minimum-seconds-to-equalize-a-circular-array

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Algorithms

You are given a 0-indexed array <code>nums</code> containing <code>n</code> integers.

At each second, you perform the following operation on the array:

<ul>
	<li>For every index <code>i</code> in the range <code>[0, n - 1]</code>, replace <code>nums[i]</code> with either <code>nums[i]</code>, <code>nums[(i - 1 + n) % n]</code>, or <code>nums[(i + 1) % n]</code>.</li>
</ul>

Note that all the elements get replaced simultaneously.

Return the minimum number of seconds needed to make all elements in the array <code>nums</code> equal.

&nbsp;
Example 1:

<pre>
Input: nums = [1,2,1,2]
Output: 1
Explanation: We can equalize the array in 1 second in the following way:
- At 1st second, replace values at each index with [nums[3],nums[1],nums[3],nums[3]]. After replacement, nums = [2,2,2,2].
It can be proven that 1 second is the minimum amount of seconds needed for equalizing the array.
</pre>

Example 2:

<pre>
Input: nums = [2,1,3,3,2]
Output: 2
Explanation: We can equalize the array in 2 seconds in the following way:
- At 1st second, replace values at each index with [nums[0],nums[2],nums[2],nums[2],nums[3]]. After replacement, nums = [2,3,3,3,3].
- At 2nd second, replace values at each index with [nums[1],nums[1],nums[2],nums[3],nums[4]]. After replacement, nums = [3,3,3,3,3].
It can be proven that 2 seconds is the minimum amount of seconds needed for equalizing the array.
</pre>

Example 3:

<pre>
Input: nums = [5,5,5,5]
Output: 0
Explanation: We don&#39;t need to perform any operations as all elements in the initial array are the same.
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= n == nums.length &lt;= 105</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 109</code></li>
</ul>

minimumSeconds

Minimum Seconds to Equalize a Circular Array

Array

Hash Table

给你一个下标从 0&nbsp;开始长度为 <code>n</code>&nbsp;的数组&nbsp;<code>nums</code>&nbsp;。

每一秒，你可以对数组执行以下操作：

<ul>
	<li>对于范围在&nbsp;<code>[0, n - 1]</code>&nbsp;内的每一个下标&nbsp;<code>i</code>&nbsp;，将&nbsp;<code>nums[i]</code> 替换成&nbsp;<code>nums[i]</code>&nbsp;，<code>nums[(i - 1 + n) % n]</code>&nbsp;或者&nbsp;<code>nums[(i + 1) % n]</code>&nbsp;三者之一。</li>
</ul>

注意，所有元素会被同时替换。

请你返回将数组 <code>nums</code>&nbsp;中所有元素变成相等元素所需要的 最少&nbsp;秒数。

&nbsp;

示例 1：

<pre>输入：nums = [1,2,1,2]
输出：1
解释：我们可以在 1 秒内将数组变成相等元素：
- 第 1 秒，将每个位置的元素分别变为 [nums[3],nums[1],nums[3],nums[3]] 。变化后，nums = [2,2,2,2] 。
1 秒是将数组变成相等元素所需要的最少秒数。
</pre>

示例 2：

<pre>输入：nums = [2,1,3,3,2]
输出：2
解释：我们可以在 2 秒内将数组变成相等元素：
- 第 1 秒，将每个位置的元素分别变为 [nums[0],nums[2],nums[2],nums[2],nums[3]] 。变化后，nums = [2,3,3,3,3] 。
- 第 2 秒，将每个位置的元素分别变为 [nums[1],nums[1],nums[2],nums[3],nums[4]] 。变化后，nums = [3,3,3,3,3] 。
2 秒是将数组变成相等元素所需要的最少秒数。
</pre>

示例 3：

<pre>输入：nums = [5,5,5,5]
输出：0
解释：不需要执行任何操作，因为一开始数组中的元素已经全部相等。
</pre>

&nbsp;

提示：

<ul>
	<li><code>1 &lt;= n == nums.length &lt;= 105</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 109</code></li>
</ul>