# Leetcode – Minimum Cost to Move Chips to The Same Position Solution

The “Minimum Cost to Move Chips to The Same Position” problem is a fascinating and slightly unconventional problem frequently seen on LeetCode and during coding interviews. This problem tests your ability to think logically, to understand the mathematical implications behind the scene, and to write clean, efficient code. It belongs to a category of problems that may seem complex at first but can be simplified with clever insights.

In this detailed article, we’ll cover the problem statement, break down the logic, and discuss various approaches to solving this problem. We’ll also analyze the time and space complexity of these approaches and present Python code implementations.

## Problem Statement

### Description

We’re given an array of integers, position, where position[i] represents the position of the ith chip. There are some chips at each of the position numbers. You need to move all the chips to the same position. In one step, you can change the position of the ith chip from position[i] to:

• position[i] + 2 with cost = 0.
• position[i] + 1 or position[i] - 1 with cost = 1.

You need to return the minimum cost needed to move all the chips to the same position.

### Constraints

• 1 <= position.length <= 100
• 1 <= position[i] <= 10^9

## Understanding the Problem

At the outset, the problem seems to deal with an array of positions that could be in any random order. The trick is to find the cheapest way to get all the chips to the same position. One key insight here is that moving the chips by 2 positions doesn’t incur any cost, while moving them by 1 position incurs a cost of 1.

## Approaches

### Approach 1: Counting Odd and Even Positions

Upon a closer look, it becomes evident that the problem can be greatly simplified. Since moving the chips by 2 positions doesn’t incur any cost, the parity (odd or even) of the positions is what really matters. If a chip starts at an even position, it can move to any other even position at zero cost, and the same goes for odd positions.

#### Algorithm

1. Count the number of chips at even and odd positions.
2. The minimum cost will be the minimum between the number of chips at odd and even positions.

#### Python Code Implementation

def minCostToMoveChips(position):
even_count = 0
odd_count = 0

for pos in position:
if pos % 2 == 0:
even_count += 1
else:
odd_count += 1

return min(even_count, odd_count)

# Test the function
print(minCostToMoveChips([1, 2, 3]))  # Output should be 1
print(minCostToMoveChips([2, 2, 2, 3, 3]))  # Output should be 2

#### Time Complexity

This approach runs in O(n) time, where n is the length of the position array, since we only traverse the array once.

#### Space Complexity

The space complexity is O(1) as we only use constant extra space to keep track of the counts.

### Approach 2: Using Python’s Counter

We can make the code more Pythonic by using the collections.Counter class to count the occurrences of each type of position (even or odd).

#### Algorithm

1. Use collections.Counter to count the occurrences of even and odd positions.
2. Return the minimum count.

#### Python Code Implementation

from collections import Counter

def minCostToMoveChips(position):
counter = Counter(pos % 2 for pos in position)
return min(counter, counter)

# Test the function
print(minCostToMoveChips([1, 2, 3]))  # Output should be 1
print(minCostToMoveChips([2, 2, 2, 3, 3]))  # Output should be 2

#### Time Complexity

The time complexity remains O(n).

#### Space Complexity

The space complexity is also O(1) since the counter object will have at most 2 keys.

## Conclusion

The “Minimum Cost to Move Chips to The Same Position” problem might seem complicated initially, but with the right insight, it becomes a simple counting exercise. Understanding the importance of parity in this problem is crucial to arriving at the optimal, simplified solution. This problem serves as a good reminder that sometimes complicated problems can have straightforward solutions if approached from the right angle.