Arrays

To count no. of inversions possible in an array

CountInversion.cpp

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+#include<bits/stdc++.h>
+using namespace std;
+
+//Algorithm used is Merge Sort
+//Time Complexity is O(n log n)
+
+long long int merge(int a[], int low, int mid, int high);
+
+int mergeSort(int a[], int low, int high)
+{
+    long long int count=0;
+    if(low<high)
+    {
+        int mid=(low+high)/2;
+        count+= mergeSort(a,low,mid); //counts inversions in left half
+        count+= mergeSort(a,mid+1,high); //counts inversions in right half
+
+        count+= merge(a,low,mid,high); //counts inversions when two halves are merged
+    }
+    return count;
+}
+
+long long int merge(int arr[], int low, int mid, int high)
+{
+    long long int count=0;
+    int n1=mid-low+1; //size of left half
+    int n2=high-mid; //size of right half (high-(mid+1)+1)==high-mid
+    int a[n1],b[n2];
+
+    for(int i=0;i<n1;i++)
+    a[i]=arr[low+i];//storing sorted left half in a temporary array
+
+    for(int i=0;i<n2;i++)
+    b[i]=arr[mid+i+1];//storing sorted right half in a temporary array
+
+    int i=0,j=0,k=low;
+    while((i<n1)&&(j<n2))
+    {
+        if(a[i]<=b[j]) //No inversion occurs
+        arr[k++]=a[i++];
+        else {
+        arr[k++]=b[j++]; //Inversion occurs! 
+        count+=(n1-i); //if a[i] is greater, a[i+1], a[i+2] etc will also be greater. So all inversions are counted here
+        }
+    }
+    while(i<n1)
+    arr[k++]=a[i++];
+    while(j<n2)
+    arr[k++]=b[j++];
+
+    return count;
+}
+
+int main()
+{   
+    int n;
+    cin>>n;
+    int a[n];
+    for(int i=0;i<n;i++)
+    cin>>a[i];
+    long long int ans=mergeSort(a,0,n-1);
+    cout<<ans<<endl;
+    return 0;
+}