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 {
    private int[] arr;

    public int[] threeEqualParts(int[] arr) {
        this.arr = arr;
        int cnt = 0;
        int n = arr.length;
        for (int v : arr) {
            cnt += v;
        }
        if (cnt % 3 != 0) {
            return new int[] {-1, -1};
        }
        if (cnt == 0) {
            return new int[] {0, n - 1};
        }
        cnt /= 3;

        int i = find(1), j = find(cnt + 1), k = find(cnt * 2 + 1);
        for (; k < n && arr[i] == arr[j] && arr[j] == arr[k]; ++i, ++j, ++k) {
        }
        return k == n ? new int[] {i - 1, j} : new int[] {-1, -1};
    }

    private int find(int x) {
        int s = 0;
        for (int i = 0; i < arr.length; ++i) {
            s += arr[i];
            if (s == x) {
                return i;
            }
        }
        return 0;
    }
}