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

Sequence Equation

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1548 Discussions

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  • wifivop476
     + 0 comments

    I totally agree! When I had just read the problem, I thought that I probably could get it in one line of code. Focusing on reducing equipment failure as a mindset, I broke the entire problem into just a few steps and was able to get a solution that only took one line of code!

  • akulbachandra67
     + 0 comments

    **Mine Python Solution **

    def permutationEquation(p):
    # Write your code here
    a = [p.index(i)+1 for i in range(min(p),max(p)+1)]
    b = [p.index(i)+1 for i in a ]
return b

  • troytjh
     + 0 comments

    C++ solution

    vector<int> permutationEquation(vector<int> p)
{
    vector<int> o(p.size());
    int j,k=0;
    for (int i = 1; i <= p.size(); ++i)
    {
        while (p[j] != i && j < p.size()) {++j;};
        while (p[k] != j+1 && k < p.size()) {++k;};
        
        o.at(i-1) = k+1;
        
        j = 0;
        k = 0;
    }
    
    return o;
}

  • afzal_saiyed931
     + 0 comments
    def permutationEquation(p):
    # Write your code here
    temp = {value: index + 1 for index, value in enumerate(p)}
    result = []
    for x in range(1, len(p)+1):
        i = temp.get(x)
        j = temp.get(i)
        result.append(j)
        
    return result

  • spriyamalar
     + 0 comments
    public static List<Integer> permutationEquation(List<Integer> p) {
    Map<Integer, Integer> map = IntStream.
            range(0, p.size()).
            boxed().
            collect(Collectors.toMap(s-> p.get(s), s-> s+1));
    List<Integer> result = IntStream.range(1, p.size()+1).
        boxed().
        map(map::get).
        map(map::get).
        collect(Collectors.toList());
    return result;
}