Rotate List Left by K - Amazon Top Interview Questions


Problem Statement :


Write a function that rotates a list of numbers to the left by k elements. Numbers should wrap around.

Note: The list is guaranteed to have at least one element, and k is guaranteed to be less than or equal to the length of the list.

Bonus: Do this without creating a copy of the list. How many swap or move operations do you need?

Constraints

n ≤ 100,000 where n is the length of nums


Example 1

Input

nums = [1, 2, 3, 4, 5, 6]
k = 2

Output

[3, 4, 5, 6, 1, 2]



Example 2

Input

nums = [1, 2, 3, 4, 5, 6]
k = 6


Output
[1, 2, 3, 4, 5, 6]



Example 3

Input

nums = [1]
k = 0


Output
[1]


Solution :



title-img 




                        Solution in C++ :

vector<int> solve(vector<int>& nums, int k) {
    int n = nums.size();
    int start = 0, end = k - 1;
    while (start < end) {
        swap(nums[start], nums[end]);
        start++;
        end--;
    }
    start = k, end = n - 1;
    while (start < end) {
        swap(nums[start], nums[end]);
        start++;
        end--;
    }
    start = 0, end = n - 1;
    while (start < end) {
        swap(nums[start], nums[end]);
        start++;
        end--;
    }
    return nums;
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    public int[] solve(int[] nums, int k) {
        if (nums == null || nums.length < 2) {
            return new int[] {nums[0]};
        }

        k %= nums.length;

        reverse(nums, 0, k - 1);
        reverse(nums, k, nums.length - 1);
        reverse(nums, 0, nums.length - 1);

        return nums;
    }

    private void reverse(int[] nums, int i, int j) {
        while (i < j) {
            swap(nums, i++, j--);
        }
    }

    private void swap(int[] nums, int i, int j) {
        int temp = nums[i];
        nums[i] = nums[j];
        nums[j] = temp;
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def reverse(self, nums, i, j):
        for idx in range((j - i + 1) // 2):
            nums[idx + i], nums[j - idx] = nums[j - idx], nums[idx + i]

    def solve(self, nums, k):
        self.reverse(nums, 0, len(nums) - 1)
        self.reverse(nums, 0, len(nums) - k - 1)
        self.reverse(nums, len(nums) - k, len(nums) - 1)
        return nums


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