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

Circular Array Rotation

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3087 Discussions

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  • dineshgangadhar2
     + 0 comments

    Python Solution

    def circularArrayRotation(a, k, queries):
    for i in range(k):
        a.insert(0, a.pop())
    return (a[i] for i in queries)

  • 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;
}

  • munikarthik222
     + 0 comments

    public static List circularArrayRotation(List a, int k, List queries) {

        for(int i=0 ; i<k ; i++){
        rotateArray(a);
    }
    List<Integer> res = new ArrayList<>();
    for(int q : queries){
        res.add(a.get(q));
    }
    return res;
}

public static void rotateArray(List<Integer> a){

    a.add(0,a.get(a.size()-1));
    a.remove(a.size()-1);   

}

  • richakesharvani2
     + 0 comments

    Python 3

    def circularArrayRotation(a, k, queries):

    n = len(a)
b = [0]*n
for i in range(n):
    b[(i+k)%n] = a[i]
k = []
for i in queries:
    k.append(b[i])
return k

  • somaditya
     + 0 comments

    Java solution using modular arithmetic:

        public static List<Integer> circularArrayRotation(List<Integer> a, int k, List<Integer> queries) {
        int n = a.size();
        int[] rotated = new int[n];
        
        for (int i = 0; i < n; i++) {
            rotated[(i+k) % n] = a.get(i);
        }

        List<Integer> result = new ArrayList<>();
        
        for (int q : queries) {
            result.add(rotated[q]);
        }
        
        return result;
    }