sum-of-distances

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cangjie

Algorithms

You are given a 0-indexed integer array <code>nums</code>. There exists an array <code>arr</code> of length <code>nums.length</code>, where <code>arr[i]</code> is the sum of <code>|i - j|</code> over all <code>j</code> such that <code>nums[j] == nums[i]</code> and <code>j != i</code>. If there is no such <code>j</code>, set <code>arr[i]</code> to be <code>0</code>.

Return the array <code>arr</code>.

&nbsp;
Example 1:

<pre>
Input: nums = [1,3,1,1,2]
Output: [5,0,3,4,0]
Explanation: 
When i = 0, nums[0] == nums[2] and nums[0] == nums[3]. Therefore, arr[0] = |0 - 2| + |0 - 3| = 5. 
When i = 1, arr[1] = 0 because there is no other index with value 3.
When i = 2, nums[2] == nums[0] and nums[2] == nums[3]. Therefore, arr[2] = |2 - 0| + |2 - 3| = 3. 
When i = 3, nums[3] == nums[0] and nums[3] == nums[2]. Therefore, arr[3] = |3 - 0| + |3 - 2| = 4. 
When i = 4, arr[4] = 0 because there is no other index with value 2. 

</pre>

Example 2:

<pre>
Input: nums = [0,5,3]
Output: [0,0,0]
Explanation: Since each element in nums is distinct, arr[i] = 0 for all i.
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= nums.length &lt;= 105</code></li>
	<li><code>0 &lt;= nums[i] &lt;= 109</code></li>
</ul>

&nbsp;
Note: This question is the same as <a href="https://leetcode.com/problems/intervals-between-identical-elements/description/" target="_blank"> 2121: Intervals Between Identical Elements.</a>

distance

Sum of Distances

Remove Duplicates from Sorted Array

remove-duplicates-from-sorted-array

删除有序数组中的重复项

Find All Duplicates in an Array

find-all-duplicates-in-an-array

数组中重复的数据

Minimum Operations to Make All Array Elements Equal

minimum-operations-to-make-all-array-elements-equal

使数组元素全部相等的最少操作次数

Array

Hash Table

Prefix Sum

给你一个下标从 0 开始的整数数组 <code>nums</code> 。现有一个长度等于 <code>nums.length</code> 的数组 <code>arr</code> 。对于满足 <code>nums[j] == nums[i]</code> 且 <code>j != i</code> 的所有 <code>j</code> ，<code>arr[i]</code> 等于所有 <code>|i - j|</code> 之和。如果不存在这样的 <code>j</code> ，则令 <code>arr[i]</code> 等于 <code>0</code> 。

返回数组 <code>arr</code> 。

&nbsp;

示例 1：

<pre>
输入：nums = [1,3,1,1,2]
输出：[5,0,3,4,0]
解释：
i = 0 ，nums[0] == nums[2] 且 nums[0] == nums[3] 。因此，arr[0] = |0 - 2| + |0 - 3| = 5 。 
i = 1 ，arr[1] = 0 因为不存在值等于 3 的其他下标。
i = 2 ，nums[2] == nums[0] 且 nums[2] == nums[3] 。因此，arr[2] = |2 - 0| + |2 - 3| = 3 。
i = 3 ，nums[3] == nums[0] 且 nums[3] == nums[2] 。因此，arr[3] = |3 - 0| + |3 - 2| = 4 。 
i = 4 ，arr[4] = 0 因为不存在值等于 2 的其他下标。
</pre>

示例 2：

<pre>
输入：nums = [0,5,3]
输出：[0,0,0]
解释：因为 nums 中的元素互不相同，对于所有 i ，都有 arr[i] = 0 。
</pre>

&nbsp;

提示：

<ul>
	<li><code>1 &lt;= nums.length &lt;= 105</code></li>
	<li><code>0 &lt;= nums[i] &lt;= 109</code></li>
</ul>