Solving LeetCode Problem 189: Rotate Array (Part 3)

Welcome back to the second part of our series on solving Leetcode Problem 189, Rotate Array. In this video, we’ll introduce a space-optimized solution with a time complexity of O(n) and space complexity of O(1).

Problem Description

Given an integer array nums, rotate the array to the right by k steps, where k is non-negative.

Example 1:

Input: nums = [1,2,3,4,5,6,7], k = 3
Output: [5,6,7,1,2,3,4]
Explanation:
rotate 1 steps to the right: [7,1,2,3,4,5,6]
rotate 2 steps to the right: [6,7,1,2,3,4,5]
rotate 3 steps to the right: [5,6,7,1,2,3,4]

Example 2:

Input: nums = [-1,-100,3,99], k = 2
Output: [3,99,-1,-100]
Explanation: 
rotate 1 steps to the right: [99,-1,-100,3]
rotate 2 steps to the right: [3,99,-1,-100]

Constraints:

1 <= nums.length <= 105
-231 <= nums[i] <= 231 - 1
0 <= k <= 105

Solution

class Solution {
    public void rotate(int[] nums, int k) {
        // Ensure k is within array bounds
        k %= nums.length;
        // Reverse entire array
        reverse(nums, 0, nums.length - 1);
        // Reverse first k elements
        reverse(nums, 0, k - 1);
        // Reverse remaining elements
        reverse(nums, k, nums.length - 1);
    }

    public void reverse(int[] nums, int from, int to) {
        int swap = nums[from];
        while (from < to) {
            swap = nums[from];
            nums[from] = nums[to];
            nums[to] = swap;
            from++;
            to--;
        }
    }
}

In this YouTube video, I was explained how to solve this problem step by step.