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    <p>Best case - finding the max element as the first (<code>O(1)</code>), worst case - it is the last element checked (<code>O(n)</code>).</p> <p>The tricky part is the average case. <br>To find the average case - we need the <a href="http://en.wikipedia.org/wiki/Expected_value" rel="noreferrer">expected</a> number of iterations!</p> <p>Since you can stop after you find the maximum, we can split the problem into two parts:</p> <ol> <li><code>[0,n-1)</code>: Since on average (assuming uniform independent distribution for each element) - the number <code>n</code> has probability 1/n to be in each place, then the expected number of iterations for this part is <code>1/n + 2*((n-1)/n)/n + 3 * ((n-1)/n)^2/n + ... + (n-1) * ((n-1)/n)^(n-2)/n</code> <sup>1</sup> <br><a href="http://www.wolframalpha.com/input/?i=sum%20j*%28%28n-1%29/n%29%5E%28j-1%29%20*%20%281/n%29%20j%20from%201%20to%20n-1%20" rel="noreferrer">The above formula yields an ugly formula which is <code>O(n)</code></a></li> <li>The last element is going to be checked if the first n-1 elements did not contain the value <code>n</code>: so you need to add to the above <code>n* ((n-1)/n)^(n-1)</code>, which is <code>O(n)</code> as well (lim to infinity is <code>1/e * n</code>).</li> </ol> <p><strong>This totals in <code>O(n)</code> average time solution.</strong></p> <hr> <p>(1): The formula for each element is <code>j*((n-1)/n)^(j-1) * (1/n)</code> because:</p> <ul> <li><code>j</code> - for the number of elements checked (number of iterations)</li> <li><code>((n-1)/n)^(j-1)</code> - Probability that there is no <code>n</code> in the previous elements</li> <li><code>(1/n)</code> - Probability this number is <code>n</code>.</li> </ul>
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    1. POSimple "maximum value in array" and complexity calculations
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    1. COWow @amit I have no words. So nicely put. Really appreciating this.
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    2. COThe "with duplicates" bit says that it is not a random permutation. Merely n random integers in the range `1..n`.
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    3. CO@btilly: The term permutation was wrong, but note that the caclulation did not assume permutation - it assumed uniform distribution for each element - so it (each element) has `1/n` chance to be the max, and `(n-1)/n` not to be it. I'll correct the terminology. Thank you for your comment.
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