You are given a 0-indexed integer array nums
.
- The low score of
nums
is the minimum value of|nums[i] - nums[j]|
over all0 <= i < j < nums.length
. - The high score of
nums
is the maximum value of|nums[i] - nums[j]|
over all0 <= i < j < nums.length
. - The score of
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 of 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
// OJ: https://leetcode.com/problems/minimum-score-by-changing-two-elements
// Author: github.com/lzl124631x
// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
int minimizeSum(vector<int>& A) {
sort(begin(A), end(A));
return min({A.back() - A[2], A[A.size() - 3] - A[0], A[A.size() - 2] - A[1]});
}
};