<p>Given an array of integers <code>arr</code>, return <code>true</code> if we can partition the array into three <strong>non-empty</strong> parts with equal sums.</p>

<p>Formally, we can partition the array if we can find indexes <code>i + 1 &lt; j</code> with <code>(arr[0] + arr[1] + ... + arr[i] == arr[i + 1] + arr[i + 2] + ... + arr[j - 1] == arr[j] + arr[j + 1] + ... + arr[arr.length - 1])</code></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> arr = [0,2,1,-6,6,-7,9,1,2,0,1]
<strong>Output:</strong> true
<strong>Explanation: </strong>0 + 2 + 1 = -6 + 6 - 7 + 9 + 1 = 2 + 0 + 1
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> arr = [0,2,1,-6,6,7,9,-1,2,0,1]
<strong>Output:</strong> false
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> arr = [3,3,6,5,-2,2,5,1,-9,4]
<strong>Output:</strong> true
<strong>Explanation: </strong>3 + 3 = 6 = 5 - 2 + 2 + 5 + 1 - 9 + 4
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>3 &lt;= arr.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>-10<sup>4</sup> &lt;= arr[i] &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution {
public:
    bool canThreePartsEqualSum(vector<int>& arr) {
     int sum = 0;
     for(auto x : arr) sum+=x;
     if(sum%3!=0) return false;
     sum/=3;
     int x = 0;int a = 0;
     for(int i=0;i<arr.size();i++){
        a+=arr[i];
        if(a==sum){
          x++;
          a=0;
        }
     }
     if(x>=3) return true;
     else return false;   
    }
};
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