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() ;
}

}

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;

}

Here is the optimal approach by using recursion and reversing the array.

//swap array elements void swap(int &a, int &b) { int temp = a; a = b; b = temp; }

void reverse(vector &arr, int i, int j) { if (i >= j) return; // Base condition to stop recursion swap(arr[i], arr[j]); reverse(arr, i + 1, j - 1); }

vector rotLeft(vector a, int d) { int n = a.size(); d = d % n; // Handle cases where d >= n reverse(a, 0, d - 1); // Reverse the first part reverse(a, d, n - 1); // Reverse the second part reverse(a, 0, n - 1); // Reverse the whole array return a; }

Java

public static List rotLeft(List a, int d) { while(d>0) { Integer removed = a.remove(0); a.add(removed); d--; } return a; } }

C#

            var newlist = new List<int>(a);

    for(int i = 0; i< a.Count; i++){
        var index = i - d%a.Count;


        if(index >=0){
            newlist[index] = a[i];
            continue;
        }

        index += a.Count;
        newlist[index] = a[i];


    return newlist;
}