count inversion.cpp

Count_Inversion/count inversion.cpp

@@ -0,0 +1,106 @@
+#include<iostream>
+using namespace std;
+
+
+int mergefn(int arr[],int temp[],int left,int mid,int right)
+{
+// temp[] is the auxillary array to store the sorted elements
+
+
+
+    int i=left;
+    int j=mid;
+    int k=left;
+    int inv_cnt=0;
+
+
+
+    while((i<=mid-1 ) && (j<=right))
+    {
+
+        if(arr[i]<=arr[j])
+            temp[k++]=arr[i++];
+        else
+        {
+
+            temp[k++]=arr[j++];
+            inv_cnt+=(mid-i);
+
+            /* 
+             if arr[i] is greater than arr[j], then there are (mid – i) inversions. because left and right subarrays are sorted,
+             so all the remaining elements in left-subarray (arr[i+1], arr[i+2] … arr[mid]) will be greater than arr[j]*/
+        }
+
+
+
+    }
+
+
+    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_cnt;
+
+
+}
+
+
+int mergesort(int arr[],int temp[],int left,int right)
+{
+
+ int inv_cnt=0;
+
+    if(right>left)
+    {
+
+        int mid=(left+right)/2;
+
+
+        inv_cnt+=mergesort(arr,temp,left,mid);// counting inversions in left subarray
+        inv_cnt+=mergesort(arr,temp,mid+1,right);//counting inversions in the right subarray 
+        inv_cnt+=mergefn(arr,temp,left,mid+1,right);// count overall inversions while merging the left subarray and right subarray
+
+    }
+    return inv_cnt;
+}
+
+
+
+
+int main()
+{
+int t;
+cin>>t;
+while(t--){
+    int n;
+    cin>>n;
+    int arr[n];
+    for(int i=0;i<n;i++)
+        cin>>arr[i];
+
+    int temp[n];
+    cout<<mergesort(arr,temp,0,n-1)<<endl;
+    }
+    return 0;
+
+}
+
+
+/* 
+Sample Input
+2
+4
+1 2 4 3
+3
+3 2 1
+Sample Output
+1
+3
+*/