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

Circular Array Rotation

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3109 Discussions

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  • madhusudhankulk1
     + 0 comments

    List res=new ArrayList<>(); int n=a.size(); k=k%n; for(int q:queries){ int newIndex=(q-k+n)%n; res.add(a.get(newIndex)); } return res;

  • anarod_val25
     + 0 comments

    Test failed succesfully???

    Test case 4 says success but it's marked red, as if it failed?

  • abyssal_cojode
     + 0 comments
    def circularArrayRotation(a, k, queries):
    # Example: For array [1, 2, 3, 4, 5] (n=5):
    # k=0: [1, 2, 3, 4, 5] (no rotation)
    # k=1: [5, 1, 2, 3, 4] (rotated right once)
    # k=2: [4, 5, 1, 2, 3] (rotated right twice)
    # ...
    # k=5: [1, 2, 3, 4, 5] (full rotation, back to original)
    
    # To rotate efficiently without multiple single shifts:
    # 1. Calculate the effective rotation offset using modulo arithmetic
    #    (since rotating by n positions brings array back to original)
    # 2. Split the array at the offset and swap the two parts
    
    # Calculate the rotation point
    offset = len(a) - k % len(a)
    
    # Perform the rotation by concatenating the two slices
    rotated = a[offset:] + a[:offset]
    
    # Return the elements at the requested indices
    return [rotated[q] for q in queries]

  • JanekPython09
     + 0 comments

    My python solution:

    def swap(arr):
    el = arr.pop(-1)
    arr.insert(0, el)
    return arr

def circularArrayRotation(a, k, queries):
    for i in range(k%len(a)):
        a = swap(a)
        
    return [a[x] for x in queries]

  • rezl_jaws
     + 0 comments

    My C++ solution

    vector circularArrayRotation(vector & arr, int k, vector queries) { vector res; // k % arr.size() enusre that we are not unnecessarily rotating back to original std::rotate(arr.rbegin(), arr.rbegin() + (k % arr.size()), arr.rend());

    for (auto a : queries)
    res.push_back(arr[a]);

return res;

    }