2401. Longest Nice Subarray
Description
You are given an array nums
consisting of positive integers.
We call a subarray of nums
nice if the bitwise AND of every pair of elements that are in different positions in the subarray is equal to 0
.
Return the length of the longest nice subarray.
A subarray is a contiguous part of an array.
Note that subarrays of length 1
are always considered nice.
Example 1:
Input: nums = [1,3,8,48,10] Output: 3 Explanation: The longest nice subarray is [3,8,48]. This subarray satisfies the conditions: - 3 AND 8 = 0. - 3 AND 48 = 0. - 8 AND 48 = 0. It can be proven that no longer nice subarray can be obtained, so we return 3.
Example 2:
Input: nums = [3,1,5,11,13] Output: 1 Explanation: The length of the longest nice subarray is 1. Any subarray of length 1 can be chosen.
Constraints:
1 <= nums.length <= 105
1 <= nums[i] <= 109
Solutions
Solution 1: Two Pointers
According to the problem description, the position of the binary $1$ in each element of the subarray must be unique to ensure that the bitwise AND result of any two elements is $0$.
Therefore, we can use two pointers, $l$ and $r$, to maintain a sliding window such that the elements within the window satisfy the problem's conditions.
We use a variable $\textit{mask}$ to represent the bitwise OR result of the elements within the window. Next, we traverse each element of the array. For the current element $x$, if the bitwise AND result of $\textit{mask}$ and $x$ is not $0$, it means that the current element $x$ has overlapping binary bits with the elements in the window. At this point, we need to move the left pointer $l$ until the bitwise AND result of $\textit{mask}$ and $x$ is $0$. Then, we assign the bitwise OR result of $\textit{mask}$ and $x$ to $\textit{mask}$ and update the answer $\textit{ans} = \max(\textit{ans}, r - l + 1)$.
After the traversal, return the answer $\textit{ans}$.
The time complexity is $O(n)$, where $n$ is the length of the array $\textit{nums}$. The space complexity is $O(1)$.
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