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  1. Arrays: Left Rotation
  2. Discussions

Arrays: Left Rotation

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  • jayantnishad34
     + 0 comments

    public static List rotLeft(List a, int d) { // Write your code here int n = a.size(); d = d % n; // Handle cases where d > n

    // Use sublists to rearrange in O(n) time
List<Integer> rotated = new ArrayList<>(a.subList(d, n));
rotated.addAll(a.subList(0, d));

return rotated;
}

  • biren_yadav114
     + 1 comment

    Python code -

  • wilson155079
     + 0 comments
        public static List<Integer> rotLeft(List<Integer> a, int d) {     
        List<Integer> rotations = new ArrayList<>();
        // Elements from index d to the end
        rotations.addAll(a.subList(d, a.size())); 
        // Elements from the start to index d
        rotations.addAll(a.subList(0, d)); 
        
        return rotations;
    }

  • vickxnegi
     + 0 comments

    C# this is with basic knowledge

    public static List rotLeft(List a, int d) {

            int length=a.Count()-1;
        int[] newArr = new int[length+1];

        for(int i = length;i>-1;i--){

            if(i-d<0)
            {                   
                int j=i-d;                   
                newArr[length+1+j]=a[i];                      
            }
            else
            {
                newArr[i-d]=a[i];
            }
        }
        return   newArr.ToList() ;
}

    }

  • air_mvp_jordan
     + 0 comments

    Solution for c#.

    public static List<int> rotLeft(List<int> a, int d)
{
	int n = a.Count;
	int rotateBy = d % n;

	List<int> rotatedPart = a.GetRange(0, rotateBy);
	a.RemoveRange(0, rotateBy);
	a.AddRange(rotatedPart);

	return a;

}