Medium
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 <= 105
1 <= nums1[i], nums2[i] <= 105
public class Solution {
public int minAbsoluteSumDiff(int[] nums1, int[] nums2) {
int min = Integer.MAX_VALUE;
int max = Integer.MIN_VALUE;
for (int i = 0; i < nums1.length; i++) {
min = Math.min(min, Math.min(nums1[i], nums2[i]));
max = Math.max(max, Math.max(nums1[i], nums2[i]));
}
int[] less = new int[max - min + 1];
int[] more = new int[max - min + 1];
less[0] = -max - 1;
more[more.length - 1] = ((max + 1) << 1);
for (int num : nums1) {
less[num - min] = num;
more[num - min] = num;
}
for (int i = 1; i < less.length; i++) {
if (less[i] == 0) {
less[i] = less[i - 1];
}
}
for (int i = more.length - 2; i >= 0; i--) {
if (more[i] == 0) {
more[i] = more[i + 1];
}
}
int total = 0;
int preSave = 0;
for (int i = 0; i < nums1.length; i++) {
int current = Math.abs(nums1[i] - nums2[i]);
total += current;
int save =
current
- Math.min(
Math.abs(less[nums2[i] - min] - nums2[i]),
Math.abs(more[nums2[i] - min] - nums2[i]));
if (save > preSave) {
total = total + preSave - save;
preSave = save;
}
total = total % 1_000_000_007;
}
return total;
}
}