You are given an array 'A' of 'N' integers.

The beauty of the array is defined as the difference between the maximum and minimum elements of the array.

Your task is to return the maximum number of elements you can remove from the back of the array such that the beauty of the array remains unchanged.

Note: that you cannot remove all the elements from the array, as beauty is not defined for an empty array.

Example:

N=4

A = [6, 2, 4, 3]

The initial beauty of the array is '6-2=4'.

We can remove the last two elements from the array, so the resultant array is [6, 2], which also has beauty equal to '6-2 = 4'.

We can see that we can not remove more than two elements from the array without affecting the beauty of the array, Therefore, the answer for this case is 2.

Detailed explanation (Input/output format, Notes, Images) The first line of each test case contains an integer 'N', denoting the length of the array 'A'.

The second line of each test case contains 'N' space-separated integers, denoting the elements of the array 'A'.

Output format:

Return the maximum number of elements you can remove from the back of the array such that the beauty of the array doesn't change.

Note:

You don't need to print anything. It has already been taken care of. Just implement the given function. Sample input 1:

2 3 2 2 2 2 1 4 Sample output 1:

2 0

Explanation of sample input 1:

For the test case 1:

The initial beauty of the array is '2-2' = '0'.

Here, we can remove the last two elements from the array, so the resultant array is [2], which also has beauty equal to '0'.

We can see that we can only remove up to two elements from the array, as we need at least one element in the array. Therefore, the answer for this case is '2'.

For the test case 2:

The initial beauty of the array is '41' = '3'. ='3'.

If we remove the last element, we will have the array with only one element whose beauty is '0'. So, we can not remove any elements from the array 'A'. So, the answer for this case is '0'.

def maximumRemovals(n,arr):