class Solution:
def minimumOperations(self, nums: List[int]) -> int:
res = 0
# 初始时先比较一次
s = set(nums)
if len(s) == len(nums):
return res
while len(nums) >= 3:
s = set(nums)
if len(s) != len(nums):
res += 1
nums = nums[3:]
# 剩下元素少于3的情形
# 1个直接得res
# 2个还要判断是否相等来决定是否还需操作一次
if len(nums) == 2 and nums[0] == nums[1]:
res += 1
return res