1
0
mirror of https://gitee.com/coder-xiaomo/leetcode-problemset synced 2025-01-10 18:48:13 +08:00
Code Issues Projects Releases Wiki Activity GitHub Gitee
leetcode-problemset/leetcode/problem/smallest-rotation-with-highest-score.html

40 lines
1.8 KiB
HTML
Raw Normal View History

2022-03-27 18:35:17 +08:00
<p>You are given an array <code>nums</code>. You can rotate it by a non-negative integer <code>k</code> so that the array becomes <code>[nums[k], nums[k + 1], ... nums[nums.length - 1], nums[0], nums[1], ..., nums[k-1]]</code>. Afterward, any entries that are less than or equal to their index are worth one point.</p>
<ul>
<li>For example, if we have <code>nums = [2,4,1,3,0]</code>, and we rotate by <code>k = 2</code>, it becomes <code>[1,3,0,2,4]</code>. This is worth <code>3</code> points because <code>1 &gt; 0</code> [no points], <code>3 &gt; 1</code> [no points], <code>0 &lt;= 2</code> [one point], <code>2 &lt;= 3</code> [one point], <code>4 &lt;= 4</code> [one point].</li>
</ul>
<p>Return <em>the rotation index </em><code>k</code><em> that corresponds to the highest score we can achieve if we rotated </em><code>nums</code><em> by it</em>. If there are multiple answers, return the smallest such index <code>k</code>.</p>
<p>&nbsp;</p>
<p><strong>Example 1:</strong></p>
<pre>
<strong>Input:</strong> nums = [2,3,1,4,0]
<strong>Output:</strong> 3
<strong>Explanation:</strong> Scores for each k are listed below:
k = 0, nums = [2,3,1,4,0], score 2
k = 1, nums = [3,1,4,0,2], score 3
k = 2, nums = [1,4,0,2,3], score 3
k = 3, nums = [4,0,2,3,1], score 4
k = 4, nums = [0,2,3,1,4], score 3
So we should choose k = 3, which has the highest score.
</pre>
<p><strong>Example 2:</strong></p>
<pre>
<strong>Input:</strong> nums = [1,3,0,2,4]
<strong>Output:</strong> 0
<strong>Explanation:</strong> nums will always have 3 points no matter how it shifts.
So we will choose the smallest k, which is 0.
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>
<ul>
<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
<li><code>0 &lt;= nums[i] &lt; nums.length</code></li>
</ul>