Count Paths With the Given XOR Value

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ArraysDynamic ProgrammingBit Manipulation

You are given a 2D integer array grid with size m x n. You are also given an integer k.

Your task is to calculate the number of paths you can take from the top-left cell (0, 0) to the bottom-right cell (m - 1, n - 1) satisfying the following constraints:

You can either move to the right or down. Formally, from the cell (i, j) you may move to the cell (i, j + 1) or to the cell (i + 1, j) if the target cell exists.
The XOR of all the numbers on the path must be equal to k.

Return the total number of such paths.

Since the answer can be very large, return the result modulo 10⁹ + 7.

Example 1:

Input: grid = [[2, 1, 5], [7, 10, 0], [12, 6, 4]], k = 11

Output: 3

Explanation:

The 3 paths are:

(0, 0) → (1, 0) → (2, 0) → (2, 1) → (2, 2)
(0, 0) → (1, 0) → (1, 1) → (1, 2) → (2, 2)
(0, 0) → (0, 1) → (1, 1) → (2, 1) → (2, 2)

Example 2:

Input: grid = [[1, 3, 3, 3], [0, 3, 3, 2], [3, 0, 1, 1]], k = 2

Output: 5

Explanation:

The 5 paths are:

(0, 0) → (1, 0) → (2, 0) → (2, 1) → (2, 2) → (2, 3)
(0, 0) → (1, 0) → (1, 1) → (2, 1) → (2, 2) → (2, 3)
(0, 0) → (1, 0) → (1, 1) → (1, 2) → (1, 3) → (2, 3)
(0, 0) → (0, 1) → (1, 1) → (1, 2) → (2, 2) → (2, 3)
(0, 0) → (0, 1) → (0, 2) → (1, 2) → (2, 2) → (2, 3)

Example 3:

Input: grid = [[1, 1, 1, 2], [3, 0, 3, 2], [3, 0, 2, 2]], k = 10

Output: 0

Constraints:

1 <= m == grid.length <= 300
1 <= n == grid[r].length <= 300
0 <= grid[r][c] < 16
0 <= k < 16

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of values for the node values in the binary tree and the target value? Can they be negative or zero?
What should be returned if no path exists with the given XOR value?
Is a 'path' strictly defined as from the root to a leaf, or can it start and end at any two nodes?
What is the maximum depth of the binary tree, and what is the expected distribution of node values to avoid potential integer overflow issues when calculating XOR values along a path?
Can I assume the input tree is a valid binary tree, or do I need to handle cases where the tree is malformed (e.g., a node with more than two children)?

Brute Force Solution

Approach

The brute force approach to counting paths with a specific XOR value involves exploring every possible path through a structure. For each path, we calculate its XOR value and check if it matches the desired value. We increment a counter each time we find a path with the matching XOR value, ensuring we've considered all possibilities.

Here's how the algorithm would work step-by-step:

Start at the beginning and consider every possible way to move through the structure, like exploring every road on a map.
As you follow each possible path, keep track of a running XOR value. This value changes as you move along the path.
When you reach the end of a path, check if the path's XOR value matches the target XOR value you're looking for.
If the path's XOR value matches the target, increase a counter to indicate that you found a valid path.
Repeat these steps, exploring every single possible path from start to finish.
Once you've checked absolutely every path, the counter will tell you the total number of paths with the target XOR value.

Code Implementation

def count_paths_with_xor_brute_force(grid, target_xor):
    number_of_rows = len(grid)
    number_of_columns = len(grid[0])
    path_count = 0

    def explore_path(row_index, column_index, current_xor):
        nonlocal path_count

        # Update XOR as we traverse
        current_xor ^= grid[row_index][column_index]

        # We've reached the end of a path
        if row_index == number_of_rows - 1 and column_index == number_of_columns - 1:
            # Check for XOR match
            if current_xor == target_xor:
                path_count += 1
            return

        # Explore down
        if row_index + 1 < number_of_rows:
            explore_path(row_index + 1, column_index, current_xor)

        # Explore right
        if column_index + 1 < number_of_columns:
            # Only explore right when within bounds
            explore_path(row_index, column_index + 1, current_xor)

    # Start the recursive exploration
    explore_path(0, 0, 0)

    return path_count

Big(O) Analysis

Time Complexity

O(2^n) – The brute force approach explores every possible path. In the worst case, if the structure allows branching at each step (e.g., a binary tree or a graph where each node can lead to multiple other nodes), the number of paths to explore grows exponentially. Specifically, we explore all possible paths leading to 2 to the power of n, which is equivalent to O(2^n).

Space Complexity

O(N) – The brute force approach, as described, explores every possible path, likely using recursion. Each recursive call stores a new path's XOR value and the current position in the structure on the call stack. In the worst-case scenario, the maximum depth of the recursion could be proportional to N, where N is the number of nodes/elements in the structure, leading to N function calls being stored on the stack. Therefore, the auxiliary space complexity is O(N) due to the recursion depth.

Optimal Solution

Approach

The challenge is to efficiently count paths in a structure where each path's exclusive OR (XOR) result equals a target value. Instead of checking every path individually, we use a clever trick to store and reuse XOR results we've already calculated. This avoids repeating calculations and dramatically speeds up the process.

Here's how the algorithm would work step-by-step:

Start at the beginning of the structure.
As you move along a path, calculate the XOR value from the starting point to the current location.
Keep track of how many times you've seen each XOR value so far. Use a 'memory' to store these counts.
For each location, check how many paths exist that, when XORed with the current path's XOR value, would result in the target XOR value. You can find this in your 'memory'.
Add that count to your total number of paths that meet the target XOR value.
Continue exploring the structure, updating the 'memory' and counting paths as you go.
By remembering XOR values you've already calculated and how often they appear, you can quickly find paths that meet the target without checking every single path.

Code Implementation

def count_paths_with_xor(matrix, target_xor):
    rows = len(matrix)
    cols = len(matrix[0])
    path_count = 0
    xor_counts = {0: 1}
    current_xor = 0

    def traverse_matrix(row_index, column_index, current_xor):
        nonlocal path_count

        if row_index < 0 or row_index >= rows or column_index < 0 or column_index >= cols:
            return

        # Update XOR value with current cell.
        current_xor ^= matrix[row_index][column_index]

        # Check how many paths lead to the target.
        required_xor = current_xor ^ target_xor
        if required_xor in xor_counts:
            path_count += xor_counts[required_xor]

        # Store the current XOR value to memory.
        if current_xor in xor_counts:
            xor_counts[current_xor] += 1

        else:
            xor_counts[current_xor] = 1

        traverse_matrix(row_index + 1, column_index, current_xor)
        traverse_matrix(row_index, column_index + 1, current_xor)

        # Backtrack: remove the current XOR value after traversal.
        xor_counts[current_xor] -= 1
        if xor_counts[current_xor] == 0:
            del xor_counts[current_xor]

    traverse_matrix(0, 0, current_xor)

    return path_count

Big(O) Analysis

Time Complexity

O(n) – The algorithm traverses a data structure (like a tree or graph) of size n. During this traversal, at each node, it performs a constant-time operation: calculating the XOR value from the starting node, updating the count of XOR values encountered in a memory (hash map), and checking if a specific XOR value exists in the memory. Since these operations at each node take constant time, the overall time complexity is proportional to the number of nodes visited, leading to O(n) time complexity.

Space Complexity

O(N) – The algorithm utilizes a 'memory' (likely a hash map or dictionary) to store the counts of XOR values encountered so far. In the worst-case scenario, where all paths from the starting point to each location have unique XOR values, the size of this memory could grow linearly with the number of locations visited. Thus, the auxiliary space required is proportional to N, where N represents the number of locations in the structure explored during the path counting.

Edge Cases

Case	How to Handle
Null or empty tree	Return 0 immediately as there are no paths.
Single node tree where node value equals the target	Return 1 as the root-to-leaf path (just the root) matches the target.
Tree with all nodes having value 0 and target is 0	Every path from root to leaf will have XOR of 0, count all paths.
Tree with large depth, potentially causing stack overflow in recursive solutions	Consider iterative DFS or BFS approach to avoid stack overflow.
Nodes with large integer values that could overflow when XORed	Ensure the integer type used for XOR operation has sufficient bits or handle potential overflows gracefully.
No path exists with the target XOR value	Return 0 after traversing the entire tree as the count remains 0.
Target is 0, but all nodes on a path have non-zero values	The algorithm should correctly not count this path.
Tree with negative node values	XOR operation works with negative numbers, but ensure the algorithm correctly handles negative values if applicable in the implementation language.

Null or empty tree

How to Handle:

Return 0 immediately as there are no paths.

Single node tree where node value equals the target

How to Handle:

Return 1 as the root-to-leaf path (just the root) matches the target.

Tree with all nodes having value 0 and target is 0

How to Handle:

Every path from root to leaf will have XOR of 0, count all paths.

Tree with large depth, potentially causing stack overflow in recursive solutions

How to Handle:

Consider iterative DFS or BFS approach to avoid stack overflow.

Nodes with large integer values that could overflow when XORed

How to Handle:

Ensure the integer type used for XOR operation has sufficient bits or handle potential overflows gracefully.

No path exists with the target XOR value

How to Handle:

Return 0 after traversing the entire tree as the count remains 0.

Target is 0, but all nodes on a path have non-zero values

How to Handle:

The algorithm should correctly not count this path.

Tree with negative node values

How to Handle:

XOR operation works with negative numbers, but ensure the algorithm correctly handles negative values if applicable in the implementation language.