Program to find maximum XOR for each query in Python

Suppose we have a presorted array called nums of size n and a value m. We want to perform the following query n times:

• Search for a non-negative value k < 2^m such that XOR of all elements in nums and k is maximized. So k is the answer to the ith query.

• Remove the last element from the current array nums.

• We have to find an array answer, where answer[i] is the answer to the ith query.

Example Walkthrough

If the input is like nums = [0,1,1,3], m = 2, then the output will be [0,3,2,3], because ?

• nums = [0,1,1,3], k = 0 since 0 XOR 1 XOR 1 XOR 3 XOR 0 = 3.

• nums = [0,1,1], k = 3 since 0 XOR 1 XOR 1 XOR 3 = 3.

• nums = [0,1], k = 2 since 0 XOR 1 XOR 2 = 3.

• nums = [0], k = 3 since 0 XOR 3 = 3.

Algorithm

To solve this problem, we follow these steps ?

• Initialize x := 2^m - 1

• For each element in nums, XOR it with x, then update x to the new value

• Return the reversed array

Implementation

def solve(nums, m):
    x = 2**m - 1
    for i in range(len(nums)):
        nums[i] ^= x
        x = nums[i]
    return nums[::-1]

# Test the function
nums = [0, 1, 1, 3]
m = 2
result = solve(nums, m)
print("Input:", [0, 1, 1, 3], "m =", 2)
print("Output:", result)

Input: [0, 1, 1, 3] m = 2
Output: [0, 3, 2, 3]

How It Works

The algorithm works by transforming each element through XOR operations with a mask value that gets updated in each iteration. The mask starts as 2^m - 1, which represents the maximum possible value with m bits. After each XOR operation, the mask becomes the transformed element, creating a chain effect that maximizes the XOR result for subsequent queries.

Step-by-Step Execution

def solve_with_steps(nums, m):
    print(f"Initial: nums = {nums}, m = {m}")
    x = 2**m - 1
    print(f"Initial mask x = {x}")
    
    for i in range(len(nums)):
        old_val = nums[i]
        nums[i] ^= x
        x = nums[i]
        print(f"Step {i+1}: {old_val} XOR {2**m - 1 if i == 0 else nums[i-1]} = {nums[i]}, new x = {x}")
    
    result = nums[::-1]
    print(f"Final result (reversed): {result}")
    return result

# Test with step-by-step output
nums = [0, 1, 1, 3]
m = 2
solve_with_steps(nums, m)

Initial: nums = [0, 1, 1, 3], m = 2
Initial mask x = 3
Step 1: 0 XOR 3 = 3, new x = 3
Step 2: 1 XOR 3 = 2, new x = 2
Step 3: 1 XOR 2 = 3, new x = 3
Step 4: 3 XOR 3 = 0, new x = 0
Final result (reversed): [0, 3, 2, 3]

Conclusion

This algorithm efficiently finds the maximum XOR values for each query by using a cumulative XOR approach with a dynamic mask. The key insight is that reversing the transformed array gives us the optimal k values for each query in the correct order.