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  1. Left Rotation
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Left Rotation

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150 Discussions

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  • vinayak25nair
     + 0 comments

    Java Easy solution

    public static List<Integer> rotateLeft(int d, List<Integer> arr) {

    if(d==arr.size()){
        return arr;
    }
    int i=0;
    for(;i<d;d--){
        int temp = arr.get(i);
        arr.remove(i);
        arr.add(temp);
    }
    return arr;
}

  • victor_barrios_1
     + 0 comments

    Swift solution:

    func rotateLeft(d: Int, arr: [Int]) -> [Int] {
    let n = arr.count
    
    // 'mod' operation avoids unnecessary rotations
    let rotations = d % n
    
    return Array(arr[rotations...] + arr[..<rotations])
}

  • victor_reial
     + 0 comments

    Rust best solution

    If you’re looking for solutions to the 3-month preparation kit in either Python or Rust, you can find them below: my solutions 

    fn left_rotation(d: i32, arr: &[i32]) -> Vec<i32> {
    //Time complexity: O(n)
    //Space complexity (ignoring input): O(n)
    let mut new_array = Vec::new();
    for index in 0..arr.len() {
        new_array.push(arr[(index + d as usize) % arr.len()]);
    }

    new_array
}

  • victor_reial
     + 0 comments

    Python best solution

    If you’re looking for solutions to the 3-month preparation kit in either Python or Rust, you can find them below: my solutions 

    def left_rotation(d, arr):
    # Time complexity: O(n)
    # Space complexity (ignoring input): O(n)
    new_array = [0] * len(arr)
    for index in range(0, len(arr)):
        new_array[(index - d) % len(arr)] = arr[index]

    return new_array


def left_rotation_in_place(d, arr):
    # Time complexity: O(n)
    # Space complexity (ignoring input): O(1)
    n = len(arr)
    start_range = 0
    end_range = d
    for index in range(0, (end_range - start_range) // 2):
        arr[index + start_range], arr[end_range - index - 1] = (
            arr[end_range - index - 1],
            arr[index + start_range],
        )

    start_range = d
    end_range = n
    for index in range(0, (end_range - start_range) // 2):
        arr[index + start_range], arr[end_range - index - 1] = (
            arr[end_range - index - 1],
            arr[index + start_range],
        )

    start_range = 0
    end_range = n
    for index in range(0, (end_range - start_range) // 2):
        arr[index + start_range], arr[end_range - index - 1] = (
            arr[end_range - index - 1],
            arr[index + start_range],
        )

    return arr

  • shiaramoon
     + 0 comments
    #python solution
def rotateLeft(d, arr):
    return arr[d:]+arr[:d]