main idea: there is no cost to move even steps, so to weigh coins number in odd and even position. Whichever has the less ammount will move to the opposite position.

# time O(N) space O(1)
class Solution:
    def minCostToMoveChips(self, position: List[int]) -> int:
        odd = 0
        even = 0 
        for pos in position:
            if pos%2 == 0:
                even += 1
            else:
                odd += 1
        return min(odd,even)