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Kotlin-based LeetCode algorithm problem solutions, regularly updated
2317\. Maximum XOR After Operations
Medium
You are given a **0-indexed** integer array `nums`. In one operation, select **any** non-negative integer `x` and an index `i`, then **update** `nums[i]` to be equal to `nums[i] AND (nums[i] XOR x)`.
Note that `AND` is the bitwise AND operation and `XOR` is the bitwise XOR operation.
Return _the **maximum** possible bitwise XOR of all elements of_ `nums` _after applying the operation **any number** of times_.
**Example 1:**
**Input:** nums = [3,2,4,6]
**Output:** 7
**Explanation:** Apply the operation with x = 4 and i = 3, num[3] = 6 AND (6 XOR 4) = 6 AND 2 = 2.
Now, nums = [3, 2, 4, 2] and the bitwise XOR of all the elements = 3 XOR 2 XOR 4 XOR 2 = 7.
It can be shown that 7 is the maximum possible bitwise XOR.
Note that other operations may be used to achieve a bitwise XOR of 7.
**Example 2:**
**Input:** nums = [1,2,3,9,2]
**Output:** 11
**Explanation:** Apply the operation zero times.
The bitwise XOR of all the elements = 1 XOR 2 XOR 3 XOR 9 XOR 2 = 11.
It can be shown that 11 is the maximum possible bitwise XOR.
**Constraints:**
* 1 <= nums.length <= 105
* 0 <= nums[i] <= 108