How do you solve Unique Paths III on LeetCode?

Unique Paths III (LeetCode #980) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(4^(m*n)) time complexity and O(m*n) space complexity. Key insight: Understanding the grid structure is crucial for pathfinding.

What is the brute force solution for Unique Paths III?

The brute force approach for Unique Paths III is Brute Force, with O(4^(m*n)) time complexity and O(m*n) space complexity. The brute force approach explores all possible paths from the starting square to the ending square, checking each path to see if it visits every non-obstacle square exactly once. This method is straightforward but can be inefficient due to the exponential number of paths. The optimal Optimal Solution approach improves this to O(4^(m*n)).

What algorithm or data structure is used to solve Unique Paths III?

Unique Paths III uses the following concepts: Array, Backtracking, Bit Manipulation, Matrix. The recommended patterns to study are: Backtracking, Recursion.

What are the interview tips for Unique Paths III?

Explain your thought process clearly while coding. Consider edge cases, like grids with no valid paths. Practice backtracking problems to get comfortable with the recursion.

What are common mistakes when solving Unique Paths III?

Not marking cells as visited properly, leading to infinite loops. Failing to check if all non-obstacle squares are visited before reaching the end.

#980

Unique Paths III

Hard

ArrayBacktrackingBit ManipulationMatrixBacktrackingRecursion

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(4^(m*n))	O(4^(m*n))
Space	O(m*n)	O(m*n)

💡

Intuition

Time O(4^(m*n))Space O(m*n)

The optimal solution uses backtracking with pruning to efficiently explore paths. By keeping track of the number of empty squares to visit and marking squares as visited, we can avoid unnecessary paths and reduce the search space significantly.

⚙️

Algorithm

3 steps

1Step 1: Count the number of empty squares and identify the starting and ending positions.
2Step 2: Implement a backtracking function that explores all four directions recursively.
3Step 3: Use a visited set to track which squares have been visited and ensure all empty squares are covered before reaching the end.

solution.py20 lines

1def uniquePathsIII(grid):
2    m, n = len(grid), len(grid[0])
3    start_x = start_y = empty_count = 0
4    for i in range(m):
5        for j in range(n):
6            if grid[i][j] == 1:
7                start_x, start_y = i, j
8            if grid[i][j] == 0:
9                empty_count += 1
10    def backtrack(x, y, visited):
11        if (x < 0 or x >= m or y < 0 or y >= n or grid[x][y] == -1 or visited[x][y]):
12            return 0
13        if grid[x][y] == 2:
14            return 1 if visited.sum() == empty_count + 1 else 0
15        visited[x][y] = True
16        paths = backtrack(x + 1, y, visited) + backtrack(x - 1, y, visited) + backtrack(x, y + 1, visited) + backtrack(x, y - 1, visited)
17        visited[x][y] = False
18        return paths
19    visited = [[False] * n for _ in range(m)]
20    return backtrack(start_x, start_y, visited)

ℹ

Complexity note: The time complexity remains exponential due to the recursive nature of the backtracking, but the use of pruning (marking visited) helps reduce the number of paths explored. The space complexity is O(m*n) due to the recursion stack and visited grid.

1Understanding the grid structure is crucial for pathfinding.
2Backtracking is a powerful technique for exploring all possible solutions.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.