time-based-key-value-store

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Algorithms

Design a time-based key-value data structure that can store multiple values for the same key at different time stamps and retrieve the key&#39;s value at a certain timestamp.

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

<ul>
	<li><code>TimeMap()</code> Initializes the object of the data structure.</li>
	<li><code>void set(String key, String value, int timestamp)</code> Stores the key <code>key</code> with the value <code>value</code> at the given time <code>timestamp</code>.</li>
	<li><code>String get(String key, int timestamp)</code> Returns a value such that <code>set</code> was called previously, with <code>timestamp_prev &lt;= timestamp</code>. If there are multiple such values, it returns the value associated with the largest <code>timestamp_prev</code>. If there are no values, it returns <code>&quot;&quot;</code>.</li>
</ul>

&nbsp;
Example 1:

<pre>
Input
[&quot;TimeMap&quot;, &quot;set&quot;, &quot;get&quot;, &quot;get&quot;, &quot;set&quot;, &quot;get&quot;, &quot;get&quot;]
[[], [&quot;foo&quot;, &quot;bar&quot;, 1], [&quot;foo&quot;, 1], [&quot;foo&quot;, 3], [&quot;foo&quot;, &quot;bar2&quot;, 4], [&quot;foo&quot;, 4], [&quot;foo&quot;, 5]]
Output
[null, null, &quot;bar&quot;, &quot;bar&quot;, null, &quot;bar2&quot;, &quot;bar2&quot;]

Explanation
TimeMap timeMap = new TimeMap();
timeMap.set(&quot;foo&quot;, &quot;bar&quot;, 1); // store the key &quot;foo&quot; and value &quot;bar&quot; along with timestamp = 1.
timeMap.get(&quot;foo&quot;, 1); // return &quot;bar&quot;
timeMap.get(&quot;foo&quot;, 3); // return &quot;bar&quot;, since there is no value corresponding to foo at timestamp 3 and timestamp 2, then the only value is at timestamp 1 is &quot;bar&quot;.
timeMap.set(&quot;foo&quot;, &quot;bar2&quot;, 4); // store the key &quot;foo&quot; and value &quot;bar2&quot; along with timestamp = 4.
timeMap.get(&quot;foo&quot;, 4); // return &quot;bar2&quot;
timeMap.get(&quot;foo&quot;, 5); // return &quot;bar2&quot;
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= key.length, value.length &lt;= 100</code></li>
	<li><code>key</code> and <code>value</code> consist of lowercase English letters and digits.</li>
	<li><code>1 &lt;= timestamp &lt;= 107</code></li>
	<li>All the timestamps <code>timestamp</code> of <code>set</code> are strictly increasing.</li>
	<li>At most <code>2 * 105</code> calls will be made to <code>set</code> and <code>get</code>.</li>
</ul>

Time Based Key-Value Store

Stock Price Fluctuation 

stock-price-fluctuation

股票价格波动

Design

Hash Table

String

Binary Search

设计一个基于时间的键值数据结构，该结构可以在不同时间戳存储对应同一个键的多个值，并针对特定时间戳检索键对应的值。

实现 <code>TimeMap</code> 类：

<ul>
	<li><code>TimeMap()</code> 初始化数据结构对象</li>
	<li><code>void set(String key, String value, int timestamp)</code> 存储给定时间戳&nbsp;<code>timestamp</code>&nbsp;时的键&nbsp;<code>key</code>&nbsp;和值&nbsp;<code>value</code>。</li>
	<li><code>String get(String key, int timestamp)</code>&nbsp;返回一个值，该值在之前调用了 <code>set</code>，其中&nbsp;<code>timestamp_prev &lt;= timestamp</code>&nbsp;。如果有多个这样的值，它将返回与最大 &nbsp;<code>timestamp_prev</code>&nbsp;关联的值。如果没有值，则返回空字符串（<code>""</code>）。</li>
</ul>
&nbsp;

示例 1：

<pre>
输入：
["TimeMap", "set", "get", "get", "set", "get", "get"]
[[], ["foo", "bar", 1], ["foo", 1], ["foo", 3], ["foo", "bar2", 4], ["foo", 4], ["foo", 5]]
输出：
[null, null, "bar", "bar", null, "bar2", "bar2"]

解释：
TimeMap timeMap = new TimeMap();
timeMap.set("foo", "bar", 1); // 存储键 "foo" 和值 "bar" ，时间戳 timestamp = 1 &nbsp; 
timeMap.get("foo", 1); // 返回 "bar"
timeMap.get("foo", 3); // 返回 "bar", 因为在时间戳 3 和时间戳 2 处没有对应 "foo" 的值，所以唯一的值位于时间戳 1 处（即 "bar"） 。
timeMap.set("foo", "bar2", 4); // 存储键 "foo" 和值 "bar2" ，时间戳 timestamp = 4&nbsp; 
timeMap.get("foo", 4); // 返回 "bar2"
timeMap.get("foo", 5); // 返回 "bar2"
</pre>

&nbsp;

提示：

<ul>
	<li><code>1 &lt;= key.length, value.length &lt;= 100</code></li>
	<li><code>key</code> 和 <code>value</code> 由小写英文字母和数字组成</li>
	<li><code>1 &lt;= timestamp &lt;= 107</code></li>
	<li><code>set</code> 操作中的时间戳 <code>timestamp</code> 都是严格递增的</li>
	<li>最多调用&nbsp;<code>set</code> 和 <code>get</code> 操作 <code>2 * 105</code> 次</li>
</ul>