Minimum Cost to Move Chips to The Same Position (E)

題目

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 poistion 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

題意

有若干個位置，每個位置上有若干個硬幣，現在要將所有硬幣移到一個位置。將一個硬幣移動一個位置的cost為1，移動兩個位置的cost為0，求需要的最小cost。

思路

因為移動兩個位置的cost為0，所以所有奇數位的硬幣都可以移到位置1，所有偶數位的硬幣都可以移到位置2，最後只要考慮從1移到2或從2移到1即可。

程式碼實現

Java

class Solution {
    public int minCostToMoveChips(int[] position) {
        int odd = 0, even = 0;

        for (int i : position) {
            if (i % 2 == 0) {
                even++;
            } else {
                odd++;
            }
        }

        return Math.min(odd, even);
    }
}