Formatted question description: https://leetcode.ca/all/1217.html

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 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

Related Topics:
Array, Math, Greedy

Solution 1.

// OJ: https://leetcode.com/problems/minimum-cost-to-move-chips-to-the-same-position/
// 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);
    }
};

Java

  • class Solution {
    public int minCostToMoveChips(int[] chips) {
        int evenCount = 0, oddCount = 0;
        int length = chips.length;
        for (int i = 0; i < length; i++) {
            int chip = chips[i];
            if (chip % 2 == 0)
                evenCount++;
            else
                oddCount++;
        }
        return Math.min(evenCount, oddCount);
    }
}

  • // OJ: https://leetcode.com/problems/minimum-cost-to-move-chips-to-the-same-position/
// 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);
    }
};

  • # 1217. Minimum Cost to Move Chips to The Same Position
# https://leetcode.com/problems/minimum-cost-to-move-chips-to-the-same-position/

class Solution:
    def minCostToMoveChips(self, position: List[int]):
        odd = even = 0
        
        for p in position:
            if p % 2 == 0: even += 1
            else: odd += 1
        
        return min(odd,even)

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