**C++ : **

```
vector<int> permutationEquation(vector<int> p) {
vector<int> permuted;
int indexAtI = -1, x = 1, indexAtJ = -1;
while(x <= p.size()){
for(int i = 0; i < p.size(); i++){
if(x == p[i]){
indexAtI = i + 1;
cout << indexAtI << endl;
for(int j = 0; j < p.size(); j++){
if(p[j] == indexAtI){
indexAtJ = j + 1;
break;
}
}
}
}
x++;
permuted.push_back(indexAtJ);
}
return permuted;
}
```

*Explanation:*

**Note: This is a terrible, terrible solution. Look at the number of nested loops! Yes, it works, but loopsss!**

*The question was not easy to understand, so the breakdown is this. Given a vector, find the index of index of a number. For example, given a vector p of 3 digits, [2, 3, 1], assuming p[1] is 2, p[2] is 3, p[3] is 1, get the index of 2, 3 and 1 respectively, which is p[p[2]] = 3, p[p[3]] = 1, and p[p[1]] = 2.*

*The outermost while loop goes through the inner loops for p.size() times, and x is set to x++ after each round. The middle for loop (looping with i) finds the first index of each x, starting from x = 1, and stores the found number in indexAtI before going into the next for loop (looping with j), to find index j of number indexAtI. This number is then stored in indexAtJ, and after breaking out of the for loop, push it into permuted vector.*

*A better solution without this many loops would be to find the p((i + 1) + 1). Probably would attempt to do this in another language.*