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1217. Minimum Cost to Move Chips to The Same Position

Easy

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

Solution

class Solution:
    def minCostToMoveChips(self, position: List[int]) -> int:
        evens, odds = 0, 0

        for p in position:
            if p % 2 == 0:
                evens += 1
            else:
                odds += 1

        return min(evens, odds)