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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.

• Java
• C++
• Python
• Go
• Javascript

• 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);
      }
  }
  
  ############
  
  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);
      }
  }
  

• // 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);
      }
  };
  

• 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)
  
  ############
  
  # 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)
  

• 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);
  };
  
  

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