题目
We have n chips, where the position of the ith chip is position[i].
We need to move all the chips to the same position. In one step, we can change the position of the ith chip from position[i] to:
- position[i] + 2 or position[i] - 2 with cost = 0.
- position[i] + 1 or position[i] - 1 with cost = 1. Return the minimum cost needed to move all the chips to the same position.
解题思路
奇数移动到奇数，偶数移动到偶数花费是0，奇数偶数相互移动花费是1，结果是全移动到奇数，或全移动到偶数，所以结果为，奇数和偶数中少的
代码
class Solution:
def minCostToMoveChips(self, position: List[int]) -> int:
even = 0
for c in position:
if c % 2 == 0:
even += 1
return min(even, len(position) - even)
