Maximum XOR for Each Query | Leetcode 1829

class Solution {
public:
    vector<int> getMaximumXor(vector<int>& nums, int maximumBit) {
        int n=nums.size();

        vector<int> prefix_xor(n);
        prefix_xor[0]=nums[0];
        for(int i=1;i<n;++i)
            prefix_xor[i] = prefix_xor[i-1] ^ nums[i];
        
        vector<int> ans;
        int mask = (1<<maximumBit)-1;
        for(int i=n-1;i>=0;--i){
            ans.push_back(prefix_xor[i] ^ mask);
        }
        return ans;
    }
};

/*
//JAVA
import java.util.ArrayList;
import java.util.List;

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

#Python
from typing import List

class Solution:
    def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:
        n = len(nums)
        prefix_xor = [0] * n
        prefix_xor[0] = nums[0]
        
        for i in range(1, n):
            prefix_xor[i] = prefix_xor[i - 1] ^ nums[i]
        
        ans = []
        mask = (1 << maximumBit) - 1
        
        for i in range(n - 1, -1, -1):
            ans.append(prefix_xor[i] ^ mask)
        
        return ans

*/