leetcode-problemset/leetcode-cn/problem (English)/区间内查询数字的频率(English) [range-frequency-queries].html

<p>Design a data structure to find the <strong>frequency</strong> of a given value in a given subarray.</p>

<p>The <strong>frequency</strong> of a value in a subarray is the number of occurrences of that value in the subarray.</p>

<p>Implement the <code>RangeFreqQuery</code> class:</p>

<ul>
	<li><code>RangeFreqQuery(int[] arr)</code> Constructs an instance of the class with the given <strong>0-indexed</strong> integer array <code>arr</code>.</li>
	<li><code>int query(int left, int right, int value)</code> Returns the <strong>frequency</strong> of <code>value</code> in the subarray <code>arr[left...right]</code>.</li>
</ul>

<p>A <strong>subarray</strong> is a contiguous sequence of elements within an array. <code>arr[left...right]</code> denotes the subarray that contains the elements of <code>nums</code> between indices <code>left</code> and <code>right</code> (<strong>inclusive</strong>).</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input</strong>
[&quot;RangeFreqQuery&quot;, &quot;query&quot;, &quot;query&quot;]
[[[12, 33, 4, 56, 22, 2, 34, 33, 22, 12, 34, 56]], [1, 2, 4], [0, 11, 33]]
<strong>Output</strong>
[null, 1, 2]

<strong>Explanation</strong>
RangeFreqQuery rangeFreqQuery = new RangeFreqQuery([12, 33, 4, 56, 22, 2, 34, 33, 22, 12, 34, 56]);
rangeFreqQuery.query(1, 2, 4); // return 1. The value 4 occurs 1 time in the subarray [33, 4]
rangeFreqQuery.query(0, 11, 33); // return 2. The value 33 occurs 2 times in the whole array.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= arr.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= arr[i], value &lt;= 10<sup>4</sup></code></li>
	<li><code>0 &lt;= left &lt;= right &lt; arr.length</code></li>
	<li>At most <code>10<sup>5</sup></code> calls will be made to <code>query</code></li>
</ul>