Add Solution to LC problem 980

dsa-solutions/lc-solutions/0900-0999/980 - Unique Paths III.md

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+---
+id: unique-paths-iii
+title: Unique Paths III
+sidebar_label: Unique Paths III
+tags: [Backtracking, DFS, Grid, C++, Python, Java]
+description: Find the number of unique paths from the starting square to the ending square in a grid, walking over every non-obstacle square exactly once.
+---
+
+## Problem Statement
+
+### Problem Description
+
+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
+
+**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:
+
+(0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(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:
+
+(0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2),(2,3)
+(0,0),(0,1),(1,1),(1,0),(2,0),(2,1),(2,2),(1,2),(0,2),(0,3),(1,3),(2,3)
+(0,0),(1,0),(2,0),(2,1),(2,2),(1,2),(1,1),(0,1),(0,2),(0,3),(1,3),(2,3)
+(0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2),(2,3)
+
+
+### 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.
+
+## Solution
+
+### Intuition
+
+To solve this problem, we can use a backtracking approach with Depth-First Search (DFS). We will start from the starting square and try to explore all possible paths to the ending square while ensuring we visit every non-obstacle square exactly once. We will backtrack whenever we hit a dead end or revisit a square.
+
+### Time Complexity and Space Complexity Analysis
+
+- **Time Complexity**: $O(4^{m \times n})$, where $m \times n$ is the total number of cells. This is because, in the worst case, we might explore all possible paths.
+- **Space Complexity**: $O(m \times n)$, for the recursion stack and the `visited` array.
+
+### Code
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    int uniquePathsIII(vector<vector<int>>& grid) {
+        int rows = grid.size();
+        int cols = grid[0].size();
+        int emptySquares = 1; // including the starting point
+        int startX, startY;
+
+        for (int i = 0; i < rows; ++i) {
+            for (int j = 0; j < cols; ++j) {
+                if (grid[i][j] == 0) {
+                    ++emptySquares;
+                } else if (grid[i][j] == 1) {
+                    startX = i;
+                    startY = j;
+                }
+            }
+        }
+
+        return dfs(grid, startX, startY, emptySquares);
+    }
+
+private:
+    int dfs(vector<vector<int>>& grid, int x, int y, int emptySquares) {
+        if (x < 0 || x >= grid.size() || y < 0 || y >= grid[0].size() || grid[x][y] == -1) {
+            return 0;
+        }
+        if (grid[x][y] == 2) {
+            return emptySquares == 0 ? 1 : 0;
+        }
+
+        grid[x][y] = -1;
+        int paths = dfs(grid, x + 1, y, emptySquares - 1) +
+                    dfs(grid, x - 1, y, emptySquares - 1) +
+                    dfs(grid, x, y + 1, emptySquares - 1) +
+                    dfs(grid, x, y - 1, emptySquares - 1);
+        grid[x][y] = 0;
+
+        return paths;
+    }
+};
+```
+
+#### Python
+```python
+class Solution:
+    def uniquePathsIII(self, grid: List[List[int]]) -> int:
+        rows, cols = len(grid), len(grid[0])
+        empty_squares = 1  # including the starting point
+        start_x, start_y = 0, 0
+
+        for i in range(rows):
+            for j in range(cols):
+                if grid[i][j] == 0:
+                    empty_squares += 1
+                elif grid[i][j] == 1:
+                    start_x, start_y = i, j
+
+        def dfs(x, y, empty_squares):
+            if x < 0 or x >= rows or y < 0 or y >= cols or grid[x][y] == -1:
+                return 0
+            if grid[x][y] == 2:
+                return 1 if empty_squares == 0 else 0
+
+            grid[x][y] = -1
+            paths = (dfs(x + 1, y, empty_squares - 1) +
+                     dfs(x - 1, y, empty_squares - 1) +
+                     dfs(x, y + 1, empty_squares - 1) +
+                     dfs(x, y - 1, empty_squares - 1))
+            grid[x][y] = 0
+
+            return paths
+
+        return dfs(start_x, start_y, empty_squares)
+```
+#### Java
+```java
+class Solution {
+    public int uniquePathsIII(int[][] grid) {
+        int rows = grid.length;
+        int cols = grid[0].length;
+        int emptySquares = 1; // including the starting point
+        int startX = 0, startY = 0;
+
+        for (int i = 0; i < rows; i++) {
+            for (int j = 0; j < cols; j++) {
+                if (grid[i][j] == 0) {
+                    emptySquares++;
+                } else if (grid[i][j] == 1) {
+                    startX = i;
+                    startY = j;
+                }
+            }
+        }
+
+        return dfs(grid, startX, startY, emptySquares);
+    }
+
+    private int dfs(int[][] grid, int x, int y, int emptySquares) {
+        if (x < 0 || x >= grid.length || y < 0 || y >= grid[0].length || grid[x][y] == -1) {
+            return 0;
+        }
+        if (grid[x][y] == 2) {
+            return emptySquares == 0 ? 1 : 0;
+        }
+
+        grid[x][y] = -1;
+        int paths = dfs(grid, x + 1, y, emptySquares - 1) +
+                    dfs(grid, x - 1, y, emptySquares - 1) +
+                    dfs(grid, x, y + 1, emptySquares - 1) +
+                    dfs(grid, x, y - 1, emptySquares - 1);
+        grid[x][y] = 0;
+
+        return paths;
+    }
+}
+```