leetcode 1217) Minimum Cost to Move Chips to The Same Position

LV. Easy 😎

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

Minimum Cost to Move Chips to The Same Position - LeetCode

 

문제

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

 

문제 해결법

짝수에서 짝수, 홀수에서 홀수로 옮기는 건 cost가 들지 않기 때문에 position의 짝수 개수와 홀수 개수를 알면 문제 해결 가능!

짝수 개수 > 홀수 개수이면 임의의 짝수에 홀수를 옮기는 게 cost가 제일 적게 듦으로 minimum cost는 홀수 개수.

짝수 개수 < 홀수 개수이면 임의의 홀수에 짝수를 옮기는 게 cost가 제일 적게 듦으로 minimum cost는 짝수 개수.

 

해결 코드

#include <iostream>
#include <vector>

using namespace std;

class Solution {
public:
	int minCostToMoveChips(vector<int> &position) {
		int even_cost = 0;
		int odd_cost = 0;

		for (int i = 0; i < position.size(); ++i) {
			if (position[i] % 2 == 0)
				odd_cost++;
			else
				even_cost++;
		}
		return odd_cost > even_cost ? even_cost : odd_cost;
	}
};

int main() {
	Solution sol;
	vector<int>a;
	a.push_back(1);
	a.push_back(100000);
	// a.push_back(3);
	// a.push_back(3);
	// a.push_back(3);

	cout << sol.minCostToMoveChips(a) << endl;
}