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1829. Maximum XOR for Each Query

中文文档

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

Solutions

Python3

class Solution:
    def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:
        n = len(nums)
        s = [0] * (n + 1)
        for i, x in enumerate(nums):
            s[i + 1] = s[i] ^ x
        ans = []
        for x in s[:0:-1]:
            t = 0
            for i in range(maximumBit):
                if ((x >> i) & 1) == 0:
                    t |= 1 << i
            ans.append(t)
        return ans

Java

class Solution {
    public int[] getMaximumXor(int[] nums, int maximumBit) {
        int n = nums.length;
        int[] s = new int[n + 1];
        for (int i = 0; i < n; ++i) {
            s[i + 1] = s[i] ^ nums[i];
        }
        int[] ans = new int[n];
        for (int i = n; i > 0; --i) {
            int t = 0, x = s[i];
            for (int j = 0; j < maximumBit; ++j) {
                if (((x >> j) & 1) == 0) {
                    t |= (1 << j);
                }
            }
            ans[n - i] = t;
        }
        return ans;
    }
}

C++

class Solution {
public:
    vector<int> getMaximumXor(vector<int>& nums, int maximumBit) {
        int n = nums.size();
        vector<int> s(n + 1);
        for (int i = 0; i < n; ++i) s[i + 1] = s[i] ^ nums[i];
        vector<int> ans;
        for (int i = n; i > 0; --i)
        {
            int t = 0, x = s[i];
            for (int j = 0; j < maximumBit; ++j) {
                if (((x >> j) & 1) == 0) t |= (1 << j);
            }
            ans.push_back(t);
        }
        return ans;
    }
};

Go

func getMaximumXor(nums []int, maximumBit int) []int {
	n := len(nums)
	s := make([]int, n+1)
	for i, v := range nums {
		s[i+1] = s[i] ^ v
	}
	var ans []int
	for i := n; i > 0; i-- {
		t, x := 0, s[i]
		for j := 0; j < maximumBit; j++ {
			if ((x >> j) & 1) == 0 {
				t |= (1 << j)
			}
		}
		ans = append(ans, t)
	}
	return ans
}

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