<p>You are given a <strong>1-indexed</strong> array of <strong>distinct</strong> integers <code>nums</code> of length <code>n</code>.</p>

<p>You need to distribute all the elements of <code>nums</code> between two arrays <code>arr1</code> and <code>arr2</code> using <code>n</code> operations. In the first operation, append <code>nums[1]</code> to <code>arr1</code>. In the second operation, append <code>nums[2]</code> to <code>arr2</code>. Afterwards, in the <code>i<sup>th</sup></code> operation:</p>

<ul>
	<li>If the last element of <code>arr1</code> is<strong> greater</strong> than the last element of <code>arr2</code>, append <code>nums[i]</code> to <code>arr1</code>. Otherwise, append <code>nums[i]</code> to <code>arr2</code>.</li>
</ul>

<p>The array <code>result</code> is formed by concatenating the arrays <code>arr1</code> and <code>arr2</code>. For example, if <code>arr1 == [1,2,3]</code> and <code>arr2 == [4,5,6]</code>, then <code>result = [1,2,3,4,5,6]</code>.</p>

<p>Return <em>the array</em> <code>result</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,1,3]
<strong>Output:</strong> [2,3,1]
<strong>Explanation:</strong> After the first 2 operations, arr1 = [2] and arr2 = [1].
In the 3<sup>rd</sup> operation, as the last element of arr1 is greater than the last element of arr2 (2 &gt; 1), append nums[3] to arr1.
After 3 operations, arr1 = [2,3] and arr2 = [1].
Hence, the array result formed by concatenation is [2,3,1].
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [5,4,3,8]
<strong>Output:</strong> [5,3,4,8]
<strong>Explanation:</strong> After the first 2 operations, arr1 = [5] and arr2 = [4].
In the 3<sup>rd</sup> operation, as the last element of arr1 is greater than the last element of arr2 (5 &gt; 4), append nums[3] to arr1, hence arr1 becomes [5,3].
In the 4<sup>th</sup> operation, as the last element of arr2 is greater than the last element of arr1 (4 &gt; 3), append nums[4] to arr2, hence arr2 becomes [4,8].
After 4 operations, arr1 = [5,3] and arr2 = [4,8].
Hence, the array result formed by concatenation is [5,3,4,8].
</pre>

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

<ul>
	<li><code>3 &lt;= n &lt;= 50</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 100</code></li>
	<li>All elements in <code>nums</code> are distinct.</li>
</ul>

class Solution:
    def resultArray(self, nums: List[int]) -> List[int]:
        arr1 = [nums[0]]
        arr2 = [nums[1]]
        nums = nums[2:]
        while True:
            if nums == []:
                break
            else:
                if arr1[-1] > arr2[-1]:
                    arr1.append(nums[0])
                else:
                    arr2.append(nums[0])
                
                nums = nums[1:]
        
        return arr1+arr2
            
        
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