Minimum Cost to Move Chips to The Same Position

Problem Statement

In a given scenario, we have a number of chips each placed at various, possibly distinct, positions on a line. Each chip's position is defined in an array where position[i] indicates the location of the i-th chip.

The objective is to move every chip such that they all share the same position. We can move a chip two positions either left or right at no cost. Alternatively, moving a chip one position left or right costs us a single unit.

We need to determine the minimal cost required to achieve the uniform positioning of all chips.

Examples

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

Approach and Intuition

When presented with the task of moving chips to a single position with minimal cost, a key observation about the cost structure significantly simplifies the problem:

• Moving a chip to an adjacent even or odd position has no cost. Thus, transitions between even positions or between odd positions are free.
• Moving a chip from an odd to an even position (or vice versa) incurs a cost of 1.

Given these rules, the strategy revolves around leveraging these cost-free moves to minimize the expensive moves. Here’s how to approach it:

1. First, differentiate chips based on their positions being odd or even:

   • Count the number of chips in even positions.
   • Count the number of chips in odd positions.

2. To minimize cost, aim to gather all chips at the position type (odd or even) where there are more chips initially:

   • If there are more chips at even positions, move chips from odd positions to an even position with a cost increment of 1 per move.
   • Conversely, if there are more chips at odd positions, move chips from even positions to an odd position with a similar cost.

The cost to unify all chips into one position will therefore be equal to the lesser count among chips at odd or even positions, as this represents the minimum number of necessary moves involving a cost.

By aligning this understanding with the constraints and boundaries of this problem, a clear, efficient solution emerges. Utilizing this strategy assures the lowest cost due to the minimal number of costly moves involved.

Solutions

class Solution {
    public int minimumChipCost(int[] chipsPosition) {
        int countEven = 0;
        int countOdd = 0;
        for (int pos : chipsPosition) {
            if (pos % 2 == 0) {
                countEven++;
            } else {
                countOdd++;
            }
        }
        return Math.min(countOdd, countEven);
    }
}

class Solution:
    def minimumCostForChips(self, chips_positions: List[int]) -> int:
        count_even = 0
        count_odd = 0
        for chip in chips_positions:
            if chip % 2 == 0:
                count_even += 1
            else:
                count_odd += 1
        return min(count_even, count_odd)