How do you solve Number of Alternating XOR Partitions on LeetCode?

Number of Alternating XOR Partitions (LeetCode #3811) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Use prefix XORs for efficient block XOR calculations.

What is the time complexity of Number of Alternating XOR Partitions?

The optimal solution for Number of Alternating XOR Partitions runs in O(n) time and uses O(n) space. We compute prefix XORs in O(n) and use a DP array to track valid partitions, leading to O(n) overall complexity.

What is the brute force solution for Number of Alternating XOR Partitions?

The brute force approach for Number of Alternating XOR Partitions is Brute Force, with O(n²) time complexity and O(1) space complexity. We can generate all possible partitions of the array and check if they alternate between target1 and target2. This approach is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Number of Alternating XOR Partitions?

Number of Alternating XOR Partitions uses the following concepts: Array, Hash Table, Dynamic Programming, Bit Manipulation. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Number of Alternating XOR Partitions?

Explain your thought process clearly. Discuss edge cases and constraints. Practice coding under time constraints.

What are common mistakes when solving Number of Alternating XOR Partitions?

Not considering all possible block sizes. Forgetting to use modulo in results.

#3811

Number of Alternating XOR Partitions

Medium

ArrayHash TableDynamic ProgrammingBit ManipulationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

Using dynamic programming and prefix XORs allows us to efficiently compute the XOR of any subarray, reducing the problem to checking valid partitions in linear time.

⚙️

Algorithm

3 steps

1Step 1: Compute prefix XORs for the array.
2Step 2: Use dynamic programming to count valid partitions based on the prefix XORs.
3Step 3: Check if the current prefix XOR matches target1 or target2 and update the count accordingly.

solution.py14 lines

1def countPartitions(nums, target1, target2):
2    MOD = 10**9 + 7
3    n = len(nums)
4    pref = [0] * (n + 1)
5    for i in range(n):
6        pref[i + 1] = pref[i] ^ nums[i]
7    dp = [0] * (n + 1)
8    dp[0] = 1
9    for i in range(1, n + 1):
10        if pref[i] == target1:
11            dp[i] = (dp[i] + dp[i - 1]) % MOD
12        if i > 1 and pref[i] == target2:
13            dp[i] = (dp[i] + dp[i - 2]) % MOD
14    return dp[n]

ℹ

Complexity note: We compute prefix XORs in O(n) and use a DP array to track valid partitions, leading to O(n) overall complexity.

1Use prefix XORs for efficient block XOR calculations.
2Dynamic programming helps in counting valid partitions.

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