<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 {
public:
    vector<int> threeEqualParts(vector<int>& arr) {
        int N = arr.size();
        vector<int> result(2, -1);
        int totalOnes = count(arr.begin(), arr.end(), 1);
        if (totalOnes % 3 != 0) return result; // Not divisible by 3
        if (totalOnes == 0) return {0, N-1}; // All zeros case
        int k = totalOnes / 3;
        int first = -1, second = -1, third = -1, oneCount = 0;

        for (int i = 0; i < N; ++i) {
            if (arr[i] == 1) {
                if (oneCount == 0) first = i;
                else if (oneCount == k) second = i;
                else if (oneCount == 2 * k) third = i;
                ++oneCount;
            }
        }

        while (third < N && arr[first] == arr[second] && arr[second] == arr[third]) {
            first++;
            second++;
            third++;
        }

        if (third == N) return {first - 1, second};
        return result;
    }
};
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