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 alli % 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)
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