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

Circular Array Rotation

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3096 Discussions

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  • joosx2000
     + 0 comments

    Java solution 100% time O(n):

    public static List<Integer> circularArrayRotation(List<Integer> a, int k, List<Integer> queries) {
    int n = a.size(), shift = k % n;
    return queries.stream()
                .map(query -> a.get((query - shift + n) % n))
                .collect(Collectors.toList());
}

  • 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;
}

  • gschandhru10
     + 0 comments
    function circularArrayRotation(a, k, queries) {
    // Normalize k to avoid unnecessary rotations if k is larger than the array length
    k = k % a.length;

    // Rotate the array
    let slicedElements = a.slice(a.length - k);  // Get the last k elements
    let remainingElements = a.slice(0, a.length - k);  // Get the first part of the array
    let rotatedArray = slicedElements.concat(remainingElements);  // Concatenate to get the rotated array

    // Initialize an array to store the results of the queries
    let results = [];

    // For each query, push the corresponding element from the rotated array
    for (let i = 0; i < queries.length; i++) {
        results.push(rotatedArray[queries[i]]);
    }

    return results;
}

  • sayeedahmad06061
     + 0 comments
    k=k%len(a)
A=a[-k:]+a[:-k]
ans=[A[i] for i in queries]
return ans

  • akshayshinde1729
     + 0 comments
    def circularArrayRotation(a, r, q):
    x = r%len(a)
    z = a[-x:]+a[:-x]
    return [z[i] for i in q]