Added count inversions in cpp

Array/Count Inversions/Count Inversions - Question.txt

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+Count Inversions:
+Given an array of integers. Find the Inversion Count in the array. 
+
+Inversion Count: For an array, inversion count indicates how far (or close) the array is from being sorted. If array is already sorted then the inversion count is 0. 
+If an array is sorted in the reverse order then the inversion count is the maximum. 
+Formally, two elements a[i] and a[j] form an inversion if a[i] > a[j] and i < j.
+
+
+Example 1:
+
+Input: N = 5, arr[] = {2, 4, 1, 3, 5}
+Output: 3
+Explanation: The sequence 2, 4, 1, 3, 5 
+has three inversions (2, 1), (4, 1), (4, 3).

Array/Count Inversions/count-inversions.cpp

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+#include <bits/stdc++.h>
+using namespace std;
+
+
+ // } Driver Code Ends
+class Solution{
+  public:
+    // arr[]: Input Array
+    // N : Size of the Array arr[]
+    // Function to count inversions in the array.
+    long long merge(long long arr[], long long temp[], long long left, long long mid, long long right){
+        long long i, j, k;
+        long long inv_count = 0; 
+
+        i = left; 
+        j = mid;
+        k = left;
+
+        while(( i <= mid - 1)&&( j <= right )){
+            if(arr[i] <= arr[j]){
+                temp[k++] = arr[i++];
+            }
+            else{ 
+                temp[k++] = arr[j++];
+
+                inv_count = inv_count + (mid - i);
+            }
+        }
+
+        while( i <= mid - 1)
+            temp[k++] = arr[i++];
+        while( j <= right )
+            temp[k++] = arr[j++];
+
+        for(i = left; i <= right; i++ ){
+            arr[i] = temp[i];
+        }    
+
+        return inv_count;
+    }
+
+    long long mergesort( long long arr[], long long temp[], long long left, long long right){
+        long long mid, inv_count = 0;
+
+        if(right > left){
+
+            mid = (left + right)/2;
+
+            inv_count += mergesort(arr, temp, left, mid);
+            inv_count += mergesort(arr, temp, mid + 1, right);
+
+            inv_count += merge(arr, temp, left, mid + 1, right);
+        }
+
+        return inv_count;
+    }
+
+    long long inversionCount(long long arr[], long long N)
+    {
+        long long temp[N];
+        long long ct = mergesort(arr, temp, 0, N - 1);
+        return ct;
+    }
+
+};
+
+// { Driver Code Starts.
+
+int main() {
+
+    long long T;
+    cin >> T;
+
+    while(T--){
+        long long N;
+        cin >> N;
+
+        long long A[N];
+        for(long long i = 0;i<N;i++){
+            cin >> A[i];
+        }
+        Solution obj;
+        cout << obj.inversionCount(A,N) << endl;
+    }
+
+    return 0;
+}
+  // } Driver Code Ends