Leetcode #980: Unique Paths III
In this guide, we solve Leetcode #980 Unique Paths III in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Bit Manipulation, Array, Backtracking, Matrix
Intuition
We must explore combinations of choices, but many branches can be pruned early.
Backtracking enumerates valid candidates while keeping the search space under control.
Approach
Use DFS to build candidates step by step, and backtrack when constraints are violated.
Pruning keeps the exploration practical for typical constraints.
Steps:
- Define the decision tree.
- DFS through choices and backtrack.
- Prune invalid paths early.
Example
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)
Python Solution
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)
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.