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Problem solution · Java

Count Paths With the Given XOR Value

Count Paths With the Given XOR Value: a Java solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Direct simulation
Source
walkccc LeetCode Solutions
Length
27 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Direct simulation

For Count Paths With the Given XOR Value, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 27 lines of Java from the credited upstream file 3393.java.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeCount Paths With the Given XOR Value · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int countPathsWithXorValue(int[][] grid, int k) {    final int MAX = 15;    final int m = grid.length;    final int n = grid[0].length;    Integer[][][] mem = new Integer[m][n][MAX + 1];    return count(grid, 0, 0, 0, k, mem);  }   private static final int MOD = 1_000_000_007;   // Return the number of paths from (i, j) to (m - 1, n - 1) with XOR value  // `xors`.  private int count(int[][] grid, int i, int j, int xors, int k, Integer[][][] mem) {    if (i == grid.length || j == grid[0].length)      return 0;    xors ^= grid[i][j];    if (i == grid.length - 1 && j == grid[0].length - 1)      return xors == k ? 1 : 0;    if (mem[i][j][xors] != null)      return mem[i][j][xors];    final int right = count(grid, i, j + 1, xors, k, mem) % MOD;    final int down = count(grid, i + 1, j, xors, k, mem) % MOD;    return mem[i][j][xors] = (right + down) % MOD;  }}

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