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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Circular Array Rotation
  5. Discussions

Circular Array Rotation

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3077 Discussions

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  • alban_tyrex
     + 0 comments

    Here is my c++ solution, you can watch the explanation here : https://youtu.be/MaT-4dnozJI

    Solution 1 O(n + m) : 

    vector<int> circularArrayRotation(vector<int> a, int k, vector<int> queries) {
    vector<int> newArray(a.size()), result;
    for(int i = 0; i < a.size(); i++){
        newArray[(i + k) % a.size()] = a[i];
    }
    for(int i = 0; i < queries.size(); i++) result.push_back(newArray[queries[i]]);
    return result;
}

    Solution 2 O(m) :

    vector<int> circularArrayRotation(vector<int> a, int k, vector<int> queries) {
    vector<int> result;
    for(int i = 0; i < queries.size(); i++) {
        int previousIndex = (queries[i] - k + a.size() * 100000 ) % a.size();
        result.push_back(a[previousIndex]);
    }
    return result;
}

  • da_estrella
     + 0 comments
    def circularArrayRotation(a, k, queries):
    return list(map(lambda x: a[x - (k % len(a))], queries))

  • dhruvbhagat98
     + 0 comments

    JS/Javascript solutiuon:-

    function circularArrayRotation(a, k, queries) {
  const actualRotation = k % a.length;
  let newArr = a;
  if (actualRotation) {
    const rightPart = a.splice(a.length - actualRotation);
    newArr = [...rightPart, ...a];
  }
  return queries.map((query) => newArr[query]);
}

  • voronin_ilia
     + 0 comments

    C++ Solution

    vector circularArrayRotation(vector a, int k, vector queries) { vector res; for(auto& q : queries) { int index = (q + a.size() - k%a.size()); if(index >= a.size()) index = index % a.size(); res.push_back(a[index]); } return res; }

  • Tusenka
     + 0 comments

    My solution

    def _rotate(a, k): return a[-k:]+a[0:k]

    def circularArrayRotation(a, k, queries): # Write your code here res=_rotate(a, k) return [res[i] for i in queries]