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    <p><strong>Edit</strong>: just figured out the <code>O(n)</code> solution for the unsorted case :) see below!</p> <h1>Sorted:</h1> <p>This can be solved in <code>O(log N</code>) for a sorted array, by means of a binary search on <code>n</code>. I'll be using OP's notation here, where <code>N = # of elements</code> and <code>n</code> is the answer we are looking for.</p> <p>If the array is sorted, it basically means we need to find a position <code>[N - n]</code> so that such position in the array contains a value greater than <code>n</code> - if it does, then there are at least <code>n</code> values greater than it, regardless of repeated values.</p> <p>Note an answer is always possible, as in the worst case the answer would be <code>0</code>, and there are always <em>at least</em> 0 elements greater than it. The answer always gets "easier" for lower values, obviously, as it is easier to find 1 element greater than 1, than 10 elements greater than 10. But more importantly, this function follows a monotonic (non-decreasing) behavior which allows us to use a binary search on it.</p> <p>The idea is as follows:</p> <pre><code>int N = 9; int arr[10] = {0,2,5,7,8,9,19,23,45}; int lo = 0, hi = N+1, mid; while(hi-lo &gt; 1){ mid = (hi+lo)/2; if(arr[N-mid] &gt; mid) lo = mid; else hi = mid; } n = lo; //highest value that worked </code></pre> <p><em>Breakdown:</em> The array has size <code>9</code>. A binary search may begin trying value <code>n = 5</code>, so we just check whether the 5th element from the end of the array is greater than 5. In this case, <code>8 &gt; 5</code> so we can try a better answer. The search would then attempt <code>7</code>, but the element at position <code>[N-7]</code> is <code>5</code>, which is lower than 7 and does not satisfy our constraints. Thus the search's last attempt is the value <code>6</code>, which returns true as <code>7 &gt; 6</code>.</p> <h1>Unsorted:</h1> <p>For the unsorted case, the idea is incredibly very similar! We can solve it in <code>O(n)</code> by using a <a href="http://en.wikipedia.org/wiki/Selection_algorithm" rel="noreferrer">Selection Algorithm</a> to identify the [N-n]th element, and at each step divide the search space in the same manner as the binary search.</p> <p>We begin by searching from <code>[0]</code> to <code>[N-1]</code> to find the median <em><code>(N/2 th)</code></em> element, and we can rearrange the array in another <code>O(N)</code> step such that the median element is placed in its correct position, and every element before it has a value <code>&lt;= median</code>, while every element after it has a value <code>&gt;=median</code>.</p> <p>Now, if that value is <em>greater</em> than <code>n</code> (in this case <code>N/2</code>), we showed above there are at least <code>n</code> elements greater than <code>n</code>, and <strong>thus we only need to search further in the lower half of the array</strong>. (If the median value is lower than <code>n</code>, we instead consider only the greater half of the array)</p> <p>Now, assuming <code>median &gt;= N/2</code> we will repeat the same process from index <code>[0]</code> to <code>[N/2]</code>, using a selection "sort" in <code>O(N/2)</code>, and so on, each time dividing the search space by 2. </p> <p>C++ code is as follows:</p> <pre><code>int N = 9; int arr[9] = {0,2,7,8,19,5,45,9,23}; int lo = 0, hi = N, mid; while(hi-lo &gt; 1){ mid = (hi+lo)/2; std::nth_element(arr+lo, arr+mid, arr+hi); if(arr[mid] &gt; N-mid) hi = mid; else lo = mid; } n = N-hi; </code></pre> <p>In the end, we achieve a complexity of <code>O(N) + O(N/2) + O(N/4) + ... = O(2*N) = O(N)</code></p>
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    1. COgood answer.. :D +1
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    2. COConsider bucket sort or radix sort as a simple hack to get ~linear complexity on unsorted arrays.
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