How do you solve Minimum XOR Path in a Grid on LeetCode?

Minimum XOR Path in a Grid (LeetCode #3882) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m * n * k) time complexity and O(m * n * k) space complexity. Key insight: XOR operation is cumulative and can be optimized using dynamic programming.

What is the time complexity of Minimum XOR Path in a Grid?

The optimal solution for Minimum XOR Path in a Grid runs in O(m * n * k) time and uses O(m * n * k) space. k is the number of unique XOR values, leading to a manageable complexity.

What is the brute force solution for Minimum XOR Path in a Grid?

The brute force approach for Minimum XOR Path in a Grid is Brute Force, with O(2^(m+n)) time complexity and O(m+n) space complexity. Explore all possible paths from the top-left to the bottom-right cell. Calculate the XOR for each path and track the minimum value. The optimal Optimal Solution approach improves this to O(m * n * k).

What algorithm or data structure is used to solve Minimum XOR Path in a Grid?

Minimum XOR Path in a Grid uses the following concepts: Array, Dynamic Programming, Bit Manipulation, Matrix. The recommended patterns to study are: Dynamic Programming, Graph Traversal.

What are the interview tips for Minimum XOR Path in a Grid?

Explain your thought process clearly. Discuss edge cases and how your solution handles them. Practice similar problems to build confidence.

What are common mistakes when solving Minimum XOR Path in a Grid?

Not considering all possible paths in the brute force approach. Overlooking the need to manage XOR values in the DP approach.

#3882

Minimum XOR Path in a Grid

Medium

ArrayDynamic ProgrammingBit ManipulationMatrixDynamic ProgrammingGraph Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(2^(m+n))	O(m * n * k)
Space	O(m+n)	O(m * n * k)

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Intuition

Time O(m * n * k)Space O(m * n * k)

Use dynamic programming to store possible XOR values at each cell, reducing redundant calculations.

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Algorithm

3 steps

1Step 1: Initialize a DP array to track possible XOR values for each cell.
2Step 2: Iterate through the grid, updating the DP array based on previous cell values.
3Step 3: Return the minimum XOR value at the bottom-right cell.

solution.py13 lines

1def minXorPath(grid):
2    m, n = len(grid), len(grid[0])
3    dp = [[set() for _ in range(n)] for _ in range(m)]
4    dp[0][0].add(grid[0][0])
5    for i in range(m):
6        for j in range(n):
7            if i > 0:
8                for x in dp[i-1][j]:
9                    dp[i][j].add(x ^ grid[i][j])
10            if j > 0:
11                for x in dp[i][j-1]:
12                    dp[i][j].add(x ^ grid[i][j])
13    return min(dp[m-1][n-1])

ℹ

Complexity note: k is the number of unique XOR values, leading to a manageable complexity.

1XOR operation is cumulative and can be optimized using dynamic programming.
2Tracking multiple XOR values allows for efficient path evaluation.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.