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


View More Similar Problems

Self Balancing Tree

An AVL tree (Georgy Adelson-Velsky and Landis' tree, named after the inventors) is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; if at any time they differ by more than one, rebalancing is done to restore this property. We define balance factor for each node as : balanceFactor = height(left subtree) - height(righ

View Solution →

Array and simple queries

Given two numbers N and M. N indicates the number of elements in the array A[](1-indexed) and M indicates number of queries. You need to perform two types of queries on the array A[] . You are given queries. Queries can be of two types, type 1 and type 2. Type 1 queries are represented as 1 i j : Modify the given array by removing elements from i to j and adding them to the front. Ty

View Solution →

Median Updates

The median M of numbers is defined as the middle number after sorting them in order if M is odd. Or it is the average of the middle two numbers if M is even. You start with an empty number list. Then, you can add numbers to the list, or remove existing numbers from it. After each add or remove operation, output the median. Input: The first line is an integer, N , that indicates the number o

View Solution →

Maximum Element

You have an empty sequence, and you will be given N queries. Each query is one of these three types: 1 x -Push the element x into the stack. 2 -Delete the element present at the top of the stack. 3 -Print the maximum element in the stack. Input Format The first line of input contains an integer, N . The next N lines each contain an above mentioned query. (It is guaranteed that each

View Solution →

Balanced Brackets

A bracket is considered to be any one of the following characters: (, ), {, }, [, or ]. Two brackets are considered to be a matched pair if the an opening bracket (i.e., (, [, or {) occurs to the left of a closing bracket (i.e., ), ], or }) of the exact same type. There are three types of matched pairs of brackets: [], {}, and (). A matching pair of brackets is not balanced if the set of bra

View Solution →

Equal Stacks

ou have three stacks of cylinders where each cylinder has the same diameter, but they may vary in height. You can change the height of a stack by removing and discarding its topmost cylinder any number of times. Find the maximum possible height of the stacks such that all of the stacks are exactly the same height. This means you must remove zero or more cylinders from the top of zero or more of

View Solution →