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

Circular Array Rotation

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3074 Discussions

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  • 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]

  • Tusenka
     + 0 comments
    k = int(first_multiple_input[1])
        ~~~~~~~~~~~~~~~~~~~~^^^

    IndexError: list index out of range

  • 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;
}

  • ali313xA
     + 1 comment
    `function circularArrayRotation(a, k, queries) {
    let n = a.length; 
    k = k % n; 
    
    let res = []; 
    
    for(let q of queries) {
        res.push( a[ (q - k + n) % n]); 
    }
    
    return res;
}

    `

  • safee7
     + 0 comments

    def reverse(a,l,r): i=l j=r while i

    def circularArrayRotation(a, k, queries): k=k%len(a) reverse(a,0,len(a)-1) reverse(a,0,k-1) reverse(a,k,len(a)-1) l=[] for i in queries: l.append(a[i]) return l