The main observation here is that moving chips from position p to p+2 or p-2 incurs no cost. Hence, the cost to align all chips into one position is essentially the cost of moving all odd-indexed chips to even positions or vice versa.
To achieve this, count the number of chips at odd positions and the number of chips at even positions. The minimum cost to move all chips to the same position is the lesser of the two counts.
Time Complexity: O(n), where n is the number of chips.
Space Complexity: O(1) because only a fixed amount of extra space is used.
1using System;
2using System.Linq;
3
4public class Solution {
5 public int MinCostToMoveChips(int[] position) {
6 int oddCount = position.Count(p => p % 2 == 1);
7 int evenCount = position.Length - oddCount;
8 return Math.Min(oddCount, evenCount);
9 }
10
11 public static void Main() {
12 Solution sol = new Solution();
13 Console.WriteLine(sol.MinCostToMoveChips(new int[] {1, 2, 3}));
14 }
15}In this C# solution, we use LINQ to count odd and even positions and return the minimum count, which determines the minimal cost to realign the chips.
A similar but slightly varied approach leverages modular arithmetic to classify indices. As the cost is incurred when moving by an odd number of steps, the strategy remains to count parity of indices and work from there.
Time Complexity: O(n) for iterating through the list of positions.
Space Complexity: O(1), with fixed space used to track counts.
1#include <stdio.h>
2
3int minCostToMoveChips(int* position, int positionSize) {
4 int count[2] = {0};
5 for (int i = 0; i < positionSize; i++) {
6 count[position[i] % 2]++;
7 }
8 return count[0] < count[1] ? count[0] : count[1];
9}
10
11int main() {
12 int position[] = {1, 2, 3};
13 int size = sizeof(position) / sizeof(position[0]);
14 printf("%d\n", minCostToMoveChips(position, size));
15 return 0;
16}By using an array `count` to track counts of even and odd indices, this solution provides an efficient way to determine the minimum movement cost using basic arithmetic for index parity.