LeetCode 1217. Minimum Cost to Move Chips to The Same Position Solution in Java, C++, Python & More | Explanation + Code

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.

PythonJavaC++GoJavaScript

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)(code-box)

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); } }(code-box)

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); } };(code-box)

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 }(code-box)

/** * @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); };(code-box)