Let array be: index = 0 1 2 3 4 nums = | 1 | 2 | 3 | 4 | 5 | n = 5

Lets see some values of x and candidates:

x = 1, candidate = n - x => 4; nums[candidate] = 5
 
nums  = | 1 | 2 | 3 | 4 | 5 |
          ^               ^
          x              n-x

x = 2, candidate = n - x => 3; nums[candidate] = 4
 
nums  = | 1 | 2 | 3 | 4 | 5 |
              ^       ^
              x      n-x

x = 3, candidate = n - x => 2; nums[candidate] = 3
 
nums  = | 1 | 2 | 3 | 4 | 5 |
                  ^  
                  x
				 n-x

Clearly x = 3 is our answer, as n-x is the right pivot points.

[{"language":"Java","code":"class Solution {\n\n\tpublic int specialArray(int[] nums) {\n\t\tArrays.sort(nums);\n\t\tint n = nums.length;\