The answer is number of inversions if for every **i** elements , there are <=2 elements smaller than it on the right.

If we take the advantage of “at most 2 elements” smaller, we can solve it linearly easily. Notice that the max answer can’t be larger than 2xn.

Each person will be bribed by the person who is behind him and has smaller number. We can easily simulate this by keeping count of how many times someone has bribed. If someone has bribed more than times, terminate the algorithm. See setter’s code for this approach.

C++ code -

#include <bits/stdc++.h>

using namespace std;

const int maxN=2e5+5;

int n,a[maxN],ans,T,k;

int bit[maxN];

int invalid;

void del() {

for (k=0;k<maxN;k++) {

bit[k]=0;

}

ans = 0;

invalid = 0;

}

void upd(int x) {

for (k = x; k < maxN; k += (k & (-k))) {

bit[k]++;

}

}

int get_sum(int x) {

int p = 0;

for (k = x; k > 0; k -= (k & (-k)))

p += bit[k];

return p;

}

int main() {

scanf("%d", &T);

while (T–) {

del();

scanf("%d", &n);

```
for (int i = 0; i < n; i++) {
scanf("%d", &a[i]);
}
for (int i = n - 1; i >= 0; i--) {
if (get_sum(a[i]) > 2)
invalid++;
ans += get_sum(a[i]);
upd(a[i]);
}
if (invalid > 0) {
printf("Not Possible\n");
} else {
printf("%d\n", ans);
}
```

}

return 0;

}