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
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 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
Related Topics:
Array, Math, Greedy
Similar Questions:
// 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);
}
};