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  1. Prepare
  2. Data Structures
  3. Arrays
  4. Left Rotation
  5. Discussions

Left Rotation

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3554 Discussions

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  • brent60
     + 0 comments

    Quick in Python!

    def rotateLeft(num_bits: int, arr: list[int]) -> list[int]:
    """Shift bits in the list/array to the left by num_bits."""
    left_shift_bits = num_bits % len(arr)
    return [*arr[left_shift_bits:], *arr[:left_shift_bits]]

  • sudiksha_email
     + 0 comments

    include

    using namespace std;

    int main() { int n, d; cin >> n >> d; vector a(n); for (int i = 0; i < n; i++) cin >> a[i];

    d %= n;
vector<int> res(n);
for (int i = 0; i < n; i++)
    res[i] = a[(i + d) % n];  

for (int i = 0; i < n; i++)
    cout << res[i] << (i == n - 1 ? '\n' : ' ');

return 0;

    }

  • ivan_romero_dev1
     + 0 comments
    import java.io.*;
import java.util.stream.Stream;

public class ArrayLeftRotation {

    public static void main(String[] args) throws IOException {
        try (BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
             BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out))) {
            int[] firstLine = Stream.of(br.readLine().split(" ")).mapToInt(Integer::parseInt).toArray();
            int n = firstLine[0];
            int d = firstLine[1];
            int[] array = Stream.of(br.readLine().split(" ")).mapToInt(Integer::parseInt).toArray();
            for (int i = 0; i < n; i++) {
                bw.write(array[(i + d) % n] + " ");
            }
            bw.newLine();
            bw.flush();
        } catch (IOException e) {
            throw new RuntimeException(e);
        }
    }
}

  • MaryRumble1
     + 0 comments

    Data structures are the backbone of efficient programming! They help organize and store data in ways that make problem-solving faster and more effective. Betinexchange247

  • himanshuramoliya
     + 0 comments

    With out extra space , use reverse function

    vector<int> rotateLeft(int d, vector<int> arr) {
    int n = arr.size();
    d = n - d%n;
    if(d==n) return arr;
    reverse(arr.begin(), arr.end());
    reverse(arr.begin(), arr.begin()+d);
    reverse(arr.begin()+d, arr.end());
    return arr;
}