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Namespaces</h2></td></tr>
<tr class="memitem:"><td class="memItemLeft" align="right" valign="top">namespace &#160;</td><td class="memItemRight" valign="bottom"><a class="el" href="../../d5/d91/namespacesorting.html">sorting</a></td></tr>
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<tr class="memdesc:d9/d13/namespaceinversion"><td class="mdescLeft">&#160;</td><td class="mdescRight">Functions for counting inversions using Merge Sort algorithm. <br /></td></tr>
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<tr class="heading"><td colspan="2"><h2 class="groupheader"><a id="func-members" name="func-members"></a>
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<tr class="memdesc:ae97a486e14101c4822ea8dc47f0d1661"><td class="mdescLeft">&#160;</td><td class="mdescRight">Implement merge Sort and count inverions while merging. <br /></td></tr>
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<tr class="memitem:a3332498eabf6579ef059c0d0e9f4ec80"><td class="memTemplItemLeft" align="right" valign="top">uint32_t&#160;</td><td class="memTemplItemRight" valign="bottom"><a class="el" href="#a3332498eabf6579ef059c0d0e9f4ec80">sorting::inversion::countInversion</a> (T *arr, const uint32_t size)</td></tr>
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<tr class="memdesc:a3332498eabf6579ef059c0d0e9f4ec80"><td class="mdescLeft">&#160;</td><td class="mdescRight">Function <a class="el" href="#a3332498eabf6579ef059c0d0e9f4ec80" title="Function countInversion() returns the number of inversion present in the input array....">countInversion()</a> returns the number of inversion present in the input array. Inversions are an estimate of how close or far off the array is to being sorted. <br /></td></tr>
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<tr class="memdesc:a851ca6a0391d14fb49a97d55e4377497"><td class="mdescLeft">&#160;</td><td class="mdescRight">UTILITY function to print array. <br /></td></tr>
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<tr class="memdesc:aa8dca7b867074164d5f45b0f3851269d"><td class="mdescLeft">&#160;</td><td class="mdescRight">Test implementations. <br /></td></tr>
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<tr class="memdesc:ae66f6b31b5ad750f1fe042a706a4e3d4"><td class="mdescLeft">&#160;</td><td class="mdescRight">Program Body contains all main funtionality. <br /></td></tr>
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<a name="details" id="details"></a><h2 class="groupheader">Detailed Description</h2>
<a name="details" id="details"></a><h2 id="header-details" class="groupheader">Detailed Description</h2>
<div class="textblock"><p>Counting Inversions using <a href="https://en.wikipedia.org/wiki/Merge_sort" target="_blank">Merge Sort</a> </p>
<p>Program to count the number of inversions in an array using merge-sort.</p>
<p>The count of inversions help to determine how close the array is to be sorted in ASCENDING order.</p>
<p>two elements a[i] and a[j] form an inversion if <code>a[i]</code> &gt; <code>a[j]</code> and i &lt; j</p>
<p>Time Complexity --&gt; <code>O(n.log n)</code></p>
<p>Space Complexity --&gt; <code>O(n)</code> ; additional array <code>temp[1..n]</code> </p>
<h3><a class="anchor" id="autotoc_md115"></a>
<p>two elements a[i] and a[j] form an inversion if <span class="tt">a[i]</span> &gt; <span class="tt">a[j]</span> and i &lt; j</p>
<p>Time Complexity --&gt; <span class="tt">O(n.log n)</span></p>
<p>Space Complexity --&gt; <span class="tt">O(n)</span> ; additional array <span class="tt">temp[1..n]</span> </p>
<h3 class="doxsection"><a class="anchor" id="autotoc_md115"></a>
Algorithm</h3>
<ol type="1">
<li>The idea is similar to merge sort, divide the array into two equal or almost equal halves in each step until the base case is reached.</li>
<li>Create a function merge that counts the number of inversions when two halves of the array are merged, create two indices i and j, i is the index for first half and j is an index of the second half. if <code>a[i]</code> is greater than <code>a[j]</code>, then there are (mid – i) inversions, Because left and right subarrays are sorted, so all the remaining elements in left-subarray (a[i+1], a[i+2] … a[mid]) will be greater than a[j].</li>
<li>Create a function merge that counts the number of inversions when two halves of the array are merged, create two indices i and j, i is the index for first half and j is an index of the second half. if <span class="tt">a[i]</span> is greater than <span class="tt">a[j]</span>, then there are (mid – i) inversions, Because left and right subarrays are sorted, so all the remaining elements in left-subarray (a[i+1], a[i+2] … a[mid]) will be greater than a[j].</li>
<li>Create a recursive function to divide the array into halves and find the answer by summing the number of inversions is the first half, number of inversion in the second half and the number of inversions by merging the two.</li>
<li>The base case of recursion is when there is only one element in the given half.</li>
<li>Print the answer</li>
@@ -195,7 +176,7 @@ Algorithm</h3>
<dl class="section author"><dt>Author</dt><dd><a href="https://github.com/rakshitraj" target="_blank">Rakshit Raj</a> </dd></dl>
<p class="definition">Definition in file <a class="el" href="../../d2/d26/count__inversions_8cpp_source.html">count_inversions.cpp</a>.</p>
</div><h2 class="groupheader">Function Documentation</h2>
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<h2 class="memtitle"><span class="permalink"><a href="#a3332498eabf6579ef059c0d0e9f4ec80">&#9670;&#160;</a></span>countInversion()</h2>
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