<p>We have <code>n</code> chips, where the position of the <code>i<sup>th</sup></code> chip is <code>position[i]</code>.</p>

<p>We need to move all the chips to <strong>the same position</strong>. In one step, we can change the position of the <code>i<sup>th</sup></code> chip from <code>position[i]</code> to:</p>

<ul>
	<li><code>position[i] + 2</code> or <code>position[i] - 2</code> with <code>cost = 0</code>.</li>
	<li><code>position[i] + 1</code> or <code>position[i] - 1</code> with <code>cost = 1</code>.</li>
</ul>

<p>Return <em>the minimum cost</em> needed to move all the chips to the same position.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/08/15/chips_e1.jpg" style="width: 750px; height: 217px;" />
<pre>
<strong>Input:</strong> position = [1,2,3]
<strong>Output:</strong> 1
<strong>Explanation:</strong> 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.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/08/15/chip_e2.jpg" style="width: 750px; height: 306px;" />
<pre>
<strong>Input:</strong> position = [2,2,2,3,3]
<strong>Output:</strong> 2
<strong>Explanation:</strong> We can move the two chips at position  3 to position 2. Each move has cost = 1. The total cost = 2.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> position = [1,1000000000]
<strong>Output:</strong> 1
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= position.length &lt;= 100</code></li>
	<li><code>1 &lt;= position[i] &lt;= 10^9</code></li>
</ul>

from typing import List

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

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