We use cookies to ensure you have the best browsing experience on our website. Please read our cookie policy for more information about how we use cookies.
  1. Left Rotation
  2. Discussions

Left Rotation

Sort by

recency

|

152 Discussions

|

  • chreachhunneng
     + 0 comments

    I think this solution only few line in java

        public static List<Integer> rotateLeft(int d, List<Integer> arr) {
        for(int i=0; i < d; i++){
            arr.add(arr.remove(0));
        }
        return arr;
    }

  • mohapatra_shrey1
     + 0 comments

    Python Solution with minimum no of rotations remainder = minimum no. of rotations or swaps required on the arr

    def rotateLeft(d, arr):
    n=len(arr)
    rem=d%n
    if d==0 or rem==0:
        return arr
    else:
        for i in range(rem):
            arr.append(float('-inf'))
            arr[0],arr[-1]=arr[-1],arr[0]
            arr.pop(0)
        return arr

  • 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
}