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1217. Minimum Cost to Move Chips to The Same Position
Description
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 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 { 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); } };

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)

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); };