# 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 :

```                        ```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```
```

## Contacts

We're going to make our own Contacts application! The application must perform two types of operations: 1 . add name, where name is a string denoting a contact name. This must store name as a new contact in the application. find partial, where partial is a string denoting a partial name to search the application for. It must count the number of contacts starting partial with and print the co

## No Prefix Set

There is a given list of strings where each string contains only lowercase letters from a - j, inclusive. The set of strings is said to be a GOOD SET if no string is a prefix of another string. In this case, print GOOD SET. Otherwise, print BAD SET on the first line followed by the string being checked. Note If two strings are identical, they are prefixes of each other. Function Descriptio

## Cube Summation

You are given a 3-D Matrix in which each block contains 0 initially. The first block is defined by the coordinate (1,1,1) and the last block is defined by the coordinate (N,N,N). There are two types of queries. UPDATE x y z W updates the value of block (x,y,z) to W. QUERY x1 y1 z1 x2 y2 z2 calculates the sum of the value of blocks whose x coordinate is between x1 and x2 (inclusive), y coor

## Direct Connections

Enter-View ( EV ) is a linear, street-like country. By linear, we mean all the cities of the country are placed on a single straight line - the x -axis. Thus every city's position can be defined by a single coordinate, xi, the distance from the left borderline of the country. You can treat all cities as single points. Unfortunately, the dictator of telecommunication of EV (Mr. S. Treat Jr.) do

## Subsequence Weighting

A subsequence of a sequence is a sequence which is obtained by deleting zero or more elements from the sequence. You are given a sequence A in which every element is a pair of integers i.e A = [(a1, w1), (a2, w2),..., (aN, wN)]. For a subseqence B = [(b1, v1), (b2, v2), ...., (bM, vM)] of the given sequence : We call it increasing if for every i (1 <= i < M ) , bi < bi+1. Weight(B) =