Medium
You are given a 0-indexed integer array nums
.
nums
is the minimum value of |nums[i] - nums[j]|
over all 0 <= i < j < nums.length
.nums
is the maximum value of |nums[i] - nums[j]|
over all 0 <= i < j < nums.length
.nums
is the sum of the high and low scores of nums.To minimize the score of nums
, we can change the value of at most two elements of nums
.
Return _the minimum possible score after changing the value of at most two elements o_f nums
.
Note that |x|
denotes the absolute value of x
.
Example 1:
Input: nums = [1,4,3]
Output: 0
Explanation: Change value of nums[1] and nums[2] to 1 so that nums becomes [1,1,1]. Now, the value of |nums[i] - nums[j]|
is always equal to 0, so we return 0 + 0 = 0.
Example 2:
Input: nums = [1,4,7,8,5]
Output: 3
Explanation: Change nums[0] and nums[1] to be 6. Now nums becomes [6,6,7,8,5].
Our low score is achieved when i = 0 and j = 1, in which case | nums[i] - nums[j] |
= | 6 - 6 | = 0. |
Our high score is achieved when i = 3 and j = 4, in which case | nums[i] - nums[j] |
= | 8 - 5 | = 3. |
The sum of our high and low score is 3, which we can prove to be minimal.
Constraints:
3 <= nums.length <= 105
1 <= nums[i] <= 109
class Solution {
fun minimizeSum(nums: IntArray): Int {
nums.sort()
return (0 until 3).map {
nums[nums.size - 3 + it] - nums[it]
}.min()
}
}