Given an array nums of integers, a move consists of choosing any element and decreasing it by 1.
An array A is a zigzag array if either:
Every even-indexed element is greater than adjacent elements, ie. A[0] > A[1] < A[2] > A[3] < A[4] > ...
OR, every odd-indexed element is greater than adjacent elements, ie. A[0] < A[1] > A[2] < A[3] > A[4] < ...
Return the minimum number of moves to transform the given array nums into a zigzag array.
Example 1:
Input: nums = [1,2,3]
Output: 2
Explanation: We can decrease 2 to 0 or 3 to 1.
Example 2:
Input: nums = [9,6,1,6,2]
Output: 4
Constraints:
1 <= nums.length <= 1000
1 <= nums[i] <= 1000
class Solution:
    def movesToMakeZigzag(self, A: List[int]) -> int:
        res1=0
        n=len(A)
        B=A.copy()
        for i in range(1,n,2):
            if A[i-1]<=A[i]:
                t=A[i]-A[i-1]+1
                res1+=t
                A[i]=A[i-1]-1
            if i+1<n and A[i+1]<=A[i]:
                t=A[i]-A[i+1]+1
                res1+=t
                A[i]=A[i+1]-1
        res2=0
        for i in range(0,n,2):
            if i-1>=0 and B[i]>=B[i-1]:
                t=B[i]-B[i-1]+1
                res2+=t
                B[i]=B[i-1]-1
            if i+1<n and B[i+1]<=B[i]:
                t=B[i]-B[i+1]+1
                res2+=t
                B[i]=B[i+1]-1
        # print(A)
        # print(B)
        # print(res1,res2)
        return min(res1,res2)
class Solution:
    def movesToMakeZigzag(self, A: List[int]) -> int:
        res1=0
        n=len(A)
        B=A.copy()
        for i in range(1,n,2):
            if A[i-1]<=A[i]:
                t=A[i]-A[i-1]+1
                res1+=t
                A[i]=A[i-1]-1
            if i+1<n and A[i+1]<=A[i]:
                t=A[i]-A[i+1]+1
                res1+=t
                A[i]=A[i+1]-1
        res2=0
        for i in range(0,n,2):
            if i-1>=0 and B[i]>=B[i-1]:
                t=B[i]-B[i-1]+1
                res2+=t
                B[i]=B[i-1]-1
            if i+1<n and B[i+1]<=B[i]:
                t=B[i]-B[i+1]+1
                res2+=t
                B[i]=B[i+1]-1
        # print(A)
        # print(B)
        # print(res1,res2)
        return min(res1,res2)