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
orposition[i] - 2
withcost = 0
. -
position[i] + 1
orposition[i] - 1
withcost = 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
SOLUTION:
class Solution:
def minCostToMoveChips(self, position: List[int]) -> int:
minCost = float('inf')
for currpos in position:
currCost = 0
for pos in position:
currCost += abs(pos - currpos) % 2
minCost = min(minCost, currCost)
return minCost
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