Count Paths With the Given XOR Value

• Define a recursive function solve(row, col, currentXor).
• The function represents the number of valid paths from (row, col) to the destination, given the XOR sum of the path from (0,0) to the cell before (row, col) is currentXor.
• Base Case 1: If row or col is out of grid bounds, it's an invalid path. Return 0.
• Base Case 2: If (row, col) is the destination (m-1, n-1), calculate the final XOR sum finalXor = currentXor ^ grid[row][col]. If finalXor == k, return 1, otherwise return 0.
• Recursive Step: Update the XOR sum: newXor = currentXor ^ grid[row][col]. Recursively call for moving down and right: solve(row + 1, col, newXor) and solve(row, col + 1, newXor). Return the sum of their results.
• The initial call is solve(0, 0, 0).

We define a recursive function, say solve(row, col, currentXor). This function calculates the number of valid paths starting from (row, col) given that the XOR sum of the path from (0,0) up to the cell before (row, col) is currentXor. The base case for the recursion is when we reach the destination cell (m-1, n-1). At this point, we calculate the final XOR sum by XORing currentXor with grid[m-1][n-1]. If this equals k, we have found a valid path and return 1. Otherwise, we return 0. Another base case is if the current cell (row, col) is out of the grid boundaries, in which case we return 0. In the recursive step, from cell (row, col), we update the XOR sum and make two recursive calls: one for moving down solve(row + 1, col, newXor) and one for moving right solve(row, col + 1, newXor). The total number of paths is the sum of the results from these two calls. The initial call would be solve(0, 0, 0).

1class Solution {2    int m, n, k;3    int[][] grid;4    int MOD = 1_000_000_007;5 6    public int numberOfPaths(int[][] grid, int k) {7        this.m = grid.length;8        this.n = grid[0].length;9        this.k = k;10        this.grid = grid;11        // Note: This will time out. A memoization table is needed.12        return solve(0, 0, 0);13    }14 15    private int solve(int r, int c, int currentXor) {16        if (r >= m || c >= n) {17            return 0;18        }19 20        int newXor = currentXor ^ grid[r][c];21 22        if (r == m - 1 && c == n - 1) {23            return (newXor == k) ? 1 : 0;24        }25 26        int pathsDown = solve(r + 1, c, newXor);27        int pathsRight = solve(r, c + 1, newXor);28 29        return (pathsDown + pathsRight) % MOD;30    }31}

• Create a 3D DP table dp[m][n][C], where C is the maximum possible XOR value (16).
• dp[i][j][x] will store the number of paths from (0,0) to (i,j) with a path XOR sum of x.
• Initialization: Set dp[0][0][grid[0][0]] = 1. All other entries are 0.
• Transition: Iterate through each cell (i, j) from top-left to bottom-right. For each cell, calculate the number of paths reaching it.
• The number of paths to (i, j) with XOR sum x is the sum of paths from (i-1, j) and (i, j-1) that result in this XOR sum.
• The recurrence relation is: dp[i][j][x] = (dp[i-1][j][x ^ grid[i][j]] + dp[i][j-1][x ^ grid[i][j]]) % MOD.
• We iterate i from 0 to m-1, j from 0 to n-1, and for each cell, we iterate through all possible XOR values x from 0 to C-1 to apply the recurrence.
• Result: The final answer is dp[m-1][n-1][k].

We define a 3D DP table dp[m][n][16], where dp[i][j][x] stores the number of paths from the start (0, 0) to cell (i, j) such that the XOR sum of elements along the path is x. This can be implemented using a bottom-up (tabulation) approach for efficiency.

We initialize dp[0][0][grid[0][0]] = 1 as there is one way to be at the starting cell with an XOR sum equal to its own value. Then, we iterate through the grid. For any cell (i, j), the paths to it can only come from the cell above, (i-1, j), or the cell to the left, (i, j-1).

For each possible XOR sum x at cell (i, j), we find the required XOR sum at the previous cells. If we are at (i, j) with XOR sum x, we must have been at (i-1, j) or (i, j-1) with an XOR sum of x ^ grid[i][j]. So, we sum up the counts from these previous states: dp[i][j][x] = (dp[i-1][j][x ^ grid[i][j]] + dp[i][j-1][x ^ grid[i][j]]). We perform this for all possible XOR values from 0 to 15. The final answer is the number of paths to the destination cell (m-1, n-1) with an XOR sum of k, which is dp[m-1][n-1][k].

1class Solution {2    public int numberOfPaths(int[][] grid, int k) {3        int m = grid.length;4        int n = grid[0].length;5        int MOD = 1_000_000_007;6        int MAX_XOR_VAL = 16; 7 8        long[][][] dp = new long[m][n][MAX_XOR_VAL];9 10        // Base case: starting cell11        dp[0][0][grid[0][0]] = 1;12 13        for (int i = 0; i < m; i++) {14            for (int j = 0; j < n; j++) {15                if (i == 0 && j == 0) continue;16 17                int currentVal = grid[i][j];18                for (int x = 0; x < MAX_XOR_VAL; x++) {19                    int prevXor = x ^ currentVal;20                    long pathsFromUp = (i > 0) ? dp[i - 1][j][prevXor] : 0;21                    long pathsFromLeft = (j > 0) ? dp[i][j - 1][prevXor] : 0;22                    dp[i][j][x] = (pathsFromUp + pathsFromLeft) % MOD;23                }24            }25        }26 27        return (int) dp[m - 1][n - 1][k];28    }29}

• The core logic is the same as the standard DP, but we optimize space.
• Instead of a 3D table dp[m][n][C], we use a 2D table dp[n][C].
• dp[j][x] will store the number of paths to the cell in the current row i at column j with an XOR sum of x.
• Initialization: Initialize dp for the first row (i=0). dp[0][grid[0][0]] = 1. Then for j from 1 to n-1, compute dp[j] based on dp[j-1].
• Iteration: For each subsequent row i from 1 to m-1:
  • The values in dp currently represent the counts for row i-1.
  • First, update dp[0] for the current row i. This only depends on the dp[0] from the previous row.
  • Then, for j from 1 to n-1, update dp[j]. The new dp[j] is calculated based on the old dp[j] (paths from above) and the new dp[j-1] (paths from the left).
• Result: After iterating through all rows, the answer is dp[n-1][k].

The space optimization is based on the observation that the DP calculation for a cell (i, j) only depends on (i-1, j) and (i, j-1). This means when computing values for row i, we only need the results from row i-1. We can discard the results from rows i-2 and earlier.

We can use a 2D array dp[n][C] to store the path counts for the current row being processed. As we iterate from row i to i+1, the dp array is updated to reflect the counts for the new row.

Let dp[j][x] be the number of paths to (i, j) with XOR sum x. When we compute for row i, we update the dp array. The new value for dp[j] will depend on its old value (which corresponds to (i-1, j)) and the just-computed value of dp[j-1] (which corresponds to (i, j-1)).

1class Solution {2    public int numberOfPaths(int[][] grid, int k) {3        int m = grid.length;4        int n = grid[0].length;5        int MOD = 1_000_000_007;6        int MAX_XOR_VAL = 16;7 8        long[][] dp = new long[n][MAX_XOR_VAL];9 10        // Initialize for row 011        dp[0][grid[0][0]] = 1;12        for (int j = 1; j < n; j++) {13            int currentVal = grid[0][j];14            for (int x = 0; x < MAX_XOR_VAL; x++) {15                if (dp[j - 1][x] > 0) {16                    dp[j][x ^ currentVal] = (dp[j][x ^ currentVal] + dp[j - 1][x]) % MOD;17                }18            }19        }20 21        // Process subsequent rows from 1 to m-122        for (int i = 1; i < m; i++) {23            // First, update the first column of the current row24            long[] tempCol0 = new long[MAX_XOR_VAL];25            int firstColVal = grid[i][0];26            for (int x = 0; x < MAX_XOR_VAL; x++) {27                if (dp[0][x] > 0) {28                    tempCol0[x ^ firstColVal] = dp[0][x];29                }30            }31            dp[0] = tempCol0;32 33            // Update the rest of the columns for the current row34            for (int j = 1; j < n; j++) {35                long[] newCounts = new long[MAX_XOR_VAL];36                int currentVal = grid[i][j];37                for (int x = 0; x < MAX_XOR_VAL; x++) {38                    long fromUp = dp[j][x]; // from (i-1, j)39                    long fromLeft = dp[j - 1][x]; // from (i, j-1)40                    newCounts[x ^ currentVal] = (fromUp + fromLeft) % MOD;41                }42                dp[j] = newCounts;43            }44        }45 46        return (int) dp[n - 1][k];47    }48}