Problem Description

You have an integer array nums and an integer k. Your goal is to split the array into exactly k non-empty subarrays (consecutive elements).

For each subarray, you calculate the XOR of all its elements. Among all k XOR values obtained, you take the maximum value.

Your task is to find the minimum possible value of this maximum XOR across all valid ways to partition the array.

Example walkthrough:

If nums = [1, 2, 3, 4] and k = 2, you could partition it as:

• [1, 2] and [3, 4]: XOR values are 1 ^ 2 = 3 and 3 ^ 4 = 7, maximum is 7
• [1] and [2, 3, 4]: XOR values are 1 and 2 ^ 3 ^ 4 = 5, maximum is 5
• [1, 2, 3] and [4]: XOR values are 1 ^ 2 ^ 3 = 0 and 4, maximum is 4

Among all partitions, the minimum of these maximum values would be the answer.

Key points:

• Each subarray must be non-empty and consecutive
• You must use exactly k subarrays
• The entire array must be covered by the k subarrays
• XOR operation: For a subarray [a, b, c], the XOR is a ^ b ^ c
• You want to minimize the largest XOR value among all k subarrays

Intuition

When we need to partition an array into subarrays with some optimization goal, dynamic programming often comes to mind. The key insight is that we can build up the solution by considering smaller subproblems.

Let's think about what information we need to track. For any position in the array, we need to know:

1. How many elements we've processed so far
2. How many subarrays we've created so far
3. What's the minimum possible maximum XOR value achievable

This naturally leads to defining a state f[i][j] representing "the minimum possible maximum XOR when partitioning the first i elements into exactly j subarrays."

To compute f[i][j], we need to consider all possible positions where the last subarray could start. If the last subarray starts at position h+1 and ends at position i, then:

• The previous h elements must be partitioned into j-1 subarrays
• The XOR of the last subarray is the XOR of elements from h+1 to i
• The overall maximum XOR is the max of the previous maximum (from f[h][j-1]) and the current subarray's XOR

To efficiently calculate XOR values for any subarray, we use prefix XOR. If g[i] stores the XOR of the first i elements, then the XOR of elements from position h+1 to i is simply g[i] ^ g[h]. This works because XOR has the property that a ^ a = 0, so the elements before position h+1 cancel out.

The base case is f[0][0] = 0 (zero elements partitioned into zero subarrays has no XOR values, so we can consider the maximum as 0). All other states start at infinity since they haven't been computed yet.

By iterating through all positions i, all possible number of partitions j, and all possible starting positions h for the last subarray, we can fill the DP table and eventually find f[n][k] - the answer to our problem.

Learn more about Dynamic Programming and Prefix Sum patterns.

Solution Approach

We implement the dynamic programming solution using a 2D table and prefix XOR array.

Step 1: Initialize Prefix XOR Array

First, we create a prefix XOR array g where g[i] stores the XOR of the first i elements:

g = [0] * (n + 1)
for i, x in enumerate(nums, 1):
    g[i] = g[i - 1] ^ x

This allows us to compute the XOR of any subarray [h+1...i] as g[i] ^ g[h] in O(1) time.

Step 2: Initialize DP Table

Create a 2D DP table f where f[i][j] represents the minimum possible maximum XOR when partitioning the first i elements into j subarrays:

f = [[inf] * (k + 1) for _ in range(n + 1)]
f[0][0] = 0

We initialize all values to infinity except f[0][0] = 0 (base case: zero elements into zero subarrays).

Step 3: Fill the DP Table

We iterate through three nested loops:

• i from 1 to n: number of elements considered
• j from 1 to min(i, k): number of partitions (can't have more partitions than elements)
• h from j-1 to i-1: ending position of the previous subarray

For each combination, we update f[i][j] using the state transition equation:

f[i][j] = min(f[i][j], max(f[h][j - 1], g[i] ^ g[h]))

The equation works as follows:

• f[h][j-1]: the minimum maximum XOR for the first h elements partitioned into j-1 subarrays
• g[i] ^ g[h]: the XOR of the current subarray from position h+1 to i
• max(...): the maximum XOR among all j subarrays (including the new one)
• min(...): we keep the minimum among all possible ways to place the last subarray

Step 4: Return the Result

After filling the entire table, f[n][k] contains the minimum possible maximum XOR when partitioning all n elements into exactly k subarrays.

Time Complexity: O(n² × k) where n is the length of the array Space Complexity: O(n × k) for the DP table

Example Walkthrough

Let's walk through a small example with nums = [3, 9, 5] and k = 2.

Step 1: Initialize Prefix XOR Array

• g[0] = 0
• g[1] = 0 ^ 3 = 3
• g[2] = 3 ^ 9 = 10
• g[3] = 10 ^ 5 = 15

This gives us g = [0, 3, 10, 15]. Now we can compute any subarray XOR quickly:

• XOR of [3]: g[1] ^ g[0] = 3 ^ 0 = 3
• XOR of [9, 5]: g[3] ^ g[1] = 15 ^ 3 = 12
• XOR of [3, 9]: g[2] ^ g[0] = 10 ^ 0 = 10

Step 2: Initialize DP Table

f[i][j] where i = 0 to 3, j = 0 to 2
Initial state (all inf except f[0][0] = 0):
    j=0   j=1   j=2
i=0  0    inf   inf
i=1  inf  inf   inf
i=2  inf  inf   inf
i=3  inf  inf   inf

Step 3: Fill the DP Table

For i=1, j=1: First element into 1 subarray

• h can only be 0 (previous 0 elements in 0 subarrays)
• f[1][1] = min(inf, max(f[0][0], g[1] ^ g[0])) = max(0, 3) = 3

For i=2, j=1: First 2 elements into 1 subarray

• h can only be 0
• f[2][1] = min(inf, max(f[0][0], g[2] ^ g[0])) = max(0, 10) = 10

For i=2, j=2: First 2 elements into 2 subarrays

• h can only be 1 (split after first element)
• f[2][2] = min(inf, max(f[1][1], g[2] ^ g[1])) = max(3, 10 ^ 3) = max(3, 9) = 9

For i=3, j=1: All 3 elements into 1 subarray

• h can only be 0
• f[3][1] = min(inf, max(f[0][0], g[3] ^ g[0])) = max(0, 15) = 15

For i=3, j=2: All 3 elements into 2 subarrays

• h=1: Split as [3] and [9, 5]
  • f[3][2] = min(inf, max(f[1][1], g[3] ^ g[1])) = max(3, 12) = 12
• h=2: Split as [3, 9] and [5]
  • f[3][2] = min(12, max(f[2][1], g[3] ^ g[2])) = min(12, max(10, 5)) = min(12, 10) = 10

Final DP Table:

    j=0   j=1   j=2
i=0  0    inf   inf
i=1  inf   3    inf
i=2  inf  10     9
i=3  inf  15    10

Step 4: Return Result f[3][2] = 10, which means the minimum possible maximum XOR when splitting [3, 9, 5] into 2 subarrays is 10.

This corresponds to the partition [3, 9] and [5]:

• XOR of [3, 9] = 3 ^ 9 = 10
• XOR of [5] = 5
• Maximum = 10

Solution Implementation

Time and Space Complexity

Time Complexity: O(n^2 × k)

The time complexity is determined by the triple nested loop structure:

• The outer loop iterates through i from 1 to n (n iterations)
• The middle loop iterates through j from 1 to min(i, k) (at most k iterations)
• The inner loop iterates through h from j-1 to i-1 (at most n iterations)

For each combination of (i, j, h), the operations performed (XOR, min, max comparisons) are O(1). Therefore, the overall time complexity is O(n × k × n) = O(n^2 × k).

Space Complexity: O(n × k)

The space complexity comes from:

• Array g of size n+1: O(n)
• 2D DP array f of size (n+1) × (k+1): O(n × k)

The dominant space requirement is the 2D array f, giving us a total space complexity of O(n × k).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Base Case Initialization

Pitfall: A common mistake is incorrectly initializing the DP table, particularly forgetting to set dp[0][0] = 0 or mistakenly setting other base cases like dp[i][0] = 0 for all i.

Why it's wrong:

• dp[i][0] for i > 0 should remain infinity because you cannot partition a non-empty array into 0 subarrays
• Only dp[0][0] = 0 is valid (0 elements into 0 subarrays)

Solution:

# Correct initialization
dp = [[inf] * (k + 1) for _ in range(n + 1)]
dp[0][0] = 0  # Only this base case is valid

# Incorrect (don't do this)
# for i in range(n + 1):
#     dp[i][0] = 0  # Wrong! Can't partition non-empty array into 0 subarrays

2. Off-by-One Errors in Loop Boundaries

Pitfall: Using incorrect loop ranges, especially for the variable h which represents the split point.

Why it's wrong:

• If h starts from j instead of j-1, you miss valid partitions
• If h goes up to i instead of i-1, you create empty subarrays

Solution:

# Correct: h ranges from j-1 to i-1
for h in range(j - 1, i):
    current_xor = prefix_xor[i] ^ prefix_xor[h]
    # This creates subarray from index h to i-1

# Incorrect examples:
# for h in range(j, i):  # Misses valid partitions
# for h in range(j - 1, i + 1):  # Creates empty subarray when h = i

3. Prefix XOR Calculation Error

Pitfall: Computing the XOR of a subarray incorrectly by using wrong indices with the prefix XOR array.

Why it's wrong:

• The XOR of subarray from index left to right (inclusive) should be prefix_xor[right + 1] ^ prefix_xor[left]
• In our DP, the subarray from position h to i-1 should use prefix_xor[i] ^ prefix_xor[h]

Solution:

# Correct: XOR of elements from index h to i-1
current_xor = prefix_xor[i] ^ prefix_xor[h]

# Incorrect (common mistakes):
# current_xor = prefix_xor[i] ^ prefix_xor[h - 1]  # Wrong indices
# current_xor = prefix_xor[i - 1] ^ prefix_xor[h]  # Wrong indices

4. Not Handling Edge Cases

Pitfall: Failing to handle edge cases like when k = 1 (entire array as one subarray) or k = n (each element as its own subarray).

Solution:

def minXor(self, nums: List[int], k: int) -> int:
    n = len(nums)
  
    # Edge case: if k equals n, each element is its own subarray
    if k == n:
        return max(nums)
  
    # Edge case: if k equals 1, return XOR of entire array
    if k == 1:
        result = 0
        for num in nums:
            result ^= num
        return result
  
    # Continue with normal DP solution...

5. Memory Optimization Opportunity Missed

Pitfall: Not recognizing that we only need the previous row of the DP table, leading to unnecessary space usage.

Solution (Space-Optimized):

# Instead of 2D array, use two 1D arrays
prev = [inf] * (k + 1)
curr = [inf] * (k + 1)
prev[0] = 0

for i in range(1, n + 1):
    curr = [inf] * (k + 1)
    for j in range(1, min(i, k) + 1):
        for h in range(j - 1, i):
            current_xor = prefix_xor[i] ^ prefix_xor[h]
            if h == 0:
                max_xor = current_xor
            else:
                max_xor = max(prev[j - 1] if h == i - 1 else dp[h][j - 1], current_xor)
            curr[j] = min(curr[j], max_xor)
    prev = curr