Merge pull request #575 from yassin64b/cpp_count_inversions

 Add function counting the number of inversions based on merge sort

Author: matthewsamuel95

Sorting/MergeSort/cpp/CountInversions.cpp

@@ -0,0 +1,43 @@
+#include <iostream>
+#include <vector>
+
+using namespace std;
+
+long long count_inversions(int *a, int n) {
+    if (n <= 1) {
+        return 0;
+    }
+    int sz1 = n / 2, sz2 = n - sz1;
+
+    // get count of inversions in left and right half
+    long long res = count_inversions(a, sz1) + count_inversions(a + sz1, sz2);
+
+    // merge both halves (now copied into t1, resp. t2) into a
+    vector<int> t1(a, a + sz1), t2(a + sz1, a + n);
+    int k = 0, i = 0, j = 0;
+    while (i < sz1 && j < sz2) {
+        if (t1[i] <= t2[j]) {
+            a[k++] = t1[i++];
+
+            // all elements in t2[0..j-1] (which are j elements) are smaller than t1[i] 
+            // and are to the right of t1[i] in the original array --> form an inversion
+            res += j; 
+        } else {
+            a[k++] = t2[j++];
+        }
+    }
+    while (i < sz1) {
+        a[k++] = t1[i++];
+        res += j; // j == sz2 holds here
+    }
+    while (j < sz2) {
+        a[k++] = t2[j++];
+    }
+    return res;
+}
+
+int main() {
+	vector<int> v{10, 10, 20, 30, 12, 8, 9, 14, 13, 11};
+	cout << "Given array has " << count_inversions(&v[0], v.size()) << " inversions.\n";
+	return 0;
+}