1217

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

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

Companies:
Apple, Amazon

Related Topics:
Array, Math, Greedy

Similar Questions:

• Minimum Number of Operations to Move All Balls to Each Box (Medium)

Solution 1.

// OJ: https://leetcode.com/problems/minimum-cost-to-move-chips-to-the-same-position/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int minCostToMoveChips(vector<int>& A) {
        int even = 0, odd = 0;
        for (int n : A) {
            if (n % 2) ++even;
            else ++odd;
        }
        return min(even, odd);
    }
};