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Minimum_Cost_To_Move_Chips_To_The_Same_Position.py
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Minimum_Cost_To_Move_Chips_To_The_Same_Position.py
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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 position 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
# O(n) Time and O(1) space
class Solution:
def minCostToMoveChips(self, position: List[int]) -> int:
return min(len([i for i in position if i%2==0]), len([i for i in position if i%2==1]))
# O(n) Time and O(1) space
class Solution:
def minCostToMoveChips(self, position: List[int]) -> int:
even_cnt = 0
odd_cnt = 0
for i in position:
if i % 2 == 0:
even_cnt += 1
else:
odd_cnt += 1
return min(even_cnt, odd_cnt)