Question

1217. Minimum Cost to Move Chips to The Same Position

中文文档

Description

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

Solutions

Solution 1: Quick Thinking

Move all chips at even indices to position 0, and all chips at odd indices to position 1, all at a cost of 0. Then, choose the position (either 0 or 1) with fewer chips and move these chips to the other position. The minimum cost required is the smaller quantity of chips.

The time complexity is $O(n)$, and the space complexity is $O(1)$. Here, $n$ is the number of chips.

class Solution:
  def minCostToMoveChips(self, position: List[int]) -> int:
    a = sum(p % 2 for p in position)
    b = len(position) - a
    return min(a, b)

class Solution {
  public int minCostToMoveChips(int[] position) {
    int a = 0;
    for (int p : position) {
      a += p % 2;
    }
    int b = position.length - a;
    return Math.min(a, b);
  }
}

class Solution {
public:
  int minCostToMoveChips(vector<int>& position) {
    int a = 0;
    for (auto& p : position) a += p & 1;
    int b = position.size() - a;
    return min(a, b);
  }
};

func minCostToMoveChips(position []int) int {
  a := 0
  for _, p := range position {
    a += p & 1
  }
  b := len(position) - a
  if a < b {
    return a
  }
  return b
}

/**
 * @param {number[]} position
 * @return {number}
 */
var minCostToMoveChips = function (position) {
  let a = 0;
  for (let v of position) {
    a += v % 2;
  }
  let b = position.length - a;
  return Math.min(a, b);
};