Minimum Cost to Move Chips to The Same Position

The idea for this problem is to understand that moving the chip at an even index to another even index or moving a chip at an odd index to another odd index costs nothing. So the solution is to simply count the number of chips at even indices and the number of chips at odd indices. The minimum cost to move all chips to the same position will be the minimum of these two counts, as it would cost 1 per move to move a chip from an even index to an odd index or vice versa.

Python solution:

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class Solution:
    def minCostToMoveChips(self, position: List[int]) -> int:
        even = 0
        odd = 0
        for i in position:
            if i % 2 == 0:
                even += 1
            else:
                odd += 1
        return min(even, odd)

We use a generator expression to count the number of chips at even indices. Then, we subtract this count from the total number of chips to get the number of chips at odd indices. The minimum cost to move all chips to the same position will be the smaller of these two counts.

Abstract Representation of the Problem

You are given a collection of objects. These objects are divided into two categories based on a certain property they possess. You can transfer any object from one category to the other at a cost. However, moving an object within the same category is free. Your goal is to minimize the total cost while ensuring all objects end up in the same category.

In this abstract problem:

• The “objects” represent the chips.
• The “categories” are the positions of the chips, which are classified as either even or odd.
• “Transfer” corresponds to moving a chip from an odd position to an even one, or vice versa.
• The “cost” is associated with moving a chip from an odd position to an even one, or vice versa.
• The condition to move an object within the same category at no cost corresponds to moving a chip by 2 positions.
• The objective to minimize the total cost to have all objects in the same category corresponds to moving all chips to either an even or an odd position.

Q&A

Initially I thought it was a DP problem and tried some cases where I realized that there are no overlapping subproblems. So , I tried the greedy approach. I took the following lists initially:

[2,3] cost=1 2.[2,4] cost=0 3.[2,5] cost=1 4.[2,6] cost=0 Then I understood that an even number can be reached by any other even number with a cost=0 because even_no+2 or even_no-2 leads to even_no. Similarly, I realized odd_no+2 or odd_no-2 leads to odd_no. So, my thinking was if I can form 1 cluster of all even numbers given and 1 cluster of all odd numbers given, then still the resultant cost will be 0!!! Hence, the problem reduces to finding minimum of total number of odd and total number of even numbers.

Your thinking is correct and aligns perfectly with the nature of this problem. The key insight is that moving between positions of the same parity (odd to odd or even to even) costs nothing, while moving between different parities has a cost of 1.

Because of these properties, the optimal strategy is indeed to move all chips to the position of the smallest group, either odd or even, because the cost to move a chip is 1 regardless of the distance. This way, we minimize the total cost by minimizing the number of costly moves.

In simpler terms, if there are fewer odd chips than even chips, we move all chips to an odd position. If there are fewer even chips than odd chips, we move all chips to an even position.

Therefore, counting the number of odd and even chips and returning the minimum is a correct and efficient solution to this problem.

In code, this would look something like:

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class Solution:
    def minCostToMoveChips(self, position: List[int]) -> int:
        even_count = 0
        odd_count = 0
        for i in position:
            if i % 2 == 0:
                even_count += 1
            else:
                odd_count += 1
        return min(even_count, odd_count)

This solution runs in O(n) time where n is the length of the input list position, which is efficient for the given constraints.

The key observation here is that it’s free to move a chip from an even number to another even number, and it is free to move a chip from an odd number to another odd number. It only costs to move an even to an odd, or an odd to an even. Therefore, we want to minimise such moves.

All chips must be on the same position once we’re done, which is either even or odd. Therefore, we want to calculate whether it is cheaper to move all the odd chips to even or all the even chips to odd. This will simply come down to how many even chips we start with, and how many odd chips. Each chip that has to be moved will cost exactly 1.

To determine the cost, we need to count how many are even, and how many are odd, and then take the minimum of these two values.

Imagine you’re playing a game where you have some coins placed on a number line. Now, the game allows you to move the coins according to specific rules. If a coin is on an even number, you can move it to another even number without any cost. Similarly, if a coin is on an odd number, you can move it to another odd number freely.

However, if you want to move a coin from an even number to an odd number or vice versa, you have to pay one unit. Your goal is to gather all the coins at a single point on the number line while spending the least amount.

So, what’s the trick? The trick is to realize that the final gathering point will be either an even number or an odd number. Therefore, we should aim to move the lesser group of coins (even or odd) to the other group’s location.

By simply counting the number of coins on even and odd positions, we can then decide whether to move the odd coins to an even position or the even coins to an odd position. We choose the option that requires fewer moves, which will be the smaller count. That’s it!

Key Takeaways

An excellent example of how the thinking part of work can be as important as the coding part of it, if not more important.

It requires some analysis to provide short and effective solution instead of brain-dead coding for ten minutes.

Great problem for deduction.

Indeed, this problem is an excellent example of how a seemingly complex problem can become much simpler through careful observation and deduction. By identifying the key properties of the problem (in this case, the rules for moving chips), you can drastically reduce the complexity of the problem and arrive at a simple, efficient solution. This approach is a critical part of problem-solving in computer science and many other fields.