Minimum Cost to Move Chips to The Same Position - LeetCode
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.
- code
class Solution:
def minCostToMoveChips(self, position: List[int]) -> int:
count_2 = 0
for v in position:
if v % 2 == 0:
count_2 += 1
return min(len(position)-count_2, count_2)