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Decrease Elements To Make Array Zigzag in Python
Suppose we have an array nums of integers, a move operation is actually choosing any element and decreasing it by 1. An array A is a zigzag array if either 1 or 2 is satisfied −
Every even-indexed element is greater than adjacent elements, So. A[0] > A[1] < A[2] > A[3] < A[4] > ... and so on.
Every odd-indexed element is greater than adjacent elements, So. A[0] < A[1] > A[2] < A[3] > A[4] < ... and so on.
We have to find the minimum number of moves to transform the given array nums into a zigzag array.
So if the array is like [1,2,3], then the output will be 2, as we can decrease 2 to 0 or 3 to 1
To solve this, we will follow these steps −
Define a method called solve(), this will take nums and start, this will work as below −
k := 0
for i in range start to length of nums, increase by 2
left := 100000 when i – 1 < 0, otherwise nums[i - 1]
right := 100000 when i + 1 >= length of nums, otherwise nums[i + 1]
temp := (minimum of left and right) – 1 – nums[i]
if temp < 0, then k := k + |temp|
return k
In the main method, it will be
ans := solve(nums, 0)
ans := minimum of ans and solve(nums, 1)
return ans
Example(Python)
Let us see the following implementation to get a better understanding −
class Solution(object): def solve(self,nums,start): k = 0 for i in range(start,len(nums),2): left = 100000 if i-1<0 else nums[i-1] right = 10000 if i+1>=len(nums) else nums[i+1] temp= (min(left,right)-1 - nums[i]) if temp<0: k+=abs(temp) return k def movesToMakeZigzag(self, nums): ans = self.solve(nums,0) ans = min(ans,self.solve(nums,1)) return ans ob = Solution() print(ob.movesToMakeZigzag([1,2,3]))
Input
[1,2,3]
Output
2
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