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 i^{th}
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 i^{th}
chip from position[i]
to:
position[i] + 2
orposition[i]  2
withcost = 0
.position[i] + 1
orposition[i]  1
withcost = 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/minimumcosttomovechipstothesameposition/
// 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/minimumcosttomovechipstothesameposition/ // 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/minimumcosttomovechipstothesameposition/ 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)