Arrays: Left Rotation Discussions |

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1 week ago+ 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;
}

2 weeks ago+ 1 comment

Python code -

2 weeks ago+ 0 comments

def rotLeft(a, d): # Write your code here return a[d:]+a[0:d]

1 month ago+ 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;
    }

3 months ago+ 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() ;
}

}

4 months ago+ 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;

}

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