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.
Example 1:
Input: position = [1,2,3]
Output: 1
Explanation: First step: Move the chip at position 3 to position 1 with cost = 0.
Second step: Move the chip at position 2 to position 1 with cost = 1.
Total cost is 1.
Example 2:
Input: position = [2,2,2,3,3]
Output: 2
Explanation: We can move the two chips at poistion 3 to position 2. Each move has cost = 1. The total cost = 2.
Example 3:
Input: position = [1,1000000000]
Output: 1
Constraints:
1 <= position.length <= 100
1 <= position[i] <= 10^9
class Solution:
    def minCostToMoveChips(self, position: List[int]) -> int:
        even = 0
        odd = 0
        for i in position:
            if i%2 == 0:
                even += 1
            else:
                odd += 1
        return  min(even, odd)
class Solution:
    def minCostToMoveChips(self, position: List[int]) -> int:
        even = 0
        odd = 0
        for i in position:
            if i%2 == 0:
                even += 1
            else:
                odd += 1
        return  min(even, odd)