What is the time complexity of Minimum XOR Sum of Two Arrays?

The optimal solution has a time complexity of O(n * 2^n).

What is the space complexity of Minimum XOR Sum of Two Arrays?

The space complexity is O(2^n).

What data structures are used to solve Minimum XOR Sum of Two Arrays?

This problem uses Array, Dynamic Programming, Bit Manipulation, Bitmask. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Minimum XOR Sum of Two Arrays?

Insight 1: Since we can rearrange `nums2`, we need to find the optimal pairing between elements of `nums1` and `nums2` to minimize the total XOR sum. Insight 2: Dynamic programming with bitmasking is a suitable approach for solving this optimization problem, especially given the constraint `1 <= n <= 14`. The bitmask represents which elements of `nums2` have been used in pairings.

Which companies ask LeetCode 1879. Minimum XOR Sum of Two Arrays in interviews?

This problem is frequently asked at Media.net.

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Minimum XOR Sum of Two Arrays - LeetCode 1879 Solution

Q: How to solve Minimum XOR Sum of Two Arrays?

The solution uses dynamic programming with bitmasking to explore all possible pairings between `nums1` and `nums2`. The `dp` array stores the minimum XOR sum achievable when using a specific subset of `nums2` elements. The bitmask represents the subset of `nums2` elements used, where the i-th bit being set means that the i-th element of `nums2` has been used. We iterate through all possible subsets of `nums2` and, for each subset, try pairing the next available element of `nums1` with each unused element of `nums2` in the current subset. We update the `dp` array with the minimum XOR sum found so far.

Q: Which companies ask LeetCode 1879. Minimum XOR Sum of Two Arrays in interviews?

This problem is frequently asked at Media.net.

Minimum XOR Sum of Two Arrays - Complete Solution Guide

Minimum XOR Sum of Two Arrays is LeetCode problem 1879, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given two integer arrays nums1 and nums2 of length n . The XOR sum of the two integer arrays is (nums1[0] XOR nums2[0]) + (nums1[1] XOR nums2[1]) + ... + (nums1[n - 1] XOR nums2[n - 1]) ( 0-indexed ). For example, the XOR sum of [1,2,3] and [3,2,1] is equal to (1 XOR 3) + (2 XOR 2) + (3 XOR 1) = 2 + 0 + 2 = 4 . Rearrange the elements of nums2 such that the resulting XOR sum is minimized . Return the XOR sum after the rearrangement . Example 1: Input: nums1 = [1,2], nums2 = [2,3] Output:

Detailed Explanation

The problem asks us to find the minimum possible XOR sum of two integer arrays, `nums1` and `nums2`, both of length `n`. We can rearrange the elements of `nums2` in any order to achieve this minimum XOR sum. The XOR sum is calculated by taking the XOR of corresponding elements in the two arrays after rearranging `nums2`, and then summing up all these XOR results. The constraints limit the array size `n` to be between 1 and 14, and the element values to be between 0 and 10^7.

Solution Approach

The solution uses dynamic programming with bitmasking to explore all possible pairings between `nums1` and `nums2`. The `dp` array stores the minimum XOR sum achievable when using a specific subset of `nums2` elements. The bitmask represents the subset of `nums2` elements used, where the i-th bit being set means that the i-th element of `nums2` has been used. We iterate through all possible subsets of `nums2` and, for each subset, try pairing the next available element of `nums1` with each unused element of `nums2` in the current subset. We update the `dp` array with the minimum XOR sum found so far.

Step-by-Step Algorithm

Step 1: Initialize a `dp` array of size `2^n` with a large value (infinity or MAX_VALUE) to represent the minimum XOR sum for each possible subset of `nums2`. `dp[0]` is initialized to 0, representing an empty subset with XOR sum of 0.
Step 2: Iterate through all possible subsets of `nums2` using a loop from `mask = 1` to `(1 << n) - 1`. Each `mask` represents a unique subset.
Step 3: For each `mask`, determine the number of set bits `k`. `k` represents the number of elements selected from `nums2`, and also indicates which element of `nums1` is being considered (i.e., `nums1[k-1]`).
Step 4: Iterate through all elements of `nums2` using a loop from `j = 0` to `n - 1`.
Step 5: Check if the j-th element of `nums2` is in the current subset (i.e., if `(mask >> j) & 1` is true). If not, skip to the next element.
Step 6: If the j-th element is in the subset, calculate the XOR sum obtained by pairing `nums1[k - 1]` with `nums2[j]`. The `prev_mask` is calculated by removing the `j`-th element from the current `mask` (i.e., `prev_mask = mask ^ (1 << j)`).
Step 7: Update `dp[mask]` with the minimum value between its current value and `dp[prev_mask] + (nums1[k - 1] ^ nums2[j])`. This step uses the principle of optimality, finding the minimum XOR sum by considering all possible pairings for the current mask and using the previously calculated XOR sums for smaller subproblems.
Step 8: After iterating through all masks, `dp[(1 << n) - 1]` will contain the minimum XOR sum obtained by pairing all elements of `nums1` and `nums2`. Return this value.

Key Insights

Insight 1: Since we can rearrange `nums2`, we need to find the optimal pairing between elements of `nums1` and `nums2` to minimize the total XOR sum.
Insight 2: Dynamic programming with bitmasking is a suitable approach for solving this optimization problem, especially given the constraint `1 <= n <= 14`. The bitmask represents which elements of `nums2` have been used in pairings.
Insight 3: The state of the dynamic programming solution will be defined by the set of elements from `nums2` that have been paired with elements from `nums1`. We iterate through possible pairings to find the minimum XOR sum.

Complexity Analysis

Time Complexity: O(n * 2^n)

Space Complexity: O(2^n)

Topics

This problem involves: Array, Dynamic Programming, Bit Manipulation, Bitmask.

Companies

Asked at: Media.net.

Minimum XOR Sum of Two Arrays | LeetCode 1879

Hard

Array Dynamic Programming Bit Manipulation Bitmask

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

Solution Code

Available in:

class Solution {
    public int minimumXORSum(int[] nums1, int[] nums2) {
        int n = nums1.length;
        int[] dp = new int[1 << n];
        for (int i = 1; i < (1 << n); i++) {
            dp[i] = Integer.MAX_VALUE;
        }
        dp[0] = 0;

        for (int mask = 1; mask < (1 << n); mask++) {
            int k = Integer.bitCount(mask);

            for (int j = 0; j < n; j++) {
                if ((mask >> j) % 2 == 1) {
                    int prev_mask = mask ^ (1 << j);
                    int cost = nums1[k - 1] ^ nums2[j];
                    dp[mask] = Math.min(dp[mask], dp[prev_mask] + cost);
                }
            }
        }

        return dp[(1 << n) - 1];
    }
}

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Problem Info

Description

Problem 1879: Minimum XOR Sum of Two Arrays - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Bit Manipulation Bitmask

Companies

Media.net

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:1879

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Initialize a `dp` array of size `2^n` with a large value (infinity or MAX_VALUE) to represent the minimum XOR sum for each possible subset of `nums2`. `dp[0]` is initialized to 0, representing an empty subset with XOR sum of 0.
2
Step 2: Iterate through all possible subsets of `nums2` using a loop from `mask = 1` to `(1 << n) - 1`. Each `mask` represents a unique subset.
3
Step 3: For each `mask`, determine the number of set bits `k`. `k` represents the number of elements selected from `nums2`, and also indicates which element of `nums1` is being considered (i.e., `nums1[k-1]`).
4
Step 4: Iterate through all elements of `nums2` using a loop from `j = 0` to `n - 1`.
5
Step 5: Check if the j-th element of `nums2` is in the current subset (i.e., if `(mask >> j) & 1` is true). If not, skip to the next element.
6
Step 6: If the j-th element is in the subset, calculate the XOR sum obtained by pairing `nums1[k - 1]` with `nums2[j]`. The `prev_mask` is calculated by removing the `j`-th element from the current `mask` (i.e., `prev_mask = mask ^ (1 << j)`).
7
Step 7: Update `dp[mask]` with the minimum value between its current value and `dp[prev_mask] + (nums1[k - 1] ^ nums2[j])`. This step uses the principle of optimality, finding the minimum XOR sum by considering all possible pairings for the current mask and using the previously calculated XOR sums for smaller subproblems.
8
Step 8: After iterating through all masks, `dp[(1 << n) - 1]` will contain the minimum XOR sum obtained by pairing all elements of `nums1` and `nums2`. Return this value.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n * 2^n)

💾

Space Complexity

O(2^n)

📝

Why?

The time complexity arises from iterating through all possible combinations of elements from `nums2` using bitmasking. The space complexity results from storing the minimum XOR sums for each combination in the DP table.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Brute Force

Generate all permutations of `nums2` and calculate the XOR sum for each permutation. Find the minimum XOR sum across all permutations.

Time: O(n!), where n is the length of the input arrays.Space: O(n) if done in-place, O(n!) if storing all permutations.

Trade-offs: The brute-force approach is only feasible for very small values of `n` (e.g., n <= 10). For larger values, the dynamic programming approach is significantly more efficient.

✨

Key Insights

1Insight 1: Since we can rearrange `nums2`, we need to find the optimal pairing between elements of `nums1` and `nums2` to minimize the total XOR sum.
2Insight 2: Dynamic programming with bitmasking is a suitable approach for solving this optimization problem, especially given the constraint `1 <= n <= 14`. The bitmask represents which elements of `nums2` have been used in pairings.
3Insight 3: The state of the dynamic programming solution will be defined by the set of elements from `nums2` that have been paired with elements from `nums1`. We iterate through possible pairings to find the minimum XOR sum.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: Empty input arrays (n=0). The problem constraints state n >= 1, but handling n=0 would require an additional check at the beginning of the function. The given solution doesn't explicitly handle n=0, but it's not required due to the problem constraints.
Edge case 2: All elements in `nums1` and `nums2` are zero. The solution will correctly calculate the XOR sum as 0, as XORing 0 with 0 results in 0.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to initialize the `dp` array with a large value or infinity. This can lead to incorrect minimum XOR sum calculations.

✗Common mistake 2: Incorrectly implementing the bitmasking logic. Ensure that the bits are being shifted and compared correctly to represent the inclusion or exclusion of elements from `nums2`.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Resource allocation. Imagine assigning tasks (represented by `nums1`) to workers (represented by `nums2`) where the XOR sum represents a 'cost' associated with each assignment. Minimizing the XOR sum translates to finding the optimal assignment of tasks to workers to minimize the total cost.

🏦Real-world scenario 2: Data compression. XOR operations are used in some data compression algorithms. The optimal arrangement of data can potentially minimize the amount of change when using XOR for compression, similar to minimizing the XOR sum in this problem.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 19, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 14

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 7

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Minimum XOR Sum of Two Arrays"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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