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;
        }
    }
    
  • // 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;
        }
    };
    
  • # 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
    
    

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