Table of Contents

## Difficulty

Moderate

## Prerequisites

- Arrays
- Lists

## Problem

You are given a positive integer N and an array A of size N. There are N lists L1,L2…LN Initially, Li=[Ai]

You can perform the following operation any number of times as long as there are at least 2 lists:

- Select 2 (non-empty) lists Li and Lj(i≠j)
- Append Lj to Li and remove the list Lj. Note that this means Lj cannot be chosen in any future operation.

Find the minimum number of operations required to obtain a set of lists that satisfies the following conditions:

- The first element and last element of each list are equal.
- The first element of all the lists is the same.

Print −1 if it is not possible to achieve this via any sequence of operations.

### Input Format

- The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains an integer N.
- The second line of each test case contains N space-separated integers A1,A2,…,AN.

### Output Format

For each test case, print a single line containing one integer: the minimum number of operations required to obtain an array of lists that satisfies the given conditions.

Print −1 if it is impossible to achieve such an array of lists.

### Constraints

- 1≤T≤105
- 1≤N≤2⋅105
- 1≤Ai≤N
- Sum of N over all test cases doesn’t exceed 2.10
^{5}

## Input

```
3
1
1
2
1 2
3
1 1 2
```

## Output

```
0
-1
2
```

## Explanation

**Test case 1:** There is only one list [1], and it trivially satisfies the condition so no operations are required.

**Test case 2:** There are only 2 ways to do an operation – either take list [1] and append it to list [2] or take list [2] and append it to list [1]. In both cases, it is not possible to satisfy both given conditions at the same time. Hence, the answer is −1.

**Test case 3:** Here is one possible order of operations:

- Select the 3rd list [2] and append it to the 1st list [1].
- Then, select the 2nd list [1] and append it to the 1st list [1,2].

Finally, we are left with the single list [1,2,1] which satisfies the given conditions. It can be verified that it is impossible to do this using less than 2 operations.

## Solution in C++

```
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin>>t;
while(t--)
{
int n;
cin>>n;
vector<int> v(n);
set<int> s;
map<int,int> mp;
for(int i = 0;i<n;i++)
{
cin>>v[i];
s.insert(v[i]);
mp[v[i]]++;
}
if(s.size() == 1)
cout<<"0";
else if(s.size() == n)
cout<<"-1";
else{
int max = -1;
for(auto itr = mp.begin();itr!=mp.end();itr++)
{
if( itr->second > max)
max = itr->second;
}
cout<<n-max+1;
}
cout<<endl;
}
return 0;
}
```

In the above code, we have initially decremented the value of t and used a while loop. We then use a for loop to append the list. Then if s.size ==1 then it prints 0 else it prints -1.

## Conclusion

In this post, we have discussed list of lists editorial and solved it using C++.