
Problem Statement
In this challenge, you have a sorted array, nums
, composed of non-negative integers, and an integer maximumBit
. You are required to perform a series of operations on this array n
times where n
is the length of nums
. Each operation involves computing the XOR of all elements in the array with an integer k
such that the result is maximized. The integer k
should be less than 2
raised to the power of maximumBit
. Once the maximum XOR value is found for that configuration, the integer k
is recorded, and then the last element of the array nums
is removed. The routine follows until the original array is emptied. The goal is to produce an output array answer
, where each element answer[i]
corresponds to the integer k
found during the i
-th operation.
Examples
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.
Approach and Intuition
Given the sorted array nums
and the restriction k < 2^maximumBit
, the XOR operation's nature and the results presented can give us a clear path for the procedure:
Compute XOR for Maximum Number: Since XOR with the same number twice results in 0, an optimal
k
can often be found when XOR-ing the entire array and then XOR-ing its result with the number just shy of2^maximumBit
(i.e.,2^maximumBit - 1
). This number in binary will be a sequence of all 1's up to themaximumBit
-th bit, likely resulting in a high outcome when XOR-ing with any given number within the constraints. This strategy is mathematically sound due to the properties of the XOR operation and the binary number system.Iterative Reduction and Calculation: Begin with the whole array and calculate the suitable
k
using the method described. Record thek
. Remove the last element ofnums
and repeat the computation for the next iteration with the now shortened array. Continue this until all elements innums
have been exhausted.Optimization Awareness: Care should be taken as Python's handling of large integers should not be an issue within the given constraints (given the
maximumBit
limit of 20, resulting in numbers that are much smaller than what Python typically can handle efficiently). Also, the sorted nature ofnums
does not directly affect the findings from the XOR operations but could be useful if the calculation pattern needs adjusting or optimizing.
This solution strategy leverages the predictability and reversal properties of XOR to ensure a maximal value through patterned binary operation, exploiting the operation's behavior with maximal binary number under given constraints.
Solutions
- C++
- Java
- Python
class Solution {
public:
vector<int> calculateMaximumXor(vector<int>& values, int bitLevel) {
int currentXor = 0;
for (int num : values) {
currentXor ^= num;
}
vector<int> result(values.size());
int fullMask = (1 << bitLevel) - 1;
for (int index = 0; index < values.size(); index++) {
result[index] = currentXor ^ fullMask;
// Reverse through list reducing xor
currentXor ^= values[values.size() - 1 - index];
}
return result;
}
};
This solution describes a method to determine the maximum XOR for each query from a list of integers and a specified bit level using C++. The approach leverages bit manipulation to achieve the results.
Firstly, the function starts by calculating the XOR of all elements in the provided list values
. It stores this result in currentXor
.
Next, it calculates the full bit mask for the given bitLevel
. This mask is computed as (1 << bitLevel) - 1
, which effectively sets all bits less than bitLevel
to 1.
The function then iterates over each element in values
. For each index, it computes the XOR of currentXor
with fullMask
and stores this in the result list. Essentially, this operation aims to flip the bits of currentXor
up to the specified bit level.
To update currentXor
for the next iteration continuously, the algorithm reverses through the values
list while reducing the XOR operation by the current values from the end of the list. This reverse XOR operation guarantees that each query's resultant XOR reflects the loss of the specific integer's bit influence from the calculations.
Finally, the function returns the results as a list of integers where each integer represents the maximum XOR possible with the remaining integers in the list by the end of each query. This solution efficiently utilizes bitwise operations for optimal performance in scenarios involving bit manipulations in queries.
class Solution {
public int[] computeMaxXor(int[] values, int maxBit) {
int cumulativeXor = 0;
for (int v : values) {
cumulativeXor ^= v;
}
int[] result = new int[values.length];
int fullMask = (1 << maxBit) - 1;
for (int idx = 0; idx < values.length; idx++) {
result[idx] = cumulativeXor ^ fullMask;
cumulativeXor ^= values[values.length - 1 - idx];
}
return result;
}
}
The Java code provided outlines a solution for computing the maximum XOR for each query based on a given array values
and an integer maxBit
.
The method
computeMaxXor
accepts two parameters:int[] values
- an array of integers.int maxBit
- the maximum bit position to consider in the XOR calculations.
The code uses an integer
cumulativeXor
initialized to 0, to accumulate the XOR of all elements in thevalues
array, ensuring that all bits up tomaxBit
are considered.An integer array
result
with the same length asvalues
is initialized to store the result of each XOR calculation against the full mask.int fullMask
is calculated using1 << maxBit
, subtracted by 1. This operation creates a binary number that consists entirely of 1's up to themaxBit
th position.A loop runs through the
values
array:- At each index
idx
, the XOR ofcumulativeXor
withfullMask
is stored inresult[idx]
. - Then,
cumulativeXor
is updated by XORing it with the element invalues
starting from the end towards the beginning (values[values.length - 1 - idx]
).
- At each index
The method finally returns the
result
array, which contains the maximum XOR results for each query, considering the sequence and the constraints set bymaxBit
.
This approach efficiently computes the required values by progressively accumulating the XOR and applying it against a full mask of significant bits. The use of bitwise operations ensures the method is optimized for quick computation.
class Solution:
def calculateMaxXor(self, num_list: List[int], max_bits: int) -> List[int]:
cumulative_xor = 0
for num in num_list:
cumulative_xor ^= num
result = [0] * len(num_list)
all_ones_mask = (1 << max_bits) - 1
for index in range(len(num_list)):
result[index] = cumulative_xor ^ all_ones_mask
cumulative_xor ^= num_list[len(num_list) - 1 - index]
return result
The given Python code defines a method to compute the maximum XOR for each query from a list of integers using a bitwise operation approach. The function calculateMaxXor
in the Solution
class takes two parameters: num_list
, a list of integers, and max_bits
, which represents the number of bits to consider for the maximum possible value.
- Initialize
cumulative_xor
to zero. This variable will store the cumulative XOR of all numbers processed so far. - Based on
max_bits
, computeall_ones_mask
which is a number consisting of all 1s of length equal tomax_bits
. - Iterate through each number in the
num_list
:- Perform XOR between
cumulative_xor
andall_ones_mask
to store in the result list for the current index. - Update
cumulative_xor
by XORing it with the current number from the end of the list moving backward.
- Perform XOR between
- Finally, return the result list containing maximum XOR values after each query.
This algorithm effectively utilizes cumulative XOR values and a mask of all ones to find the desired result and adjusts the cumulative XOR in reverse to cater for each individual query dynamically. This method ensures that each XOR result is calculated using an efficient bitwise manipulation, achieving the task in an optimized manner.
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