,3)

Example 3:

Input: grid = [[0,1],[2,0]]
Output: 0
Explanation: There is no path that walks over every empty square exactly once.
Note that the starting and ending square can be anywhere in the grid.

Constraints:

m == grid.length
n == grid[i].length
1 <= m, n <= 20
1 <= m * n <= 20
-1 <= grid[i][j] <= 2
There is exactly one starting cell and one ending cell.

Step 01

Brute Force Baseline

Problem summary: You are given an m x n integer array grid where grid[i][j] could be: 1 representing the starting square. There is exactly one starting square. 2 representing the ending square. There is exactly one ending square. 0 representing empty squares we can walk over. -1 representing obstacles that we cannot walk over. Return the number of 4-directional walks from the starting square to the ending square, that walk over every non-obstacle square exactly once.

Baseline thinking

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

Pattern signal: Array · Backtracking · Bit Manipulation

Example 1

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

Example 2

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

Example 3

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

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

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 #980: Unique Paths III
class Solution {
    private int m;
    private int n;
    private int cnt;
    private int[][] grid;
    private boolean[][] vis;

    public int uniquePathsIII(int[][] grid) {
        m = grid.length;
        n = grid[0].length;
        this.grid = grid;
        int x = 0, y = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == 0) {
                    ++cnt;
                } else if (grid[i][j] == 1) {
                    x = i;
                    y = j;
                }
            }
        }
        vis = new boolean[m][n];
        vis[x][y] = true;
        return dfs(x, y, 0);
    }

    private int dfs(int i, int j, int k) {
        if (grid[i][j] == 2) {
            return k == cnt + 1 ? 1 : 0;
        }
        int ans = 0;
        int[] dirs = {-1, 0, 1, 0, -1};
        for (int h = 0; h < 4; ++h) {
            int x = i + dirs[h], y = j + dirs[h + 1];
            if (x >= 0 && x < m && y >= 0 && y < n && !vis[x][y] && grid[x][y] != -1) {
                vis[x][y] = true;
                ans += dfs(x, y, k + 1);
                vis[x][y] = false;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #980: Unique Paths III
func uniquePathsIII(grid [][]int) int {
	m, n := len(grid), len(grid[0])
	cnt := 0
	vis := make([][]bool, m)
	x, y := 0, 0
	for i, row := range grid {
		vis[i] = make([]bool, n)
		for j, v := range row {
			if v == 0 {
				cnt++
			} else if v == 1 {
				x, y = i, j
			}
		}
	}
	dirs := [5]int{-1, 0, 1, 0, -1}
	var dfs func(i, j, k int) int
	dfs = func(i, j, k int) int {
		if grid[i][j] == 2 {
			if k == cnt+1 {
				return 1
			}
			return 0
		}
		ans := 0
		for h := 0; h < 4; h++ {
			x, y := i+dirs[h], j+dirs[h+1]
			if x >= 0 && x < m && y >= 0 && y < n && !vis[x][y] && grid[x][y] != -1 {
				vis[x][y] = true
				ans += dfs(x, y, k+1)
				vis[x][y] = false
			}
		}
		return ans
	}
	vis[x][y] = true
	return dfs(x, y, 0)
}

# Accepted solution for LeetCode #980: Unique Paths III
class Solution:
    def uniquePathsIII(self, grid: List[List[int]]) -> int:
        def dfs(i: int, j: int, k: int) -> int:
            if grid[i][j] == 2:
                return int(k == cnt + 1)
            ans = 0
            for a, b in pairwise(dirs):
                x, y = i + a, j + b
                if 0 <= x < m and 0 <= y < n and (x, y) not in vis and grid[x][y] != -1:
                    vis.add((x, y))
                    ans += dfs(x, y, k + 1)
                    vis.remove((x, y))
            return ans

        m, n = len(grid), len(grid[0])
        start = next((i, j) for i in range(m) for j in range(n) if grid[i][j] == 1)
        dirs = (-1, 0, 1, 0, -1)
        cnt = sum(row.count(0) for row in grid)
        vis = {start}
        return dfs(*start, 0)

// Accepted solution for LeetCode #980: Unique Paths III
struct Solution;

impl Solution {
    fn unique_paths_iii(mut grid: Vec<Vec<i32>>) -> i32 {
        let mut res = 0;
        let n = grid.len();
        let m = grid[0].len();
        let mut count = 0;
        for i in 0..n {
            for j in 0..m {
                if grid[i][j] == 0 {
                    count += 1;
                }
            }
        }
        for i in 0..n {
            for j in 0..m {
                if grid[i][j] == 1 {
                    Self::dfs(i, j, count, &mut res, &mut grid, n, m);
                }
            }
        }
        res as i32
    }

    fn dfs(
        i: usize,
        j: usize,
        left: usize,
        all: &mut usize,
        grid: &mut Vec<Vec<i32>>,
        n: usize,
        m: usize,
    ) {
        match grid[i][j] {
            2 => {
                if left == 0 {
                    *all += 1;
                }
            }
            1 => {
                grid[i][j] = -1;
                for (r, c) in Self::adj(i, j, n, m) {
                    Self::dfs(r, c, left, all, grid, n, m);
                }
                grid[i][j] = 1;
            }

            0 => {
                grid[i][j] = -1;
                for (r, c) in Self::adj(i, j, n, m) {
                    Self::dfs(r, c, left - 1, all, grid, n, m);
                }
                grid[i][j] = 0;
            }

            _ => {}
        }
    }

    fn adj(i: usize, j: usize, n: usize, m: usize) -> Vec<(usize, usize)> {
        let mut res = vec![];
        if i > 0 {
            res.push((i - 1, j));
        }
        if j > 0 {
            res.push((i, j - 1));
        }
        if i + 1 < n {
            res.push((i + 1, j));
        }
        if j + 1 < m {
            res.push((i, j + 1));
        }
        res
    }
}

#[test]
fn test() {
    let grid = vec_vec_i32![[1, 0, 0, 0], [0, 0, 0, 0], [0, 0, 2, -1]];
    let res = 2;
    assert_eq!(Solution::unique_paths_iii(grid), res);
    let grid = vec_vec_i32![[1, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 2]];
    let res = 4;
    assert_eq!(Solution::unique_paths_iii(grid), res);
    let grid = vec_vec_i32![[0, 1], [2, 0]];
    let res = 0;
    assert_eq!(Solution::unique_paths_iii(grid), res);
}

// Accepted solution for LeetCode #980: Unique Paths III
function uniquePathsIII(grid: number[][]): number {
    const m = grid.length;
    const n = grid[0].length;
    let [x, y] = [0, 0];
    let cnt = 0;
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            if (grid[i][j] === 0) {
                ++cnt;
            } else if (grid[i][j] == 1) {
                [x, y] = [i, j];
            }
        }
    }
    const vis: boolean[][] = Array.from({ length: m }, () => Array(n).fill(false));
    vis[x][y] = true;
    const dirs = [-1, 0, 1, 0, -1];
    const dfs = (i: number, j: number, k: number): number => {
        if (grid[i][j] === 2) {
            return k === cnt + 1 ? 1 : 0;
        }
        let ans = 0;
        for (let d = 0; d < 4; ++d) {
            const [x, y] = [i + dirs[d], j + dirs[d + 1]];
            if (x >= 0 && x < m && y >= 0 && y < n && !vis[x][y] && grid[x][y] !== -1) {
                vis[x][y] = true;
                ans += dfs(x, y, k + 1);
                vis[x][y] = false;
            }
        }
        return ans;
    };
    return dfs(x, y, 0);
}

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(3^m × n)

Space

O(m × n)

Approach Breakdown

EXHAUSTIVE

O(nⁿ) time

O(n) space

Generate every possible combination without any filtering. At each of n positions we choose from up to n options, giving nⁿ total candidates. Each candidate takes O(n) to validate. No pruning means we waste time on clearly invalid partial solutions.

BACKTRACKING + PRUNING

O(n!) time

O(n) space

Backtracking explores a decision tree, but prunes branches that violate constraints early. Worst case is still factorial or exponential, but pruning dramatically reduces the constant factor in practice. Space is the recursion depth (usually O(n) for n-level decisions).

Shortcut: Backtracking time = size of the pruned search tree. Focus on proving your pruning eliminates most branches.

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.

Missing undo step on backtrack

Wrong move: Mutable state leaks between branches.

Usually fails on: Later branches inherit selections from earlier branches.

Fix: Always revert state changes immediately after recursive call.