How do you solve Minimum XOR Sum of Two Arrays on LeetCode?

Minimum XOR Sum of Two Arrays (LeetCode #1879) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * 2^n) time complexity and O(2^n) space complexity. Key insight: Using bitmasking allows us to efficiently explore combinations without generating all permutations.

What is the brute force solution for Minimum XOR Sum of Two Arrays?

The brute force approach for Minimum XOR Sum of Two Arrays is Brute Force, with O(n!) time complexity and O(n) space complexity. The brute-force approach involves generating all possible permutations of nums2 and calculating the XOR sum for each permutation. This guarantees finding the minimum XOR sum but is inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n * 2^n).

What algorithm or data structure is used to solve Minimum XOR Sum of Two Arrays?

Minimum XOR Sum of Two Arrays uses the following concepts: Array, Dynamic Programming, Bit Manipulation, Bitmask. The recommended patterns to study are: Dynamic Programming, Bitmasking, Subset Problems.

What are the interview tips for Minimum XOR Sum of Two Arrays?

Always clarify the constraints and edge cases before diving into the solution. Discuss your thought process and approach with the interviewer to gauge their feedback. Practice problems involving permutations and combinations to become comfortable with bitmasking.

What are common mistakes when solving Minimum XOR Sum of Two Arrays?

Failing to consider all combinations of nums2 when calculating the XOR sum. Not using bitmasking effectively, leading to unnecessary complexity in the solution.

#1879

Minimum XOR Sum of Two Arrays

Hard

ArrayDynamic ProgrammingBit ManipulationBitmaskDynamic ProgrammingBitmaskingSubset Problems

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n!)	O(n * 2^n)
Space	O(n)	O(2^n)

💡

Intuition

Time O(n * 2^n)Space O(2^n)

The optimal solution uses dynamic programming with bitmasking to efficiently explore all possible selections of nums2 without generating all permutations explicitly. This reduces the time complexity significantly.

⚙️

Algorithm

3 steps

1Step 1: Use a bitmask to represent the selection of elements from nums2.
2Step 2: For each bitmask, calculate the XOR sum by selecting corresponding elements from nums2.
3Step 3: Track the minimum XOR sum across all bitmask selections.

solution.py13 lines

1def minXORSum(nums1, nums2):
2    n = len(nums1)
3    dp = [float('inf')] * (1 << n)
4    dp[0] = 0
5    for mask in range(1 << n):
6        count = bin(mask).count('1')
7        if count > n:
8            continue
9        for j in range(n):
10            if mask & (1 << j) == 0:
11                new_mask = mask | (1 << j)
12                dp[new_mask] = min(dp[new_mask], dp[mask] + (nums1[count] ^ nums2[j]))
13    return dp[(1 << n) - 1]

ℹ

Complexity note: The time complexity is O(n * 2^n) because we iterate through all subsets of nums2 (2^n) and for each subset, we perform O(n) operations. The space complexity is O(2^n) for the dp array storing results for each subset.

1Using bitmasking allows us to efficiently explore combinations without generating all permutations.
2Dynamic programming can significantly reduce the time complexity for problems involving subsets.

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