leetcode-problemset/leetcode/problem/online-majority-element-in-subarray.html

<p>Design a data structure that efficiently finds the <strong>majority element</strong> of a given subarray.</p>

<p>The <strong>majority element</strong> of a subarray is an element that occurs <code>threshold</code> times or more in the subarray.</p>

<p>Implementing the <code>MajorityChecker</code> class:</p>

<ul>
	<li><code>MajorityChecker(int[] arr)</code> Initializes the instance of the class with the given array <code>arr</code>.</li>
	<li><code>int query(int left, int right, int threshold)</code> returns the element in the subarray <code>arr[left...right]</code> that occurs at least <code>threshold</code> times, or <code>-1</code> if no such element exists.</li>
</ul>

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

<pre>
<strong>Input</strong>
[&quot;MajorityChecker&quot;, &quot;query&quot;, &quot;query&quot;, &quot;query&quot;]
[[[1, 1, 2, 2, 1, 1]], [0, 5, 4], [0, 3, 3], [2, 3, 2]]
<strong>Output</strong>
[null, 1, -1, 2]

<strong>Explanation</strong>
MajorityChecker majorityChecker = new MajorityChecker([1, 1, 2, 2, 1, 1]);
majorityChecker.query(0, 5, 4); // return 1
majorityChecker.query(0, 3, 3); // return -1
majorityChecker.query(2, 3, 2); // return 2
</pre>

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

<ul>
	<li><code>1 &lt;= arr.length &lt;= 2 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= arr[i] &lt;= 2 * 10<sup>4</sup></code></li>
	<li><code>0 &lt;= left &lt;= right &lt; arr.length</code></li>
	<li><code>threshold &lt;= right - left + 1</code></li>
	<li><code>2 * threshold &gt; right - left + 1</code></li>
	<li>At most <code>10<sup>4</sup></code> calls will be made to <code>query</code>.</li>
</ul>