1829. Maximum XOR for Each Query

1829. Maximum XOR for Each Query

Difficulty: Medium

Topics: Array, Bit Manipulation, Prefix Sum

You are given a sorted array nums of n non-negative integers and an integer maximumBit. You want to perform the following query n times:

• Find a non-negative integer k < 2maximumBit such that nums[0] XOR nums[1] XOR ... XOR nums[nums.length-1] XOR k is maximized. k is the answer to the ith query.
• Remove the last element from the current array nums.

Return an array answer, where answer[i] is the answer to the ith query.

Example 1:

• Input: nums = [0,1,1,3], maximumBit = 2
• Output: [0,3,2,3]
• Explanation: The queries are answered as follows:
  • 1^stst query: nums = [0,1,1,3], k = 0 since 0 XOR 1 XOR 1 XOR 3 XOR 0 = 3.
  • 2^nd query: nums = [0,1,1], k = 3 since 0 XOR 1 XOR 1 XOR 3 = 3.
  • 3^rd query: nums = [0,1], k = 2 since 0 XOR 1 XOR 2 = 3.
  • 4^th query: nums = [0], k = 3 since 0 XOR 3 = 3.

Example 2:

• Input: nums = [2,3,4,7], maximumBit = 3
• Output: [5,2,6,5]
• Explanation: The queries are answered as follows:
  • 1^stst query: nums = [2,3,4,7], k = 5 since 2 XOR 3 XOR 4 XOR 7 XOR 5 = 7.
  • 2^nd query: nums = [2,3,4], k = 2 since 2 XOR 3 XOR 4 XOR 2 = 7.
  • 3^rd query: nums = [2,3], k = 6 since 2 XOR 3 XOR 6 = 7.
  • 4^th query: nums = [2], k = 5 since 2 XOR 5 = 7.

Example 3:

• Input: nums = [0,1,2,2,5,7], maximumBit = 3
• Output: [4,3,6,4,6,7]

Constraints:

• nums.length == n
• 1 <= n <= 105
• 1 <= maximumBit <= 20
• 0 <= nums[i] < 2maximumBit
• nums​​​ is sorted in ascending order.

Hint:

1. Note that the maximum possible XOR result is always 2(maximumBit) - 1
2. So the answer for a prefix is the XOR of that prefix XORed with 2(maximumBit)-1

Solution:

We need to efficiently calculate the XOR of elements in the array and maximize the result using a value k such that k is less than 2^maximumBit. Here's the approach for solving this problem:

Observations and Approach

1. Maximizing XOR:
   The maximum number we can XOR with any prefix sum for maximumBit bits is ( 2^{\text{maximumBit}} - 1 ). This is because XORing with a number of all 1s (i.e., 111...1 in binary) will always maximize the result.

2. Prefix XOR Calculation:
   Instead of recalculating the XOR for each query, we can maintain a cumulative XOR for the entire array. Since XOR has the property that A XOR B XOR B = A, removing the last element from the array can be achieved by XORing out that element from the cumulative XOR.

3. Algorithm:

   • Compute the XOR of all elements in nums initially. Let's call this currentXOR.
   • For each query (from last to first):
     • Calculate the optimal value of k for that query by XORing currentXOR with maxNum where maxNum = 2^maximumBit - 1.
     • Append k to the result list.
     • Remove the last element from nums by XORing it out of currentXOR.
   • The result list will contain the answers in reverse order, so reverse it at the end.

Let's implement this solution in PHP: 1829. Maximum XOR for Each Query

<?php
/**
 * @param Integer[] $nums
 * @param Integer $maximumBit
 * @return Integer[]
 */
function getMaximumXor($nums, $maximumBit) {
    ...
    ...
    ...
    /**
     * go to ./solution.php
     */
}

// Example usage:
$nums = [0,1,1,3];
$maximumBit = 2;
print_r(getMaximumXor($nums, $maximumBit));  // Output should be [0, 3, 2, 3]
?>

Explanation:

1. Calculate maxNum:

   • maxNum is calculated as 2^maximumBit - 1, which is the number with all 1s in binary for the specified bit length.

2. Initial XOR Calculation:

   • We XOR all elements in nums to get the cumulative XOR (currentXOR), representing the XOR of all numbers in the array.

3. Iterate Backwards:

   • We start from the last element in nums and calculate the maximum XOR for each step:
     • currentXOR ^ maxNum gives the maximum k for the current state.
     • Append k to answer.
   • We then XOR the last element of nums with currentXOR to "remove" it from the XOR sum for the next iteration.

4. Return the Answer:

   • Since we processed the list in reverse, answer will contain the values in reverse order, so the final list is already arranged correctly for our requirements.

Complexity Analysis

• Time Complexity: O(n), since we compute the initial XOR in O(n) and each query is processed in constant time.
• Space Complexity: O(n), for storing the answer.

This code is efficient and should handle the upper limits of the constraints well.

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