minimum-sum-of-squared-difference

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Algorithms

You are given two positive 0-indexed integer arrays <code>nums1</code> and <code>nums2</code>, both of length <code>n</code>.

The sum of squared difference of arrays <code>nums1</code> and <code>nums2</code> is defined as the sum of <code>(nums1[i] - nums2[i])2</code> for each <code>0 &lt;= i &lt; n</code>.

You are also given two positive integers <code>k1</code> and <code>k2</code>. You can modify any of the elements of <code>nums1</code> by <code>+1</code> or <code>-1</code> at most <code>k1</code> times. Similarly, you can modify any of the elements of <code>nums2</code> by <code>+1</code> or <code>-1</code> at most <code>k2</code> times.

Return the minimum sum of squared difference after modifying array <code>nums1</code> at most <code>k1</code> times and modifying array <code>nums2</code> at most <code>k2</code> times.

Note: You are allowed to modify the array elements to become negative integers.

&nbsp;
Example 1:

<pre>
Input: nums1 = [1,2,3,4], nums2 = [2,10,20,19], k1 = 0, k2 = 0
Output: 579
Explanation: The elements in nums1 and nums2 cannot be modified because k1 = 0 and k2 = 0. 
The sum of square difference will be: (1 - 2)2 + (2 - 10)2 + (3 - 20)2 + (4 - 19)2&nbsp;= 579.
</pre>

Example 2:

<pre>
Input: nums1 = [1,4,10,12], nums2 = [5,8,6,9], k1 = 1, k2 = 1
Output: 43
Explanation: One way to obtain the minimum sum of square difference is: 
- Increase nums1[0] once.
- Increase nums2[2] once.
The minimum of the sum of square difference will be: 
(2 - 5)2 + (4 - 8)2 + (10 - 7)2 + (12 - 9)2&nbsp;= 43.
Note that, there are other ways to obtain the minimum of the sum of square difference, but there is no way to obtain a sum smaller than 43.</pre>

&nbsp;
Constraints:

<ul>
	<li><code>n == nums1.length == nums2.length</code></li>
	<li><code>1 &lt;= n &lt;= 105</code></li>
	<li><code>0 &lt;= nums1[i], nums2[i] &lt;= 105</code></li>
	<li><code>0 &lt;= k1, k2 &lt;= 109</code></li>
</ul>

minSumSquareDiff

Minimum Sum of Squared Difference

Minimum Absolute Sum Difference

minimum-absolute-sum-difference

绝对差值和

Partition Array Into Two Arrays to Minimize Sum Difference

partition-array-into-two-arrays-to-minimize-sum-difference

将数组分成两个数组并最小化数组和的差

Greedy

Array

Binary Search

Sorting

Heap (Priority Queue)

给你两个下标从 0&nbsp;开始的整数数组&nbsp;<code>nums1</code> 和&nbsp;<code>nums2</code>&nbsp;，长度为&nbsp;<code>n</code>&nbsp;。

数组&nbsp;<code>nums1</code> 和&nbsp;<code>nums2</code>&nbsp;的 差值平方和&nbsp;定义为所有满足&nbsp;<code>0 &lt;= i &lt; n</code>&nbsp;的&nbsp;<code>(nums1[i] - nums2[i])2</code>&nbsp;之和。

同时给你两个正整数&nbsp;<code>k1</code> 和&nbsp;<code>k2</code>&nbsp;。你可以将&nbsp;<code>nums1</code>&nbsp;中的任意元素&nbsp;<code>+1</code> 或者&nbsp;<code>-1</code>&nbsp;至多&nbsp;<code>k1</code>&nbsp;次。类似的，你可以将&nbsp;<code>nums2</code>&nbsp;中的任意元素&nbsp;<code>+1</code> 或者&nbsp;<code>-1</code>&nbsp;至多&nbsp;<code>k2</code>&nbsp;次。

请你返回修改数组&nbsp;<code>nums1</code>&nbsp;至多&nbsp;<code>k1</code>&nbsp;次且修改数组&nbsp;<code>nums2</code>&nbsp;至多 <code>k2</code>&nbsp;次后的最小&nbsp;差值平方和&nbsp;。

注意：你可以将数组中的元素变成&nbsp;负&nbsp;整数。

&nbsp;

示例 1：

<pre>输入：nums1 = [1,2,3,4], nums2 = [2,10,20,19], k1 = 0, k2 = 0
输出：579
解释：nums1 和 nums2 中的元素不能修改，因为 k1 = 0 和 k2 = 0 。
差值平方和为：(1 - 2)2 + (2 - 10)2 + (3 - 20)2 + (4 - 19)2&nbsp;= 579 。
</pre>

示例 2：

<pre>输入：nums1 = [1,4,10,12], nums2 = [5,8,6,9], k1 = 1, k2 = 1
输出：43
解释：一种得到最小差值平方和的方式为：
- 将 nums1[0] 增加一次。
- 将 nums2[2] 增加一次。
最小差值平方和为：
(2 - 5)2 + (4 - 8)2 + (10 - 7)2 + (12 - 9)2&nbsp;= 43 。
注意，也有其他方式可以得到最小差值平方和，但没有得到比 43 更小答案的方案。</pre>

&nbsp;

提示：

<ul>
	<li><code>n == nums1.length == nums2.length</code></li>
	<li><code>1 &lt;= n &lt;= 105</code></li>
	<li><code>0 &lt;= nums1[i], nums2[i] &lt;= 105</code></li>
	<li><code>0 &lt;= k1, k2 &lt;= 109</code></li>
</ul>