leetcode-problemset/leetcode/originData/minimum-index-of-a-valid-split.json

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            "title": "Minimum Index of a Valid Split",
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            "content": "<p>An element <code>x</code> of an integer array <code>arr</code> of length <code>m</code> is <strong>dominant</strong> if <code>freq(x) * 2 &gt; m</code>, where <code>freq(x)</code> is the number of occurrences of <code>x</code> in <code>arr</code>. Note that this definition implies that <code>arr</code> can have <strong>at most one</strong> dominant element.</p>\n\n<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code> of length <code>n</code> with one dominant element.</p>\n\n<p>You can split <code>nums</code> at an index <code>i</code> into two arrays <code>nums[0, ..., i]</code> and <code>nums[i + 1, ..., n - 1]</code>, but the split is only <strong>valid</strong> if:</p>\n\n<ul>\n\t<li><code>0 &lt;= i &lt; n - 1</code></li>\n\t<li><code>nums[0, ..., i]</code>, and <code>nums[i + 1, ..., n - 1]</code> have the same dominant element.</li>\n</ul>\n\n<p>Here, <code>nums[i, ..., j]</code> denotes the subarray of <code>nums</code> starting at index <code>i</code> and ending at index <code>j</code>, both ends being inclusive. Particularly, if <code>j &lt; i</code> then <code>nums[i, ..., j]</code> denotes an empty subarray.</p>\n\n<p>Return <em>the <strong>minimum</strong> index of a <strong>valid split</strong></em>. If no valid split exists, return <code>-1</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [1,2,2,2]\n<strong>Output:</strong> 2\n<strong>Explanation:</strong> We can split the array at index 2 to obtain arrays [1,2,2] and [2]. \nIn array [1,2,2], element 2 is dominant since it occurs twice in the array and 2 * 2 &gt; 3. \nIn array [2], element 2 is dominant since it occurs once in the array and 1 * 2 &gt; 1.\nBoth [1,2,2] and [2] have the same dominant element as nums, so this is a valid split. \nIt can be shown that index 2 is the minimum index of a valid split. </pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [2,1,3,1,1,1,7,1,2,1]\n<strong>Output:</strong> 4\n<strong>Explanation:</strong> We can split the array at index 4 to obtain arrays [2,1,3,1,1] and [1,7,1,2,1].\nIn array [2,1,3,1,1], element 1 is dominant since it occurs thrice in the array and 3 * 2 &gt; 5.\nIn array [1,7,1,2,1], element 1 is dominant since it occurs thrice in the array and 3 * 2 &gt; 5.\nBoth [2,1,3,1,1] and [1,7,1,2,1] have the same dominant element as nums, so this is a valid split.\nIt can be shown that index 4 is the minimum index of a valid split.</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [3,3,3,3,7,2,2]\n<strong>Output:</strong> -1\n<strong>Explanation:</strong> It can be shown that there is no valid split.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>\n\t<li><code>nums</code> has exactly one dominant element.</li>\n</ul>\n",
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                "Find the dominant element of nums by using a hashmap to maintain element frequency, we denote the dominant element as x and its frequency as f.",
                "For each index in [0, n - 2], calculate f1, x’s frequency in the subarray [0, i] when looping the index. And f2, x’s frequency in the subarray [i + 1, n - 1] which is equal to f - f1. Then we can check whether x is dominant in both subarrays."
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