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1829. Maximum XOR for Each Query

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

Solution 1: Bitwise Operation + Enumeration

First, we preprocess the XOR sum xs of the array nums, i.e., xs=nums[0]nums[1]nums[n1].

Next, we enumerate each element x in the array nums from back to front. The current XOR sum is xs. We need to find a number k such that the value of xsk is as large as possible, and k<2maximumBit.

That is to say, we start from the maximumBit1 bit of xs and enumerate to the lower bit. If a bit of xs is 0, then we set the corresponding bit of k to 1. Otherwise, we set the corresponding bit of k to 0. In this way, the final k is the answer to each query. Then, we update xs to xsx and continue to enumerate the next element.

The time complexity is O(n×m), where n and m are the values of the array nums and maximumBit respectively. Ignoring the space consumption of the answer, the space complexity is O(1).

Solution 2: Enumeration Optimization

Similar to Solution 1, we first preprocess the XOR sum xs of the array nums, i.e., xs=nums[0]nums[1]nums[n1].

Next, we calculate 2maximumBit1, which is 2maximumBit minus 1, denoted as mask. Then, we enumerate each element x in the array nums from back to front. The current XOR sum is xs, then k=xsmask is the answer to each query. Then, we update xs to xsx and continue to enumerate the next element.

The time complexity is O(n), where n is the length of the array nums. Ignoring the space consumption of the answer, the space complexity is O(1).

  • class Solution {
        public int[] getMaximumXor(int[] nums, int maximumBit) {
            int n = nums.length;
            int xs = 0;
            for (int x : nums) {
                xs ^= x;
            }
            int[] ans = new int[n];
            for (int i = 0; i < n; ++i) {
                int x = nums[n - i - 1];
                int k = 0;
                for (int j = maximumBit - 1; j >= 0; --j) {
                    if (((xs >> j) & 1) == 0) {
                        k |= 1 << j;
                    }
                }
                ans[i] = k;
                xs ^= x;
            }
            return ans;
        }
    }
    
  • class Solution {
    public:
        vector<int> getMaximumXor(vector<int>& nums, int maximumBit) {
            int xs = 0;
            for (int& x : nums) {
                xs ^= x;
            }
            int n = nums.size();
            vector<int> ans(n);
            for (int i = 0; i < n; ++i) {
                int x = nums[n - i - 1];
                int k = 0;
                for (int j = maximumBit - 1; ~j; --j) {
                    if ((xs >> j & 1) == 0) {
                        k |= 1 << j;
                    }
                }
                ans[i] = k;
                xs ^= x;
            }
            return ans;
        }
    };
    
  • class Solution:
        def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:
            ans = []
            xs = reduce(xor, nums)
            for x in nums[::-1]:
                k = 0
                for i in range(maximumBit - 1, -1, -1):
                    if (xs >> i & 1) == 0:
                        k |= 1 << i
                ans.append(k)
                xs ^= x
            return ans
    
    
  • func getMaximumXor(nums []int, maximumBit int) (ans []int) {
    	xs := 0
    	for _, x := range nums {
    		xs ^= x
    	}
    	for i := range nums {
    		x := nums[len(nums)-i-1]
    		k := 0
    		for j := maximumBit - 1; j >= 0; j-- {
    			if xs>>j&1 == 0 {
    				k |= 1 << j
    			}
    		}
    		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 n = nums.length;
        const ans = new Array(n);
        for (let i = 0; i < n; ++i) {
            const x = nums[n - i - 1];
            let k = 0;
            for (let j = maximumBit - 1; j >= 0; --j) {
                if (((xs >> j) & 1) == 0) {
                    k |= 1 << j;
                }
            }
            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 n = nums.length;
        const ans = new Array(n);
        for (let i = 0; i < n; ++i) {
            const x = nums[n - i - 1];
            let k = 0;
            for (let j = maximumBit - 1; j >= 0; --j) {
                if (((xs >> j) & 1) == 0) {
                    k |= 1 << j;
                }
            }
            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 n = nums.Length;
            int[] ans = new int[n];
            for (int i = 0; i < n; ++i) {
                int x = nums[n - i - 1];
                int k = 0;
                for (int j = maximumBit - 1; j >= 0; --j) {
                    if ((xs >> j & 1) == 0) {
                        k |= 1 << j;
                    }
                }
                ans[i] = k;
                xs ^= x;
            }
            return ans;
        }
    }
    
    

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