leetcode-1217-easy

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

思路一：挨个位置统计。看了一下题解，发现只要把奇数、偶数位置的值统计后比较就行，这种算法是最快的

    public int minCostToMoveChips(int[] position) {
        int cost = Integer.MAX_VALUE;

        for (int i = 1; i < 101; i++) {
            int temp = calcCost(position, i);
            cost = Math.min(temp, cost);
        }

        return cost;
    }

    public int calcCost(int[] position, int idx) {
        int cost = 0;
        for (int i = 0; i < position.length; i++) {

            if (position[i] == idx) {
                continue;
            }

            if (Math.abs(position[i] - idx) % 2 != 0) {
                cost++;
            }
        }

        return cost;
    }