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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Sequence Equation
  5. Discussions

Sequence Equation

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  • wifivop476
     + 0 comments

    The Sequence Equation can be tricky, but once you break it down step by step, it starts making sense! It's kind of like how Concrete Estimating Services work — you need accuracy and attention to detail. If you're looking for an easier way, using capcut for pc tools can really simplify the process and save time.

  • alban_tyrex
     + 0 comments

    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;
}

  • willy756_taiwan
     + 0 comments

    Python Method : Create dict to store the element as key and the index+1 of p as value. For example, [5,2,1,3,4] we can get {5:1, 2:2, 1:3, 3:4, 4:5} and use p_dict[p_dict[1]] to get the answer 4 . It can reduce the computing of list.index().

    def permutationEquation(p):
    p_dict = {j:i+1 for i,j in enumerate(p)}
    y = []
    for x in range(1,len(p)+1):
        y.append(p_dict[p_dict[x]])
    return y

  • jared_ghobrial
     + 0 comments
    def permutationEquation(p):
    # Write your code here
    res = []
    for i in range(len(p)):
        x = p.index(i + 1)
        y = p.index(x + 1)
        res.append(y + 1)
    return res

  • kunalkawate424
     + 0 comments

    Here is the solution of this question in PYTHON

    def permutationEquation(p):
    position1 = []
    position2 = []
    for i in range(1,len(p)+1):
        for j in range(len(p)):
            if(i == p[j]):
                position1.append(j+1)
                break
    for item in position1:
        for i in range(len(position1)):
            if(item == p[i]):
                position2.append(i+1)
    return position2