find-the-maximum-sequence-value-of-array

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Algorithms

You are given an integer array <code>nums</code> and a positive integer <code>k</code>.

The value of a sequence <code>seq</code> of size <code>2 * x</code> is defined as:

<ul>
	<li><code>(seq[0] OR seq[1] OR ... OR seq[x - 1]) XOR (seq[x] OR seq[x + 1] OR ... OR seq[2 * x - 1])</code>.</li>
</ul>

Return the maximum value of any subsequence of <code>nums</code> having size <code>2 * k</code>.

&nbsp;
Example 1:

<div class="example-block">
Input: nums = [2,6,7], k = 1

Output: 5

Explanation:

The subsequence <code>[2, 7]</code> has the maximum value of <code>2 XOR 7 = 5</code>.
</div>

Example 2:

<div class="example-block">
Input: nums = [4,2,5,6,7], k = 2

Output: 2

Explanation:

The subsequence <code>[4, 5, 6, 7]</code> has the maximum value of <code>(4 OR 5) XOR (6 OR 7) = 2</code>.
</div>

&nbsp;
Constraints:

<ul>
	<li><code>2 &lt;= nums.length &lt;= 400</code></li>
	<li><code>1 &lt;= nums[i] &lt; 27</code></li>
	<li><code>1 &lt;= k &lt;= nums.length / 2</code></li>
</ul>

maxValue

Find the Maximum Sequence Value of Array

Bitwise ORs of Subarrays

bitwise-ors-of-subarrays

子数组按位或操作

Bit Manipulation

Array

Dynamic Programming

给你一个整数数组&nbsp;<code>nums</code>&nbsp;和一个 正&nbsp;整数&nbsp;<code>k</code>&nbsp;。

定义长度为 <code>2 * x</code>&nbsp;的序列 <code>seq</code>&nbsp;的 值&nbsp;为：

<ul>
	<li><code>(seq[0] OR seq[1] OR ... OR seq[x - 1]) XOR (seq[x] OR seq[x + 1] OR ... OR seq[2 * x - 1])</code>.</li>
</ul>

请你求出 <code>nums</code>&nbsp;中所有长度为 <code>2 * k</code>&nbsp;的 子序列 的 最大值&nbsp;。

&nbsp;

示例 1：

<div class="example-block">
输入：nums = [2,6,7], k = 1

输出：5

解释：

子序列&nbsp;<code>[2, 7]</code>&nbsp;的值最大，为&nbsp;<code>2 XOR 7 = 5</code>&nbsp;。
</div>

示例 2：

<div class="example-block">
输入：nums = [4,2,5,6,7], k = 2

输出：2

解释：

子序列&nbsp;<code>[4, 5, 6, 7]</code>&nbsp;的值最大，为&nbsp;<code>(4 OR 5) XOR (6 OR 7) = 2</code>&nbsp;。
</div>

&nbsp;

提示：

<ul>
	<li><code>2 &lt;= nums.length &lt;= 400</code></li>
	<li><code>1 &lt;= nums[i] &lt; 27</code></li>
	<li><code>1 &lt;= k &lt;= nums.length / 2</code></li>
</ul>