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    <p><strong>UPD2:</strong> The following solution is for a problem when it is not required that subarray is contiguous. I misunderstood the problem statement. Not deleting this, as somebody may have an idea based on mine that will work for the actual problem.</p> <hr> <p>Here's what I've come up with:</p> <p>Create an instance of a dictionary (which is implemented as hash table, giving O(1) in normal situations). Keys are integers, values are hash sets of integers (also O(1)) – <code>var D = new Dictionary&lt;int, HashSet&lt;int&gt;&gt;</code>.</p> <p>Iterate through the array <code>A</code> and for each integer <code>n</code> with index <code>i</code> do:</p> <ol> <li>Check whether keys <code>n-1</code> and <code>n+1</code> are contained in <code>D</code>. <ul> <li>if neither key exists, do <code>D.Add(n, new HashSet&lt;int&gt;)</code></li> <li>if only one of the keys exists, e.g. <code>n-1</code>, do <code>D.Add(n, D[n-1])</code></li> <li>if both keys exist, do <code>D[n-1].UnionWith(D[n+1]); D[n+1] = D[n] = D[n-1];</code></li> </ul></li> <li><code>D[n].Add(n)</code></li> </ol> <p>Now go through each key in <code>D</code> and find the hash set with the greatest length (finding length is O(1)). The greatest length will be the answer.</p> <p>To my understanding, the <em>worst case</em> complexity will be O(n*log(n)), only because of the <code>UnionWith</code> operation. I don't know how to calculate the average complexity, but it should be close to O(n). Please correct me if I am wrong.</p> <p><strong>UPD:</strong> To speak code, here's a test implementation in C# that gives the correct result in both of the OP's examples:</p> <pre><code>var A = new int[] {4, 5, 1, 5, 7, 6, 8, 4, 1}; var D = new Dictionary&lt;int, HashSet&lt;int&gt;&gt;(); foreach(int n in A) { if(D.ContainsKey(n-1) &amp;&amp; D.ContainsKey(n+1)) { D[n-1].UnionWith(D[n+1]); D[n+1] = D[n] = D[n-1]; } else if(D.ContainsKey(n-1)) { D[n] = D[n-1]; } else if(D.ContainsKey(n+1)) { D[n] = D[n+1]; } else if(!D.ContainsKey(n)) { D.Add(n, new HashSet&lt;int&gt;()); } D[n].Add(n); } int result = int.MinValue; foreach(HashSet&lt;int&gt; H in D.Values) { if(H.Count &gt; result) { result = H.Count; } } Console.WriteLine(result); </code></pre>
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    1. POLongest subarray whose elements form a continuous sequence
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    1. USDmytro Shevchenko
    1. USDmytro Shevchenko
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    1. COThe solution of the array `[10, 5, 3, 1, 4, 2, 8, 7, 0]` should be `5`, the maximum subarray is `[5, 3, 1, 4, 2]`. The additional `0` in the end does not change this! However, your code returns the result `6`, as it assumes the element `0` to be part of the (non-contiguous!) maximal subarray.
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    2. CO@blubb OP never stated the subarray must be contiguous. Only the elements when sorted must be continuous. But now I see the problem statement is ambiguous. The OP may have implied contiguosity.
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    3. COIt is implied in the question: the solution to `[4, 5, 1, 5, 7, 6, 8, 4, 1]` would be `7` instead of `5` if non-contiguity were allowed. Additionally, the number of possible subsets would be `2^n` instead of `n^2`.
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