Three Equal Parts
<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>
Array; Math
class Solution:
    def threeEqualParts(self, arr: List[int]) -> List[int]:
        def find(x):
            s = 0
            for i, v in enumerate(arr):
                s += v
                if s == x:
                    return i

        n = len(arr)
        cnt, mod = divmod(sum(arr), 3)
        if mod:
            return [-1, -1]
        if cnt == 0:
            return [0, n - 1]

        i, j, k = find(1), find(cnt + 1), find(cnt * 2 + 1)
        while k < n and arr[i] == arr[j] == arr[k]:
            i, j, k = i + 1, j + 1, k + 1
        return [i - 1, j] if k == n else [-1, -1]
