1217. Minimum Cost to Move Chips to The Same Position

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

这题很有意思，看起来是个难题，实际上是一个脑筋急转弯，想出来非常简单。

首先移动2步，则代价为0,移动1步，则代价为1。在此基础上，无论有多少硬币，我们都将其移动到两个格子上，一个格子上存奇数，一个上面存偶数。

然后我们只要看哪个小即可，就这么简单直接。

class MinimumCostToMoveChipsToTheSamePosition : public Solution {
public:
    void Exec() {

    }
    int minCostToMoveChips(vector<int>& position) {
        int poses[2] = {0};
        for (auto i : position) {
            poses[i % 2]++;
        }
        return poses[0] < poses[1] ? poses[0] : poses[1];
    }
};