LeetCode #189 — MEDIUM

Rotate Array

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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

Follow up:

Try to come up with as many solutions as you can. There are at least three different ways to solve this problem.
Could you do it in-place with O(1) extra space?

Step 01

Brute Force Baseline

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

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Math · Two Pointers

Example 1

[1,2,3,4,5,6,7]
3

Example 2

[-1,-100,3,99]
2

Core Insight

What unlocks the optimal approach

The easiest solution would use additional memory and that is perfectly fine.
The actual trick comes when trying to solve this problem without using any additional memory. This means you need to use the original array somehow to move the elements around. Now, we can place each element in its original location and shift all the elements around it to adjust as that would be too costly and most likely will time out on larger input arrays.
One line of thought is based on reversing the array (or parts of it) to obtain the desired result. Think about how reversal might potentially help us out by using an example.
The other line of thought is a tad bit complicated but essentially it builds on the idea of placing each element in its original position while keeping track of the element originally in that position. Basically, at every step, we place an element in its rightful position and keep track of the element already there or the one being overwritten in an additional variable. We can't do this in one linear pass and the idea here is based on <b>cyclic-dependencies</b> between elements.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #189: Rotate Array
class Solution {
    private int[] nums;

    public void rotate(int[] nums, int k) {
        this.nums = nums;
        int n = nums.length;
        k %= n;
        reverse(0, n - 1);
        reverse(0, k - 1);
        reverse(k, n - 1);
    }

    private void reverse(int i, int j) {
        for (; i < j; ++i, --j) {
            int t = nums[i];
            nums[i] = nums[j];
            nums[j] = t;
        }
    }
}

// Accepted solution for LeetCode #189: Rotate Array
func rotate(nums []int, k int) {
	n := len(nums)
	k %= n
	reverse := func(i, j int) {
		for ; i < j; i, j = i+1, j-1 {
			nums[i], nums[j] = nums[j], nums[i]
		}
	}
	reverse(0, n-1)
	reverse(0, k-1)
	reverse(k, n-1)
}

# Accepted solution for LeetCode #189: Rotate Array
class Solution:
    def rotate(self, nums: List[int], k: int) -> None:
        def reverse(i: int, j: int):
            while i < j:
                nums[i], nums[j] = nums[j], nums[i]
                i, j = i + 1, j - 1

        n = len(nums)
        k %= n
        reverse(0, n - 1)
        reverse(0, k - 1)
        reverse(k, n - 1)

// Accepted solution for LeetCode #189: Rotate Array
impl Solution {
    pub fn rotate(nums: &mut Vec<i32>, k: i32) {
        let n = nums.len();
        let k = (k as usize) % n;
        nums.reverse();
        nums[..k].reverse();
        nums[k..].reverse();
    }
}

// Accepted solution for LeetCode #189: Rotate Array
/**
 Do not return anything, modify nums in-place instead.
 */
function rotate(nums: number[], k: number): void {
    const n: number = nums.length;
    k %= n;
    const reverse = (i: number, j: number): void => {
        for (; i < j; ++i, --j) {
            const t: number = nums[i];
            nums[i] = nums[j];
            nums[j] = t;
        }
    };
    reverse(0, n - 1);
    reverse(0, k - 1);
    reverse(k, n - 1);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.

TWO POINTERS

O(n) time

O(1) space

Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.

Shortcut: Two converging pointers on sorted data → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.