<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 cnt = 0;
        for(int i = 0 ; i < arr.size() ; i++) if(arr[i] == 1) cnt++;
        if(cnt % 3 != 0) return {-1, -1};
        if(cnt == 0) return {0, 2};

        int k = cnt / 3, curr = 0;
        int i1 = -1, i2 = -1, i3 = -1;
        int j1 = -1, j2 = -1, j3 = -1;

        for(int i = 0 ; i < arr.size() ; i++)
        {
            if(arr[i] == 1)
            {
                curr++;
                if(curr == 1) i1 = i;
                if(curr == k + 1) i2 = i;
                if(curr == 2*k + 1) i3 = i;

                if(curr == k) j1 = i;
                if(curr == 2*k) j2 = i;
                if(curr == 3*k) j3 = i;
            }
        }
        vector < int > a, b, c;
        copy(arr.begin() + i1, arr.begin() + j1 + 1, back_inserter(a));
        copy(arr.begin() + i2, arr.begin() + j2 + 1, back_inserter(b));
        copy(arr.begin() + i3, arr.begin() + j3 + 1, back_inserter(c));

        if(a != b || a != c) return {-1, -1};

        int first = 0, second = 0, third = 0;
        first = i2 - j1 - 1;
        second = i3 - j2 - 1;
        third = arr.size() - j3 - 1;

        if(third > min(first, second)) return {-1, -1};
        
        return {j1 + third, j2 + third + 1};
    }
};
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