Construct Target Array With Multiple Sums
<p>You are given an array <code>target</code> of n integers. From a starting array <code>arr</code> consisting of <code>n</code> 1&#39;s, you may perform the following procedure :</p>

<ul>
	<li>let <code>x</code> be the sum of all elements currently in your array.</li>
	<li>choose index <code>i</code>, such that <code>0 &lt;= i &lt; n</code> and set the value of <code>arr</code> at index <code>i</code> to <code>x</code>.</li>
	<li>You may repeat this procedure as many times as needed.</li>
</ul>

<p>Return <code>true</code> <em>if it is possible to construct the</em> <code>target</code> <em>array from</em> <code>arr</code><em>, otherwise, return</em> <code>false</code>.</p>

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

<pre>
<strong>Input:</strong> target = [9,3,5]
<strong>Output:</strong> true
<strong>Explanation:</strong> Start with arr = [1, 1, 1] 
[1, 1, 1], sum = 3 choose index 1
[1, 3, 1], sum = 5 choose index 2
[1, 3, 5], sum = 9 choose index 0
[9, 3, 5] Done
</pre>

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

<pre>
<strong>Input:</strong> target = [1,1,1,2]
<strong>Output:</strong> false
<strong>Explanation:</strong> Impossible to create target array from [1,1,1,1].
</pre>

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

<pre>
<strong>Input:</strong> target = [8,5]
<strong>Output:</strong> true
</pre>

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

<ul>
	<li><code>n == target.length</code></li>
	<li><code>1 &lt;= n &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= target[i] &lt;= 10<sup>9</sup></code></li>
</ul>
Array; Heap (Priority Queue)
class Solution {
    public boolean isPossible(int[] target) {
        PriorityQueue<Long> pq = new PriorityQueue<>(Collections.reverseOrder());
        long s = 0;
        for (int x : target) {
            s += x;
            pq.offer((long) x);
        }
        while (pq.peek() > 1) {
            long mx = pq.poll();
            long t = s - mx;
            if (t == 0 || mx - t < 1) {
                return false;
            }
            long x = mx % t;
            if (x == 0) {
                x = t;
            }
            pq.offer(x);
            s = s - mx + x;
        }
        return true;
    }
}