<p>You are given an array <code>arr</code> which consists of only zeros and ones, divide the array into <strong>three non-empty parts</strong> such that all of these parts represent the same binary value.</p>

<p>If it is possible, return any <code>[i, j]</code> with <code>i + 1 &lt; j</code>, such that:</p>

<ul>
	<li><code>arr[0], arr[1], ..., arr[i]</code> is the first part,</li>
	<li><code>arr[i + 1], arr[i + 2], ..., arr[j - 1]</code> is the second part, and</li>
	<li><code>arr[j], arr[j + 1], ..., arr[arr.length - 1]</code> is the third part.</li>
	<li>All three parts have equal binary values.</li>
</ul>

<p>If it is not possible, return <code>[-1, -1]</code>.</p>

<p>Note that the entire part is used when considering what binary value it represents. For example, <code>[1,1,0]</code> represents <code>6</code> in decimal, not <code>3</code>. Also, leading zeros <strong>are allowed</strong>, so <code>[0,1,1]</code> and <code>[1,1]</code> represent the same value.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> arr = [1,0,1,0,1]
<strong>Output:</strong> [0,3]
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> arr = [1,1,0,1,1]
<strong>Output:</strong> [-1,-1]
</pre><p><strong class="example">Example 3:</strong></p>
<pre><strong>Input:</strong> arr = [1,1,0,0,1]
<strong>Output:</strong> [0,2]
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>3 &lt;= arr.length &lt;= 3 * 10<sup>4</sup></code></li>
	<li><code>arr[i]</code> is <code>0</code> or <code>1</code></li>
</ul>

class Solution:
    def threeEqualParts(self, arr: List[int]) -> List[int]:
        n = sum(arr)
        if n % 3: return [-1, -1]
        if n == 0: return [0, 2]
        k, indices = n // 3, []
        for i, num in enumerate(arr):
            if num: indices.append(i)
        i1, i2, i3, i, j = *[indices[-(j * k)] for j in (3, 2, 1)], 0 ,0
        part1, part2, part3 = arr[i1: i2], arr[i2: i3], arr[i3:]
        l1, l2, l3 = i2 - i1, i3 - i2, len(part3)
        if l3 > l2 or part2[:l3] != part3 or l3 > l1 or part1[:l3] != part3: return [-1, -1]
        i, j = i1 + l3 - 1, i2 + l3
        return [i, j]
        
        
        
18100