How do you solve Partition Array to Minimize XOR on LeetCode?

Partition Array to Minimize XOR (LeetCode #3599) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n² * k) time complexity and O(n * k) space complexity. Key insight: Understanding XOR properties can simplify calculations.

What is the time complexity of Partition Array to Minimize XOR?

The optimal solution for Partition Array to Minimize XOR runs in O(n² * k) time and uses O(n * k) space. The complexity arises from filling the DP table, iterating through n and k, leading to a quadratic relation.

What is the brute force solution for Partition Array to Minimize XOR?

The brute force approach for Partition Array to Minimize XOR is Brute Force, with O(n²) time complexity and O(1) space complexity. We can generate all possible partitions of the array into k subarrays and compute the XOR for each subarray. The goal is to find the partition that minimizes the maximum XOR value among the subarrays. The optimal Optimal Solution approach improves this to O(n² * k).

What are the interview tips for Partition Array to Minimize XOR?

Explain your thought process clearly. Consider edge cases and constraints. Practice similar problems to build confidence.

What are common mistakes when solving Partition Array to Minimize XOR?

Overlooking the need for non-empty subarrays. Miscalculating XOR values due to incorrect partitioning.

#3599

Partition Array to Minimize XOR

Medium

ArrayDynamic ProgrammingBit ManipulationPrefix SumDynamic ProgrammingPrefix Sum

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n² * k)
Space	O(1)	O(n * k)

💡

Intuition

Time O(n² * k)Space O(n * k)

We can use dynamic programming to efficiently compute the minimum possible value of the maximum XOR among k subarrays. By precomputing the XOR values, we can reduce redundant calculations.

⚙️

Algorithm

3 steps

1Step 1: Precompute the prefix XOR array where pre[i] = nums[0] ^ ... ^ nums[i].
2Step 2: Use dynamic programming to find the minimum maximum XOR for k partitions by iterating through possible partition points.
3Step 3: Update the DP table based on previously computed values and the current subarray XOR.

solution.py11 lines

1def minXor(nums, k):
2    n = len(nums)
3    pre = [0] * (n + 1)
4    for i in range(n): pre[i + 1] = pre[i] ^ nums[i]
5    dp = [[float('inf')] * (k + 1) for _ in range(n + 1)]
6    dp[0][0] = 0
7    for i in range(1, n + 1):
8        for j in range(1, k + 1):
9            for p in range(i):
10                dp[i][j] = min(dp[i][j], max(dp[p][j - 1], pre[i] ^ pre[p]))
11    return dp[n][k]

ℹ

Complexity note: The complexity arises from filling the DP table, iterating through n and k, leading to a quadratic relation.

1Understanding XOR properties can simplify calculations.
2Dynamic programming can optimize partitioning problems.

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