leetcode-problemset/leetcode-cn/problem (Chinese)/查询超过阈值频率最高元素 [threshold-majority-queries].html

<p>给你一个长度为 <code>n</code> 的整数数组 <code>nums</code> 和一个查询数组 <code>queries</code>，其中 <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>, threshold<sub>i</sub>]</code>。</p>

<p>返回一个整数数组 <code data-end="33" data-start="28">ans</code>，其中 <code data-end="48" data-start="40">ans[i]</code> 等于子数组 <code data-end="102" data-start="89">nums[l<sub>i</sub>...r<sub>i</sub>]</code> 中出现&nbsp;<strong>至少</strong> <code data-end="137" data-start="125">threshold<sub>i</sub></code> 次的元素，选择频率&nbsp;<strong>最高&nbsp;</strong>的元素（如果频率相同则选择&nbsp;<strong>最小&nbsp;</strong>的元素），如果不存在这样的元素则返回 -1。</p>

<p>&nbsp;</p>

<p><strong class="example">示例 1:</strong></p>

<div class="example-block">
<p><strong>输入：</strong> <span class="example-io">nums = [1,1,2,2,1,1], queries = [[0,5,4],[0,3,3],[2,3,2]]</span></p>

<p><strong>输出：</strong> <span class="example-io">[1,-1,2]</span></p>

<p><strong>解释：</strong></p>

<table style="border: 1px solid black;">
	<thead>
		<tr>
			<th align="left" style="border: 1px solid black;">查询</th>
			<th align="left" style="border: 1px solid black;">子数组</th>
			<th align="left" style="border: 1px solid black;">阈值</th>
			<th align="left" style="border: 1px solid black;">频率表</th>
			<th align="left" style="border: 1px solid black;">答案</th>
		</tr>
	</thead>
	<tbody>
		<tr>
			<td align="left" style="border: 1px solid black;">[0, 5, 4]</td>
			<td align="left" style="border: 1px solid black;">[1, 1, 2, 2, 1, 1]</td>
			<td align="left" style="border: 1px solid black;">4</td>
			<td align="left" style="border: 1px solid black;">1 → 4, 2 → 2</td>
			<td align="left" style="border: 1px solid black;">1</td>
		</tr>
		<tr>
			<td align="left" style="border: 1px solid black;">[0, 3, 3]</td>
			<td align="left" style="border: 1px solid black;">[1, 1, 2, 2]</td>
			<td align="left" style="border: 1px solid black;">3</td>
			<td align="left" style="border: 1px solid black;">1 → 2, 2 → 2</td>
			<td align="left" style="border: 1px solid black;">-1</td>
		</tr>
		<tr>
			<td align="left" style="border: 1px solid black;">[2, 3, 2]</td>
			<td align="left" style="border: 1px solid black;">[2, 2]</td>
			<td align="left" style="border: 1px solid black;">2</td>
			<td align="left" style="border: 1px solid black;">2 → 2</td>
			<td align="left" style="border: 1px solid black;">2</td>
		</tr>
	</tbody>
</table>
</div>

<p>&nbsp;</p>

<p><strong class="example">示例 2:</strong></p>

<div class="example-block">
<p><strong>输入：</strong><span class="example-io">nums = [3,2,3,2,3,2,3], queries = [[0,6,4],[1,5,2],[2,4,1],[3,3,1]]</span></p>

<p><strong>输出：</strong><span class="example-io">[3,2,3,2]</span></p>

<p><strong>解释：</strong></p>

<table style="border: 1px solid black;">
	<thead>
		<tr>
			<th align="left" style="border: 1px solid black;">查询</th>
			<th align="left" style="border: 1px solid black;">子数组</th>
			<th align="left" style="border: 1px solid black;">阈值</th>
			<th align="left" style="border: 1px solid black;">频率表</th>
			<th align="left" style="border: 1px solid black;">答案</th>
		</tr>
	</thead>
	<tbody>
		<tr>
			<td align="left" style="border: 1px solid black;">[0, 6, 4]</td>
			<td align="left" style="border: 1px solid black;">[3, 2, 3, 2, 3, 2, 3]</td>
			<td align="left" style="border: 1px solid black;">4</td>
			<td align="left" style="border: 1px solid black;">3 → 4, 2 → 3</td>
			<td align="left" style="border: 1px solid black;">3</td>
		</tr>
		<tr>
			<td align="left" style="border: 1px solid black;">[1, 5, 2]</td>
			<td align="left" style="border: 1px solid black;">[2, 3, 2, 3, 2]</td>
			<td align="left" style="border: 1px solid black;">2</td>
			<td align="left" style="border: 1px solid black;">2 → 3, 3 → 2</td>
			<td align="left" style="border: 1px solid black;">2</td>
		</tr>
		<tr>
			<td align="left" style="border: 1px solid black;">[2, 4, 1]</td>
			<td align="left" style="border: 1px solid black;">[3, 2, 3]</td>
			<td align="left" style="border: 1px solid black;">1</td>
			<td align="left" style="border: 1px solid black;">3 → 2, 2 → 1</td>
			<td align="left" style="border: 1px solid black;">3</td>
		</tr>
		<tr>
			<td align="left" style="border: 1px solid black;">[3, 3, 1]</td>
			<td align="left" style="border: 1px solid black;">[2]</td>
			<td align="left" style="border: 1px solid black;">1</td>
			<td align="left" style="border: 1px solid black;">2 → 1</td>
			<td align="left" style="border: 1px solid black;">2</td>
		</tr>
	</tbody>
</table>
</div>

<p>&nbsp;</p>

<p><strong>提示：</strong></p>

<ul>
	<li data-end="51" data-start="19"><code data-end="49" data-start="19">1 &lt;= nums.length == n &lt;= 10<sup>4</sup></code></li>
	<li data-end="82" data-start="54"><code data-end="80" data-start="54">1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
	<li data-end="120" data-start="85"><code data-end="118" data-start="85">1 &lt;= queries.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li data-end="195" data-start="123"><code data-end="193" data-is-only-node="" data-start="155">queries[i] = [l<sub>i</sub>, r<sub>i</sub>, threshold<sub>i</sub>]</code></li>
	<li data-end="221" data-start="198"><code data-end="219" data-start="198">0 &lt;= l<sub>i</sub> &lt;= r<sub>i</sub> &lt; n</code></li>
	<li data-end="259" data-is-last-node="" data-start="224"><code data-end="259" data-is-last-node="" data-start="224">1 &lt;= threshold<sub>i</sub> &lt;= r<sub>i</sub> - l<sub>i</sub> + 1</code></li>
</ul>