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40 lines
1.6 KiB
HTML
40 lines
1.6 KiB
HTML
<p>Design a data structure that efficiently finds the <strong>majority element</strong> of a given subarray.</p>
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<p>The <strong>majority element</strong> of a subarray is an element that occurs <code>threshold</code> times or more in the subarray.</p>
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<p>Implementing the <code>MajorityChecker</code> class:</p>
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<ul>
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<li><code>MajorityChecker(int[] arr)</code> Initializes the instance of the class with the given array <code>arr</code>.</li>
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<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>
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</ul>
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<p> </p>
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<p><strong>Example 1:</strong></p>
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<pre>
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<strong>Input</strong>
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["MajorityChecker", "query", "query", "query"]
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[[[1, 1, 2, 2, 1, 1]], [0, 5, 4], [0, 3, 3], [2, 3, 2]]
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<strong>Output</strong>
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[null, 1, -1, 2]
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<strong>Explanation</strong>
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MajorityChecker majorityChecker = new MajorityChecker([1, 1, 2, 2, 1, 1]);
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majorityChecker.query(0, 5, 4); // return 1
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majorityChecker.query(0, 3, 3); // return -1
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majorityChecker.query(2, 3, 2); // return 2
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</pre>
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<p> </p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 <= arr.length <= 2 * 10<sup>4</sup></code></li>
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<li><code>1 <= arr[i] <= 2 * 10<sup>4</sup></code></li>
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<li><code>0 <= left <= right < arr.length</code></li>
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<li><code>threshold <= right - left + 1</code></li>
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<li><code>2 * threshold > right - left + 1</code></li>
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<li>At most <code>10<sup>4</sup></code> calls will be made to <code>query</code>.</li>
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</ul>
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