You are given a?0-indexed?array of?distinct?integers?nums.

There is an element in?nums?that has the?lowest?value and an element that has the?highest?value. We call them the?minimum?and?maximum?respectively. Your goal is to remove?both?these elements from the array.

A?deletion?is defined as either removing an element from the?front?of the array or removing an element from the?back?of the array.

Return?the?minimum?number of deletions it would take to remove?both?the minimum and maximum element from the array.

?

Example 1:

Input: nums = [2,10,7,5,4,1,8,6]

Output: 5

Explanation:?

The minimum element in the array is nums[5], which is 1.

The maximum element in the array is nums[1], which is 10.

We can remove both the minimum and maximum by removing 2 elements from the front and 3 elements from the back.?

This results in 2 + 3 = 5 deletions, which is the minimum number possible.

Example 2:

Input: nums = [0,-4,19,1,8,-2,-3,5]

Output: 3

Explanation:?

The minimum element in the array is nums[1], which is -4.?

The maximum element in the array is nums[2], which is 19.?

We can remove both the minimum and maximum by removing 3 elements from the front. This results in only 3 deletions, which is the minimum number possible.

Example 3:

Input: nums = [101]

Output: 1

Explanation: ?

There is only one element in the array, which makes it both the minimum and maximum element. We can remove it with 1 deletion.

?

Constraints:

• 1 <= nums.length <= 105

• -105?<= nums[i] <= 105

• The integers in?nums?are?distinct.

其實就是分3種情況，

[....i....j]方案1的長度是j+1;

[i....j....]方案2的長度是n-i;

[....i]+[j.....]方案3的長度是i+n-j+1;

我們先遍歷整個數(shù)組，找到最大值最小值的位置，一左一右，放好，然后就是根據(jù)這3個方案求出最小值即可；

Runtime:?2 ms, faster than?100.00%?of?Java?online submissions for?Removing Minimum and Maximum From Array.

Memory Usage:?58.5 MB, less than?53.50%?of?Java?online submissions for?Removing Minimum and Maximum From Array.