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Formatted question description: https://leetcode.ca/all/1829.html

1829. Maximum XOR for Each Query

Level

Medium

Description

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:

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

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

Example 1:

Input: nums = [0,1,1,3], maximumBit = 2

Output: [0,3,2,3]

Explanation: The queries are answered as follows:

1st query: nums = [0,1,1,3], k = 0 since 0 XOR 1 XOR 1 XOR 3 XOR 0 = 3.

2nd query: nums = [0,1,1], k = 3 since 0 XOR 1 XOR 1 XOR 3 = 3.

3rd query: nums = [0,1], k = 2 since 0 XOR 1 XOR 2 = 3.

4th 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:

1st query: nums = [2,3,4,7], k = 5 since 2 XOR 3 XOR 4 XOR 7 XOR 5 = 7.

2nd query: nums = [2,3,4], k = 2 since 2 XOR 3 XOR 4 XOR 2 = 7.

3rd query: nums = [2,3], k = 6 since 2 XOR 3 XOR 6 = 7.

4th 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 <= 10^5
  • 1 <= maximumBit <= 20
  • 0 <= nums[i] < 2^maximumBit
  • nums is sorted in ascending order.

Solution

Create an array prefixXors of length n, where prefixXors[i] is the xor from nums[0] to nums[i]. Each time, we need to find the k such that prefixXors[n - 1 - i] ^ k is maximized.

Since k < 2^maximumBit and all elements in nums are less than 2^maximumBit, obviously all elements in maximumXors are less than 2^maximumBit. Therefore, the maximum possible value for prefixXors[n - 1 - i] ^ k is (1 << maximumBit) - 1. The answer to the i-th query is perfixXors[n - 1 - i] ^ ((1 << maximumBit) - 1).

Finally, return the array of the answers.

  • class Solution {
    public int[] getMaximumXor(int[] nums, int maximumBit) {
        int maximum = (1 << maximumBit) - 1;
        int n = nums.length;
        int[] prefixXors = new int[n];
        prefixXors[0] = nums[0];
        for (int i = 1; i < n; i++)
            prefixXors[i] = prefixXors[i - 1] ^ nums[i];
        int[] maximumXors = new int[n];
        for (int i = 0; i < n; i++) {
            int prefixXor = prefixXors[n - 1 - i];
            maximumXors[i] = maximum ^ (maximum & prefixXor);
        }
        return maximumXors;
    }
}

############

class Solution {
    public int[] getMaximumXor(int[] nums, int maximumBit) {
        int xs = 0;
        for (int x : nums) {
            xs ^= x;
        }
        int mask = (1 << maximumBit) - 1;
        int n = nums.length;
        int[] ans = new int[n];
        for (int i = 0; i < n; ++i) {
            int x = nums[n - i - 1];
            int k = xs ^ mask;
            ans[i] = k;
            xs ^= x;
        }
        return ans;
    }
}

  • // OJ: https://leetcode.com/problems/maximum-xor-for-each-query/
// Time: O(N)
// Space: O(1)
class Solution {
public:
    vector<int> getMaximumXor(vector<int>& A, int MB) {
        int N = A.size(), t = (1 << MB) - 1;
        vector<int> ans(N);
        for (int i = 0; i < N; ++i) {
            A[i] ^= i == 0 ? 0 : A[i - 1];
            ans[N - i - 1] = t ^ A[i];
        }
        return ans;
    }
};

  • class Solution:
    def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:
        ans = []
        xs = reduce(xor, nums)
        mask = (1 << maximumBit) - 1
        for x in nums[::-1]:
            k = xs ^ mask
            ans.append(k)
            xs ^= x
        return ans

############

# 1829. Maximum XOR for Each Query
# https://leetcode.com/problems/maximum-xor-for-each-query/

class Solution:
    def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:
        n = len(nums)
        m = (2 ** maximumBit) - 1
        res = []
        
        for i in range(1, n):
            nums[i] ^= nums[i - 1]

        for num in nums[::-1]:
            res.append(num ^ m)
            
        return res


  • func getMaximumXor(nums []int, maximumBit int) (ans []int) {
	xs := 0
	for _, x := range nums {
		xs ^= x
	}
	mask := (1 << maximumBit) - 1
	for i := range nums {
		x := nums[len(nums)-i-1]
		k := xs ^ mask
		ans = append(ans, k)
		xs ^= x
	}
	return
}

  • function getMaximumXor(nums: number[], maximumBit: number): number[] {
    let xs = 0;
    for (const x of nums) {
        xs ^= x;
    }
    const mask = (1 << maximumBit) - 1;
    const n = nums.length;
    const ans = new Array(n);
    for (let i = 0; i < n; ++i) {
        const x = nums[n - i - 1];
        let k = xs ^ mask;
        ans[i] = k;
        xs ^= x;
    }
    return ans;
}


  • /**
 * @param {number[]} nums
 * @param {number} maximumBit
 * @return {number[]}
 */
var getMaximumXor = function (nums, maximumBit) {
    let xs = 0;
    for (const x of nums) {
        xs ^= x;
    }
    const mask = (1 << maximumBit) - 1;
    const n = nums.length;
    const ans = new Array(n);
    for (let i = 0; i < n; ++i) {
        const x = nums[n - i - 1];
        let k = xs ^ mask;
        ans[i] = k;
        xs ^= x;
    }
    return ans;
};


  • public class Solution {
    public int[] GetMaximumXor(int[] nums, int maximumBit) {
        int xs = 0;
        foreach (int x in nums) {
            xs ^= x;
        }
        int mask = (1 << maximumBit) - 1;
        int n = nums.Length;
        int[] ans = new int[n];
        for (int i = 0; i < n; ++i) {
            int x = nums[n - i - 1];
            int k = xs ^ mask;
            ans[i] = k;
            xs ^= x;
        }
        return ans;
    }
}

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