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3882. Minimum XOR Path in a Grid
Description
You are given a 2D integer array grid of size m * n.
You start at the top-left cell (0, 0) and want to reach the bottom-right cell (m - 1, n - 1).
At each step, you may move either right or down.
The cost of a path is defined as the bitwise XOR of all the values in the cells along that path, including the start and end cells.
Return the minimum possible XOR value among all valid paths from (0, 0) to (m - 1, n - 1).
Example 1:
Input: grid = [[1,2],[3,4]]
Output: 6
Explanation:
There are two valid paths:
(0, 0) → (0, 1) → (1, 1)with XOR:1 XOR 2 XOR 4 = 7(0, 0) → (1, 0) → (1, 1)with XOR:1 XOR 3 XOR 4 = 6
The minimum XOR value among all valid paths is 6.
Example 2:
Input: grid = [[6,7],[5,8]]
Output: 9
Explanation:
There are two valid paths:
(0, 0) → (0, 1) → (1, 1)with XOR:6 XOR 7 XOR 8 = 9(0, 0) → (1, 0) → (1, 1)with XOR:6 XOR 5 XOR 8 = 11
The minimum XOR value among all valid paths is 9.
Example 3:
Input: grid = [[2,7,5]]
Output: 0
Explanation:
There is only one valid path:
(0, 0) → (0, 1) → (0, 2)with XOR:2 XOR 7 XOR 5 = 0
The XOR value of this path is 0, which is the minimum possible.
Constraints:
1 <= m == grid.length <= 10001 <= n == grid[i].length <= 1000m * n <= 10000 <= grid[i][j] <= 1023
Solutions
Solution 1
Solution 2: Dynamic Programming (Rolling Array)
Thinking
Solution 1 keeps every reachable XOR of every cell, so the space grows with $mn$. Cell $(i, j)$ reads only the cell above and the cell to the left, and a row-major scan finishes both first. The previous row is still complete when the current row begins, and the current row only needs the left neighbor already written, so two rolling rows are enough. The transition still looks back at $k \oplus x$ for a cell of value $x$, the answer is still the smallest reachable XOR at the lower-right cell, and the space drops to $O(n \cdot \textit{mx})$.
Define $f[i \bmod 2][j][k]$ as whether row $i$ and column $j$ are reachable from $(0, 0)$ with the bitwise XOR of every cell on the path equal to $k$. The value of $\textit{mx}$ matches Solution 1, and the start cell again has $f[0][0][\textit{grid}[0][0]] = \text{true}$.
Cells are processed in row-major order. For cell $(i, j)$ with value $x$, if it is not the start, every $k \in [0, \textit{mx})$ is considered. The row index is taken modulo $2$, so the current slot may still hold an older row. Each $k$ is therefore cleared first, then set with the same conditions as Solution 1:
- If $i > 0$ and $f[(i - 1) \bmod 2][j][k \oplus x]$ is true, the cell can be entered from above, and $f[i \bmod 2][j][k] = \text{true}$.
- If $j > 0$ and $f[i \bmod 2][j - 1][k \oplus x]$ is true, the cell can be entered from the left, and $f[i \bmod 2][j][k] = \text{true}$.
Row $i$ is stored at index $i \bmod 2$, the previous row stays at $(i - 1) \bmod 2$, and the left cell shares that row. The answer is the smallest $k$ for which $f[(m - 1) \bmod 2][n - 1][k]$ is true. At least one path from the upper-left cell to the lower-right cell exists, so this $k$ is guaranteed.
The time complexity is $O(mn \cdot \textit{mx})$ and the space complexity is $O(n \cdot \textit{mx})$, where $m$ and $n$ are the number of rows and columns, and $\textit{mx} \le 1024$.
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class Solution { public int minCost(int[][] grid) { int m = grid.length, n = grid[0].length; int mx = 0; for (int[] row : grid) { for (int x : row) { mx = Math.max(mx, x); } } mx = 1 << (Integer.SIZE - Integer.numberOfLeadingZeros(mx)); boolean[][][] f = new boolean[m][n][mx]; f[0][0][grid[0][0]] = true; for (int i = 0; i < m; i++) { for (int j = 0; j < n; j++) { int x = grid[i][j]; if (i > 0 || j > 0) { for (int k = 0; k < mx; k++) { if (i > 0 && f[i - 1][j][k ^ x]) { f[i][j][k] = true; } if (j > 0 && f[i][j - 1][k ^ x]) { f[i][j][k] = true; } } } } } for (int k = 0;; k++) { if (f[m - 1][n - 1][k]) { return k; } } } } // Solution 2 class Solution { public int minCost(int[][] grid) { int m = grid.length, n = grid[0].length; int mx = 0; for (int[] row : grid) { for (int x : row) { mx = Math.max(mx, x); } } mx = 1 << (Integer.SIZE - Integer.numberOfLeadingZeros(mx)); boolean[][][] f = new boolean[2][n][mx]; f[0][0][grid[0][0]] = true; for (int i = 0; i < m; i++) { for (int j = 0; j < n; j++) { int x = grid[i][j]; if (i > 0 || j > 0) { for (int k = 0; k < mx; k++) { f[i & 1][j][k] = false; if (i > 0 && f[(i - 1) & 1][j][k ^ x]) { f[i & 1][j][k] = true; } if (j > 0 && f[i & 1][j - 1][k ^ x]) { f[i & 1][j][k] = true; } } } } } for (int k = 0;; k++) { if (f[(m - 1) & 1][n - 1][k]) { return k; } } } } -
class Solution { public: int minCost(vector<vector<int>>& grid) { int m = grid.size(), n = grid[0].size(); int mx = 0; for (auto& row : grid) { for (int x : row) { mx = max(mx, x); } } mx = 1 << bit_width(static_cast<unsigned>(mx)); vector<vector<vector<char>>> f(m, vector<vector<char>>(n, vector<char>(mx, 0))); f[0][0][grid[0][0]] = 1; for (int i = 0; i < m; i++) { for (int j = 0; j < n; j++) { int x = grid[i][j]; if (i > 0 || j > 0) { for (int k = 0; k < mx; k++) { if (i > 0 && f[i - 1][j][k ^ x]) { f[i][j][k] = 1; } if (j > 0 && f[i][j - 1][k ^ x]) { f[i][j][k] = 1; } } } } } for (int k = 0;; k++) { if (f[m - 1][n - 1][k]) { return k; } } } }; // Solution 2 class Solution { public: int minCost(vector<vector<int>>& grid) { int m = grid.size(), n = grid[0].size(); int mx = 0; for (auto& row : grid) { for (int x : row) { mx = max(mx, x); } } mx = 1 << bit_width(static_cast<unsigned>(mx)); vector<vector<vector<char>>> f(2, vector<vector<char>>(n, vector<char>(mx, 0))); f[0][0][grid[0][0]] = 1; for (int i = 0; i < m; i++) { for (int j = 0; j < n; j++) { int x = grid[i][j]; if (i > 0 || j > 0) { for (int k = 0; k < mx; k++) { f[i & 1][j][k] = 0; if (i > 0 && f[(i - 1) & 1][j][k ^ x]) { f[i & 1][j][k] = 1; } if (j > 0 && f[i & 1][j - 1][k ^ x]) { f[i & 1][j][k] = 1; } } } } } for (int k = 0;; k++) { if (f[(m - 1) & 1][n - 1][k]) { return k; } } } }; -
class Solution: def minCost(self, grid: list[list[int]]) -> int: m, n = len(grid), len(grid[0]) mx = 1 << (max(max(row) for row in grid)).bit_length() f = [[[False] * mx for _ in range(n)] for _ in range(m)] f[0][0][grid[0][0]] = True for i, row in enumerate(grid): for j, x in enumerate(row): if i > 0 or j > 0: for k in range(mx): if i > 0 and f[i - 1][j][k ^ x]: f[i][j][k] = True if j > 0 and f[i][j - 1][k ^ x]: f[i][j][k] = True return next(k for k in range(mx) if f[m - 1][n - 1][k]) # Solution 2 class Solution: def minCost(self, grid: list[list[int]]) -> int: m, n = len(grid), len(grid[0]) mx = 1 << (max(max(row) for row in grid)).bit_length() f = [[[False] * mx for _ in range(n)] for _ in range(2)] f[0][0][grid[0][0]] = True for i, row in enumerate(grid): for j, x in enumerate(row): if i > 0 or j > 0: for k in range(mx): f[i & 1][j][k] = False if i > 0 and f[(i - 1) & 1][j][k ^ x]: f[i & 1][j][k] = True if j > 0 and f[i & 1][j - 1][k ^ x]: f[i & 1][j][k] = True return next(k for k in range(mx) if f[(m - 1) & 1][n - 1][k]) -
func minCost(grid [][]int) int { m, n := len(grid), len(grid[0]) mx := 0 for _, row := range grid { for _, x := range row { if x > mx { mx = x } } } mx = 1 << bits.Len(uint(mx)) f := make([][][]bool, m) for i := range f { f[i] = make([][]bool, n) for j := range f[i] { f[i][j] = make([]bool, mx) } } f[0][0][grid[0][0]] = true for i := 0; i < m; i++ { for j := 0; j < n; j++ { x := grid[i][j] if i > 0 || j > 0 { for k := 0; k < mx; k++ { if i > 0 && f[i-1][j][k^x] { f[i][j][k] = true } if j > 0 && f[i][j-1][k^x] { f[i][j][k] = true } } } } } for k := 0; ; k++ { if f[m-1][n-1][k] { return k } } } // Solution 2 func minCost(grid [][]int) int { m, n := len(grid), len(grid[0]) mx := 0 for _, row := range grid { for _, x := range row { if x > mx { mx = x } } } mx = 1 << bits.Len(uint(mx)) f := make([][][]bool, 2) for i := range f { f[i] = make([][]bool, n) for j := range f[i] { f[i][j] = make([]bool, mx) } } f[0][0][grid[0][0]] = true for i := 0; i < m; i++ { for j := 0; j < n; j++ { x := grid[i][j] if i > 0 || j > 0 { for k := 0; k < mx; k++ { f[i&1][j][k] = false if i > 0 && f[(i-1)&1][j][k^x] { f[i&1][j][k] = true } if j > 0 && f[i&1][j-1][k^x] { f[i&1][j][k] = true } } } } } for k := 0; ; k++ { if f[(m-1)&1][n-1][k] { return k } } } -
function minCost(grid: number[][]): number { const m = grid.length, n = grid[0].length; let mx = 0; for (const row of grid) { for (const x of row) { mx = Math.max(mx, x); } } mx = 1 << (32 - Math.clz32(mx)); const f: boolean[][][] = Array.from({ length: m }, () => Array.from({ length: n }, () => new Array<boolean>(mx).fill(false)), ); f[0][0][grid[0][0]] = true; for (let i = 0; i < m; i++) { for (let j = 0; j < n; j++) { const x = grid[i][j]; if (i > 0 || j > 0) { for (let k = 0; k < mx; k++) { if (i > 0 && f[i - 1][j][k ^ x]) { f[i][j][k] = true; } if (j > 0 && f[i][j - 1][k ^ x]) { f[i][j][k] = true; } } } } } for (let k = 0; ; k++) { if (f[m - 1][n - 1][k]) { return k; } } } // Solution 2 function minCost(grid: number[][]): number { const m = grid.length, n = grid[0].length; let mx = 0; for (const row of grid) { for (const x of row) { mx = Math.max(mx, x); } } mx = 1 << (32 - Math.clz32(mx)); const f: boolean[][][] = Array.from({ length: 2 }, () => Array.from({ length: n }, () => new Array<boolean>(mx).fill(false)), ); f[0][0][grid[0][0]] = true; for (let i = 0; i < m; i++) { for (let j = 0; j < n; j++) { const x = grid[i][j]; if (i > 0 || j > 0) { for (let k = 0; k < mx; k++) { f[i & 1][j][k] = false; if (i > 0 && f[(i - 1) & 1][j][k ^ x]) { f[i & 1][j][k] = true; } if (j > 0 && f[i & 1][j - 1][k ^ x]) { f[i & 1][j][k] = true; } } } } } for (let k = 0; ; k++) { if (f[(m - 1) & 1][n - 1][k]) { return k; } } }