#1879

Hard

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Minimum XOR Sum of Two Arrays

Minimize the XOR sum of two integer arrays by rearranging elements using dynamic programming and bit manipulation.

array dynamic programming bit manipulation bitmask

Problem Statement

You are given two integer arrays nums1 and nums2, both of length n. The XOR sum of the two arrays is calculated as the sum of (nums1[i] XOR nums2[i]) for each index i. The goal is to rearrange the elements of nums2 to minimize the resulting XOR sum for all pairs.

Since n is small (≤ 14), it is feasible to check all permutations of nums2, but a more optimal approach uses dynamic programming with bitmasking to handle the problem efficiently. The XOR operation on each pair can be leveraged to find the minimum sum.

Examples

Example 1

Input: nums1 = [1,2], nums2 = [2,3]

Output: 2

Rearrange nums2 so that it becomes [3,2]. The XOR sum is (1 XOR 3) + (2 XOR 2) = 2 + 0 = 2.

Example 2

Input: nums1 = [1,0,3], nums2 = [5,3,4]

Output: 8

Rearrange nums2 so that it becomes [5,4,3]. The XOR sum is (1 XOR 5) + (0 XOR 4) + (3 XOR 3) = 4 + 4 + 0 = 8.

Constraints

n == nums1.length
n == nums2.length
1 <= n <= 14
0 <= nums1[i], nums2[i] <= 107

Solution Approach

Dynamic Programming with Bitmask

We can solve this problem using dynamic programming with bitmasking. The state represents which elements of nums2 have been paired, and the dp array stores the minimum XOR sum for each state. We calculate the result by iterating over all subsets of nums2, trying each pairing, and updating the DP state.

Bitmasking State Transition

Using bitmasking allows us to represent the state of paired elements in nums2 efficiently. Each bit in the bitmask corresponds to whether a particular element in nums2 has been used, allowing us to transition between different pairings while keeping track of the XOR sum.

Optimization for Small n

Since n is small (≤ 14), the total number of subsets is at most 2^14, which is computationally feasible. This allows us to explore all possible pairings efficiently with bitmasking, minimizing the XOR sum with dynamic programming.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity of the solution is O(n * 2^n), where n is the length of the arrays. This is because there are 2^n possible states, and for each state, we check each element of nums1 to find the best XOR sum. The space complexity is O(2^n), as we need to store the DP table for all states.

What Interviewers Usually Probe

Check if the candidate understands the concept of dynamic programming with bitmasking.
Look for a candidate's ability to handle small n values efficiently using DP.
Assess whether the candidate can explain the trade-off of using DP over brute force permutations.

Common Pitfalls or Variants

Common pitfalls

Forgetting to properly manage the bitmask when updating the DP state.
Overlooking the fact that bitmasking can reduce time complexity exponentially, so brute-forcing is inefficient.
Incorrectly handling the transition between different DP states, leading to wrong XOR sums.

Follow-up variants

Consider alternative bitmask approaches for larger values of n (though the complexity may not scale well).
Explore greedy algorithms or heuristics for larger inputs, but note the trade-offs in terms of optimality.
Optimize for space by reducing the size of the DP table when dealing with sparse subsets.

FAQ

What is the primary pattern for the 'Minimum XOR Sum of Two Arrays' problem?

The primary pattern for this problem is dynamic programming with bitmasking. The state transitions between subsets of nums2, evaluating the XOR sum for each pairing.

How do bitmasks help in solving this problem?

Bitmasks are used to represent the state of paired elements from nums2. Each bit indicates whether a particular element has been paired, enabling efficient state transitions during the dynamic programming process.

What is the time complexity of the solution for this problem?

The time complexity is O(n * 2^n), where n is the length of the arrays. This comes from iterating through all 2^n possible states and performing calculations for each.

What is the maximum length of the arrays nums1 and nums2?

The maximum length of both arrays nums1 and nums2 is 14, which allows for feasible computation using bitmasking and dynamic programming.

What common mistakes should I avoid while solving this problem?

Common mistakes include incorrectly managing the bitmask, overlooking the efficiency of dynamic programming, and mishandling state transitions between subsets of nums2.

terminal

Solution

Solution 1

#### Python3

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class Solution:
    def minimumXORSum(self, nums1: List[int], nums2: List[int]) -> int:
        n = len(nums2)
        f = [[inf] * (1 << n) for _ in range(n + 1)]
        f[0][0] = 0
        for i, x in enumerate(nums1, 1):
            for j in range(1 << n):
                for k in range(n):
                    if j >> k & 1:
                        f[i][j] = min(f[i][j], f[i - 1][j ^ (1 << k)] + (x ^ nums2[k]))
        return f[-1][-1]

Solution 2

#### Python3

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class Solution:
    def minimumXORSum(self, nums1: List[int], nums2: List[int]) -> int:
        n = len(nums2)
        f = [[inf] * (1 << n) for _ in range(n + 1)]
        f[0][0] = 0
        for i, x in enumerate(nums1, 1):
            for j in range(1 << n):
                for k in range(n):
                    if j >> k & 1:
                        f[i][j] = min(f[i][j], f[i - 1][j ^ (1 << k)] + (x ^ nums2[k]))
        return f[-1][-1]

Solution 3

#### Python3

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class Solution:
    def minimumXORSum(self, nums1: List[int], nums2: List[int]) -> int:
        n = len(nums2)
        f = [[inf] * (1 << n) for _ in range(n + 1)]
        f[0][0] = 0
        for i, x in enumerate(nums1, 1):
            for j in range(1 << n):
                for k in range(n):
                    if j >> k & 1:
                        f[i][j] = min(f[i][j], f[i - 1][j ^ (1 << k)] + (x ^ nums2[k]))
        return f[-1][-1]

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