find-the-k-or-of-an-array

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Algorithms

You are given an integer array <code>nums</code>, and an integer <code>k</code>. Let&#39;s introduce&nbsp;K-or operation by extending the standard bitwise OR. In K-or, a bit position in the result is set to <code>1</code>&nbsp;if at least <code>k</code> numbers in <code>nums</code> have a <code>1</code> in that position.

Return the K-or of <code>nums</code>.

&nbsp;
Example 1: 

<div class="example-block" style="border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;">
Input: nums = [7,12,9,8,9,15], k = 4 

Output: 9 

Explanation: 

Represent numbers in binary:

<table style="text-indent:10px; margin-bottom=20px;">
	<tbody>
		<tr>
			<th>Number</th>
			<th>Bit 3</th>
			<th>Bit 2</th>
			<th>Bit 1</th>
			<th>Bit 0</th>
		</tr>
		<tr>
			<td>7</td>
			<td>0</td>
			<td>1</td>
			<td>1</td>
			<td>1</td>
		</tr>
		<tr>
			<td>12</td>
			<td>1</td>
			<td>1</td>
			<td>0</td>
			<td>0</td>
		</tr>
		<tr>
			<td>9</td>
			<td>1</td>
			<td>0</td>
			<td>0</td>
			<td>1</td>
		</tr>
		<tr>
			<td>8</td>
			<td>1</td>
			<td>0</td>
			<td>0</td>
			<td>0</td>
		</tr>
		<tr>
			<td>9</td>
			<td>1</td>
			<td>0</td>
			<td>0</td>
			<td>1</td>
		</tr>
		<tr>
			<td>15</td>
			<td>1</td>
			<td>1</td>
			<td>1</td>
			<td>1</td>
		</tr>
		<tr>
			<td>Result = 9</td>
			<td>1</td>
			<td>0</td>
			<td>0</td>
			<td>1</td>
		</tr>
	</tbody>
</table>

Bit 0 is set in 7, 9, 9, and 15. Bit 3 is set in 12, 9, 8, 9, and 15. 
Only bits 0 and 3 qualify. The result is <code>(1001)2 = 9</code>.
</div>

Example 2: 

<div class="example-block" style="border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;">
Input: nums = [2,12,1,11,4,5], k = 6 

Output: 0 

Explanation:&nbsp;No bit appears as 1 in all six array numbers, as required for K-or with <code>k = 6</code>. Thus, the result is 0.
</div>

Example 3: 

<div class="example-block" style="border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;">
Input: nums = [10,8,5,9,11,6,8], k = 1 

Output: 15 

Explanation: Since <code>k == 1</code>, the 1-or of the array is equal to the bitwise OR of all its elements. Hence, the answer is <code>10 OR 8 OR 5 OR 9 OR 11 OR 6 OR 8 = 15</code>.
</div>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= nums.length &lt;= 50</code></li>
	<li><code>0 &lt;= nums[i] &lt; 231</code></li>
	<li><code>1 &lt;= k &lt;= nums.length</code></li>
</ul>

findKOr

Find the K-or of an Array

Counting Bits

counting-bits

比特位计数

Sum of Values at Indices With K Set Bits

sum-of-values-at-indices-with-k-set-bits

计算 K 置位下标对应元素的和

Bit Manipulation

Array

给你一个整数数组 <code>nums</code> 和一个整数 <code>k</code> 。让我们通过扩展标准的按位或来介绍 K-or 操作。在 K-or 操作中，如果在 <code>nums</code> 中，至少存在 <code>k</code> 个元素的第 <code>i</code> 位值为 1 ，那么 K-or 中的第 <code>i</code> 位的值是 1 。

返回 <code>nums</code> 的 K-or 值。

&nbsp;

示例 1：

<pre>
输入：nums = [7,12,9,8,9,15], k = 4
输出：9
解释：
用二进制表示 numbers：
</pre>

<table style="text-indent:10px; margin-bottom=20px;">
	<tbody>
		<tr>
			<th>Number</th>
			<th>Bit 3</th>
			<th>Bit 2</th>
			<th>Bit 1</th>
			<th>Bit 0</th>
		</tr>
		<tr>
			<td>7</td>
			<td>0</td>
			<td>1</td>
			<td>1</td>
			<td>1</td>
		</tr>
		<tr>
			<td>12</td>
			<td>1</td>
			<td>1</td>
			<td>0</td>
			<td>0</td>
		</tr>
		<tr>
			<td>9</td>
			<td>1</td>
			<td>0</td>
			<td>0</td>
			<td>1</td>
		</tr>
		<tr>
			<td>8</td>
			<td>1</td>
			<td>0</td>
			<td>0</td>
			<td>0</td>
		</tr>
		<tr>
			<td>9</td>
			<td>1</td>
			<td>0</td>
			<td>0</td>
			<td>1</td>
		</tr>
		<tr>
			<td>15</td>
			<td>1</td>
			<td>1</td>
			<td>1</td>
			<td>1</td>
		</tr>
		<tr>
			<td>Result = 9</td>
			<td>1</td>
			<td>0</td>
			<td>0</td>
			<td>1</td>
		</tr>
	</tbody>
</table>

<pre>
位 0 在 7, 9, 9, 15 中为 1。位 3 在 12, 9, 8, 9, 15 中为 1。 只有位 0 和 3 满足。结果是 (1001)2 = 9。
</pre>

示例 2：

<pre>
输入：nums = [2,12,1,11,4,5], k = 6
输出：0
解释：没有位在所有 6 个数字中都为 1，如 k = 6 所需要的。所以，答案为 0。
</pre>

示例 3：

<pre>
输入：nums = [10,8,5,9,11,6,8], k = 1
输出：15
解释：因为 k == 1 ，数组的 1-or 等于其中所有元素按位或运算的结果。因此，答案为 10 OR 8 OR 5 OR 9 OR 11 OR 6 OR 8 = 15 。</pre>

&nbsp;

提示：

<ul>
	<li><code>1 &lt;= nums.length &lt;= 50</code></li>
	<li><code>0 &lt;= nums[i] &lt; 231</code></li>
	<li><code>1 &lt;= k &lt;= nums.length</code></li>
</ul>