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    <p>Since each of the element in the original array are considered unique even when they have the same value, the problem can be treated in this way: For each element in the array, you have two choices: including it or not including it in the set. Therefore each element doubles the number of solutions, resulting the overall number of solutions: 2 * 2 * 2 ... * 2 - 1 = 2^n - 1. Minus one is because you are excluding empty set from the solutions. Since the number of solutions is 2^n - 1, the complexity of your algorithm cannot be better than O(2^n). The only improvement you can do is probably write the algorithm in a way that is more compact or easier to understand.</p> <p>I have the following code that I think is more straightforward. I have not tested so there might be bugs.</p> <pre><code>// Generate unique sets in the range [k..n) of the input arr. With elements // from [0..k) already in the prefix. void Unique(int[] arr, int k, List&lt;int&gt; prefix, List&lt;int[]&gt; solutions) { // Got a solution when we reached the end of array. if (k == arr.Length) { if (prefix.Length &gt; 0) solutions.Add(prefix.ToArray()); return; } // Exclude arr[k] Unique(arr, k + 1, prefix, solutions); // Include arr[k] prefix.Add(arr[k]); Unique(arr, k + 1, prefix, solutions); prefix.Remove(arr[k]); // Restore the prefix list } List&lt;int[]&gt; Unique(int[] arr) { List&lt;int[]&gt; solutions = new List&lt;int[]&gt;(); Unique(arr, 0, new List&lt;int&gt;(), List&lt;int[]&gt; solutions); return solutions; } </code></pre>
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    1. POUnique, Unordered Sets from Collection of Elements (Power set)
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