design-an-ordered-stream

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Algorithms

There is a stream of <code>n</code> <code>(idKey, value)</code> pairs arriving in an arbitrary order, where <code>idKey</code> is an integer between <code>1</code> and <code>n</code> and <code>value</code> is a string. No two pairs have the same <code>id</code>.

Design a stream that returns the values in increasing order of their IDs by returning a chunk (list) of values after each insertion. The concatenation of all the chunks should result in a list of the sorted values.

Implement the <code>OrderedStream</code> class:

<ul>
	<li><code>OrderedStream(int n)</code> Constructs the stream to take <code>n</code> values.</li>
	<li><code>String[] insert(int idKey, String value)</code> Inserts the pair <code>(idKey, value)</code> into the stream, then returns the largest possible chunk of currently inserted values that appear next in the order.</li>
</ul>

&nbsp;
Example:

<img alt="" src="https://assets.leetcode.com/uploads/2020/11/10/q1.gif" style="width: 682px; height: 240px;" />

<pre>
Input
[&quot;OrderedStream&quot;, &quot;insert&quot;, &quot;insert&quot;, &quot;insert&quot;, &quot;insert&quot;, &quot;insert&quot;]
[[5], [3, &quot;ccccc&quot;], [1, &quot;aaaaa&quot;], [2, &quot;bbbbb&quot;], [5, &quot;eeeee&quot;], [4, &quot;ddddd&quot;]]
Output
[null, [], [&quot;aaaaa&quot;], [&quot;bbbbb&quot;, &quot;ccccc&quot;], [], [&quot;ddddd&quot;, &quot;eeeee&quot;]]

Explanation
// Note that the values ordered by ID is [&quot;aaaaa&quot;, &quot;bbbbb&quot;, &quot;ccccc&quot;, &quot;ddddd&quot;, &quot;eeeee&quot;].
OrderedStream os = new OrderedStream(5);
os.insert(3, &quot;ccccc&quot;); // Inserts (3, &quot;ccccc&quot;), returns [].
os.insert(1, &quot;aaaaa&quot;); // Inserts (1, &quot;aaaaa&quot;), returns [&quot;aaaaa&quot;].
os.insert(2, &quot;bbbbb&quot;); // Inserts (2, &quot;bbbbb&quot;), returns [&quot;bbbbb&quot;, &quot;ccccc&quot;].
os.insert(5, &quot;eeeee&quot;); // Inserts (5, &quot;eeeee&quot;), returns [].
os.insert(4, &quot;ddddd&quot;); // Inserts (4, &quot;ddddd&quot;), returns [&quot;ddddd&quot;, &quot;eeeee&quot;].
// Concatentating all the chunks returned:
// [] + [&quot;aaaaa&quot;] + [&quot;bbbbb&quot;, &quot;ccccc&quot;] + [] + [&quot;ddddd&quot;, &quot;eeeee&quot;] = [&quot;aaaaa&quot;, &quot;bbbbb&quot;, &quot;ccccc&quot;, &quot;ddddd&quot;, &quot;eeeee&quot;]
// The resulting order is the same as the order above.
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= n &lt;= 1000</code></li>
	<li><code>1 &lt;= id &lt;= n</code></li>
	<li><code>value.length == 5</code></li>
	<li><code>value</code>&nbsp;consists only of lowercase letters.</li>
	<li>Each call to <code>insert</code>&nbsp;will have a unique <code>id.</code></li>
	<li>Exactly <code>n</code> calls will be made to <code>insert</code>.</li>
</ul>

Design an Ordered Stream

Longest Uploaded Prefix

longest-uploaded-prefix

最长上传前缀

Design

Array

Hash Table

Data Stream

有 <code>n</code> 个 <code>(id, value)</code> 对，其中 <code>id</code> 是 <code>1</code> 到 <code>n</code> 之间的一个整数，<code>value</code> 是一个字符串。不存在 <code>id</code> 相同的两个 <code>(id, value)</code> 对。

设计一个流，以 任意 顺序获取 <code>n</code> 个 <code>(id, value)</code> 对，并在多次调用时 按 <code>id</code> 递增的顺序 返回一些值。

实现 <code>OrderedStream</code> 类：

<ul>
	<li><code>OrderedStream(int n)</code> 构造一个能接收 <code>n</code> 个值的流，并将当前指针 <code>ptr</code> 设为 <code>1</code> 。</li>
	<li><code>String[] insert(int id, String value)</code> 向流中存储新的 <code>(id, value)</code> 对。存储后：
	<ul>
		<li>如果流存储有 <code>id = ptr</code> 的 <code>(id, value)</code> 对，则找出从 <code>id = ptr</code> 开始的 最长 id 连续递增序列 ，并 按顺序 返回与这些 id 关联的值的列表。然后，将 <code>ptr</code> 更新为最后那个  <code>id + 1</code> 。</li>
		<li>
		否则，返回一个空列表。
		</li>
	</ul>
	</li>
</ul>

 

示例：

<img alt="" src="https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2020/11/15/q1.gif" style="width: 682px; height: 240px;" />

<pre>
输入
["OrderedStream", "insert", "insert", "insert", "insert", "insert"]
[[5], [3, "ccccc"], [1, "aaaaa"], [2, "bbbbb"], [5, "eeeee"], [4, "ddddd"]]
输出
[null, [], ["aaaaa"], ["bbbbb", "ccccc"], [], ["ddddd", "eeeee"]]

解释
OrderedStream os= new OrderedStream(5);
os.insert(3, "ccccc"); // 插入 (3, "ccccc")，返回 []
os.insert(1, "aaaaa"); // 插入 (1, "aaaaa")，返回 ["aaaaa"]
os.insert(2, "bbbbb"); // 插入 (2, "bbbbb")，返回 ["bbbbb", "ccccc"]
os.insert(5, "eeeee"); // 插入 (5, "eeeee")，返回 []
os.insert(4, "ddddd"); // 插入 (4, "ddddd")，返回 ["ddddd", "eeeee"]
</pre>

 

提示：

<ul>
	<li><code>1 <= n <= 1000</code></li>
	<li><code>1 <= id <= n</code></li>
	<li><code>value.length == 5</code></li>
	<li><code>value</code> 仅由小写字母组成</li>
	<li>每次调用 <code>insert</code> 都会使用一个唯一的 <code>id</code></li>
	<li>恰好调用 <code>n</code> 次 <code>insert</code></li>
</ul>