Java 100% O(N):

public static List<Integer> permutationEquation(List<Integer> p) {
    List<Integer> y = new ArrayList<>();
    Map<Integer, Integer> idxToValue = new HashMap<>();
    for (int i=1 ; i<=p.size() ; i++) {
        idxToValue.put(p.get(i-1), i);
    }

    for (int i=1 ; i<=p.size() ; i++) {
        y.add(idxToValue.get(idxToValue.get(i)));
    }

    return y;
}

Here is my O(N) c++ solution, you can watch the explanation here : https://youtu.be/D9nfVOmmv7Q

vector<int> permutationEquation(vector<int> p) {
    vector<int> indexes(p.size() + 1), result;
    for(int i = 1; i <= p.size(); i++) indexes[p[i-1]] = i;
    for(int i = 1; i <= p.size(); i++) result.push_back(indexes[indexes[i]]);
    return result;
}

My python3 solution

l = len(p)
r = []
for x in range(0,l):
    r.append( p.index ( (p.index(x+1))+1 ) + 1)
 return(r)

Kotlin:

fun permutationEquation(p: Array<Int>): Array<Int> {
    val size=p.size
    val indexArray=IntArray(size) { 0 }
    val result=Array(size) { 0 }
    for(i in 0 until size){
        indexArray[i]= p.indexOf(i+1)+1
    }
    for(i in 0 until size){
        result[i]=p.indexOf(indexArray[i])+1
    }
    return result
}

The Sequence Equation can be a bit tricky at first, but once you understand the pattern, it’s pretty satisfying to solve! It’s all about mapping numbers in a sequence and finding their positions. By the way, if you're looking for a stylish yet comfy choice, Women's Black Sandals are always a great go-to for any outfit! Perfect for casual days or dressing up a bit.