# HackerRank Insertion Sort Advanced Analysis problem solution

In this HackerRank Insertion Sort Advanced Analysis problem we have given an array and we need to calculate the number of shifts an insertion sort performs when sorting an array.

## Problem solution in Python programming.

```from array import array
from bisect import bisect_right

t = int(input())
for _ in range(t):
n = int(input())
arr = array('I', [int(i) for i in input().split()])
sarr = array('I', [arr[0]])
result = 0
for i in range(1, n):
e = arr[i]
j = bisect_right(sarr, e)
sarr.insert(j, e)
result += i - j
print(result)            ```

## Problem solution in Java programming.

```import java.io.*;
import java.util.*;

public class Solution {
private static final int MAXVAL = 10000000;
private static int[] array = new int[MAXVAL+1];

public static void main(String[] args) {
/* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
Scanner sc = new Scanner(System.in);
int testCaseCount = sc.nextInt();
for (int i = 0; i < testCaseCount; i++) {
int size = sc.nextInt();
long sum = 0;
Arrays.fill(array, 0);
for (int j = 0; j < size; j++) {
sum += assign(sc.nextInt(), array, j);
}
System.out.println(sum);
}
}
private static int assign(int x, int[] prefixSums, int current) {
update(prefixSums, x);
return current-n;
}

private static int read(int[] prefixSums, int x) {
int nrt=0;
while(x>0) {
nrt += prefixSums[x];
x -= (x&(-x));
}
return nrt;
}

private static void update(int[] prefixSums, int x) {
while(x <= MAXVAL) {
prefixSums[x]++;
x += (x&(-x));
}
}
}```

## Problem solution in C++ programming.

```#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std ;
#define MAXN 100002
#define MAX 1000002
int n,a[MAXN],c[MAX] ;
int main()
{
int runs ;
scanf("%d",&runs) ;
while(runs--)
{
scanf("%d",&n) ;
for(int i = 0;i < n;i++) scanf("%d",&a[i]) ;
long long ret = 1LL * n * (n - 1) / 2 ;
memset(c,0,sizeof c) ;
for(int i = 0;i < n;i++)
{
for(int j = a[i];j > 0;j -= j & -j) ret -= c[j] ;
for(int j = a[i];j < MAX;j += j & -j) c[j]++ ;
}
cout << ret << endl ;
}
return 0 ;
}```

## Problem solution in C programming.

```#include <stdio.h>
#include <string.h>
long long inv(int *A, int N)
{
if(N <= 1) return 0;
int m = N/2;
long long ans = inv(A, m) + inv(A + m, N - m);
static int tmp[100000];
int left = 0, right = m, i;
for(i = 0; i < N; i ++)
{
int min = 1000001;
if(m - left > 0 && min > A[left])
min = A[left++];
if(N - right > 0 && min > A[right])
{
if(min != 1000001) left --;
min = A[right++];
ans += m - left;
}
tmp[i] = min;
}
memcpy(A, tmp, sizeof(int)*N);
return ans;
}
void solve()
{
int N;
static int A[1000005];
scanf("%d",&N);
int i;
for(i = 0; i < N; i ++) scanf("%d",A + i);
printf("%lld\n",inv(A,N));
}
int main()
{
int K;
scanf("%d",&K);
while(K--) solve();

return 0;
}```