1217. Minimum Cost to Move Chips to The Same Position (E)

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

题目

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

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题意

有若干个位置，每个位置上有若干个硬币，现在要将所有硬币移到一个位置。将一个硬币移动一个位置的cost为1，移动两个位置的cost为0，求需要的最小cost。

思路

因为移动两个位置的cost为0，所以所有奇数位的硬币都可以移到位置1，所有偶数位的硬币都可以移到位置2，最后只要考虑从1移到2或从2移到1即可。

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代码实现

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);
    }
}