LeetCode #1217 Play with Chips 玩筹码

1217 Play with Chips 玩筹码

Description:
There are some chips, and the i-th chip is at position chips[i].

You can perform any of the two following types of moves any number of times (possibly zero) on any chip:

Move the i-th chip by 2 units to the left or to the right with a cost of 0.
Move the i-th chip by 1 unit to the left or to the right with a cost of 1.
There can be two or more chips at the same position initially.

Return the minimum cost needed to move all the chips to the same position (any position).

Example:

Example 1:

Input: chips = [1,2,3]
Output: 1
Explanation: Second chip will be moved to positon 3 with cost 1. First chip will be moved to position 3 with cost 0. Total cost is 1.

Example 2:

Input: chips = [2,2,2,3,3]
Output: 2
Explanation: Both fourth and fifth chip will be moved to position two with cost 1. Total minimum cost will be 2.

Constraints:

1 <= chips.length <= 100
1 <= chips[i] <= 10^9

题目描述:
数轴上放置了一些筹码，每个筹码的位置存在数组 chips 当中。

你可以对 任何筹码 执行下面两种操作之一（不限操作次数，0 次也可以）：

将第 i 个筹码向左或者右移动 2 个单位，代价为 0。
将第 i 个筹码向左或者右移动 1 个单位，代价为 1。
最开始的时候，同一位置上也可能放着两个或者更多的筹码。

返回将所有筹码移动到同一位置（任意位置）上所需要的最小代价。

示例 :

示例 1：

输入：chips = [1,2,3]
输出：1
解释：第二个筹码移动到位置三的代价是 1，第一个筹码移动到位置三的代价是 0，总代价为 1。

示例 2：

输入：chips = [2,2,2,3,3]
输出：2
解释：第四和第五个筹码移动到位置二的代价都是 1，所以最小总代价为 2。

提示：

1 <= chips.length <= 100
1 <= chips[i] <= 10^9

思路:

因为增加 2的倍数不需要 cost, 把数组按奇偶分开即可
时间复杂度O(n), 空间复杂度O(1)

代码:
C++:

class Solution 
{
public:
    int minCostToMoveChips(vector& chips) 
    {
        int odd = 0, even = 0;
        for (auto chip : chips)
        {
            if (chip & 1) ++odd;
            else ++even;
        }
        return min(odd, even);
    }
};

Java:

class Solution {
    public int minCostToMoveChips(int[] chips) {
        int odd = 0, even = 0;
        for (int chip : chips) {
            if ((chip & 1) == 0) ++even;
            else ++odd;
        }
        return Math.min(odd, even);
    }
}

Python:

class Solution:
    def minCostToMoveChips(self, chips: List[int]) -> int:
        return min(sum(chip & 1 for chip in chips), len(chips) - sum(chip & 1 for chip in chips))