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
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