题目
You are given two positive integer arrays nums1
and nums2
, both of length n
.
The absolute sum difference of arrays nums1
and nums2
is defined as the sum of |nums1[i] - nums2[i]|
for each 0 <= i < n
(0-indexed).
You can replace at most one element of nums1
with any other element in nums1
to minimize the absolute sum difference.
Return the *minimum absolute sum difference after replacing at most one ***element in the array nums1
. Since the answer may be large, return it modulo 109 + 7
.
|x|
is defined as:
x
if x >= 0
, or
x
if x < 0
.
Example 1:
Input: nums1 = [1,7,5], nums2 = [2,3,5]
Output: 3
Explanation:There are two possible optimal solutions:
- Replace the second element with the first: [1,7,5] => [1,1,5], or
- Replace the second element with the third: [1,7,5] => [1,5,5].
Both will yield an absolute sum difference of|1-2| + (|1-3| or |5-3|) + |5-5| =3.
Example 2:
Input: nums1 = [2,4,6,8,10], nums2 = [2,4,6,8,10]
Output: 0
Explanation:nums1 is equal to nums2 so no replacement is needed. This will result in an
absolute sum difference of 0.
Example 3:
Input: nums1 = [1,10,4,4,2,7], nums2 = [9,3,5,1,7,4]
Output: 20
Explanation:Replace the first element with the second: [1,10,4,4,2,7] => [10,10,4,4,2,7].
This yields an absolute sum difference of|10-9| + |10-3| + |4-5| + |4-1| + |2-7| + |7-4| = 20
Constraints:
n == nums1.length
n == nums2.length
1 <= n <= 10^5
1 <= nums1[i], nums2[i] <= 10^5
题目大意
给你两个正整数数组 nums1 和 nums2 ,数组的长度都是 n 。数组 nums1 和 nums2 的 绝对差值和 定义为所有 |nums1[i] - nums2[i]|(0 <= i < n)的 总和(下标从 0 开始)。你可以选用 nums1 中的 任意一个 元素来替换 nums1 中的 至多 一个元素,以 最小化 绝对差值和。在替换数组 nums1 中最多一个元素 之后 ,返回最小绝对差值和。因为答案可能很大,所以需要对 10^9 + 7 取余 后返回。
解题思路
$$\sum \left | nums1[i] - nums2[i] \right |$$
$$\begin{aligned}&\sum \left | nums1[i] - nums2[i] \right | - \left ( \left | nums1[i] - nums2[i] \right | - \left | nums1[j] - nums2[i] \right |\right )\= &\sum \left | nums1[i] - nums2[i] \right | - \Delta \end{aligned}$$
题目要求返回最小绝对差值和,即求
$$\Delta $$
的最大值。暴力枚举 nums1 和 nums2 中两两差值,找到 maxdiff。
代码
package leetcode
func minAbsoluteSumDiff(nums1 []int, nums2 []int) int {
diff := 0
maxDiff := 0
for i, n2 := range nums2 {
d := abs(nums1[i] - n2)
diff += d
if maxDiff < d {
t := 100001
for _, n1 := range nums1 {
maxDiff = max(maxDiff, d-min(t, abs(n1-n2)))
}
}
}
return (diff - maxDiff) % (1e9 + 7)
}
func max(a, b int) int {
if a > b {
return a
}
return b
}
func abs(a int) int {
if a > 0 {
return a
}
return -a
}
func min(a, b int) int {
if a > b {
return b
}
return a
}