## DEV Community

Abhishek Chaudhary

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# Minimum Deletions to Make Array Beautiful

You are given a 0-indexed integer array `nums`. The array `nums` is beautiful if:

• `nums.length` is even.
• `nums[i] != nums[i + 1]` for all `i % 2 == 0`.

Note that an empty array is considered beautiful.

You can delete any number of elements from `nums`. When you delete an element, all the elements to the right of the deleted element will be shifted one unit to the left to fill the gap created and all the elements to the left of the deleted element will remain unchanged.

Return the minimum number of elements to delete from `nums` to make it beautiful.

Example 1:

Input: nums = [1,1,2,3,5]
Output: 1
Explanation: You can delete either `nums[0]` or `nums[1]` to make `nums` = [1,2,3,5] which is beautiful. It can be proven you need at least 1 deletion to make `nums` beautiful.

Example 2:

Input: nums = [1,1,2,2,3,3]
Output: 2
Explanation: You can delete `nums[0]` and `nums[5]` to make nums = [1,2,2,3] which is beautiful. It can be proven you need at least 2 deletions to make nums beautiful.

Constraints:

• `1 <= nums.length <= 105`
• `0 <= nums[i] <= 105`

SOLUTION:

``````class Solution:
def minDel(self, nums, i = 0, switch = False, dels = 0):
n = len(nums)
if i >= n:
if (n - dels) % 2 == 0:
return 0
return 1
if switch:
if i < n - 1 and i % 2 == 1 and nums[i] == nums[i + 1]:
return 1 + self.minDel(nums, i = i + 1, switch = False, dels = dels + 1)
return self.minDel(nums, i = i + 1, switch = True, dels = dels)
else:
if i < n - 1 and i % 2 == 0 and nums[i] == nums[i + 1]:
return 1 + self.minDel(nums, i = i + 1, switch = True, dels = dels + 1)
return self.minDel(nums, i = i + 1, switch = False, dels = dels)

def minDeletion(self, nums: List[int]) -> int:
return self.minDel(nums, i = 0, switch = False, dels = 0)
``````