Coding problem · Easy · Basics

Rotate an Array

Easy Algorithms & data structures

Return nums rotated to the right by k steps: the last k elements move to the front. k may be larger than the array's length.

Example 1

Input: nums = [1,2,3,4,5], k = 2
Output: [4,5,1,2,3]

Example 2

Input: nums = [1,2], k = 3
Output: [2,1]
Explanation: Rotating by 3 is the same as rotating by 1.

Hints

Hint 1

Rotating by the array's length gives the same array, so use k % n.

Hint 2

The answer is the last k elements followed by the rest.

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Solution and explanation

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Time O(n) · Space O(n)

Normalise k = k % n, then concatenate the last k elements with the first n - k. (An in-place version reverses the whole array, then each part.)

Reference solution

function rotateRight(nums, k) {
  const n = nums.length;
  if (n === 0) return [];
  k %= n;
  return [...nums.slice(n - k), ...nums.slice(0, n - k)];
}

function rotateRight(nums: number[], k: number): number[] {
  const n = nums.length;
  if (n === 0) return [];
  k %= n;
  return [...nums.slice(n - k), ...nums.slice(0, n - k)];
}

def rotate_right(nums: list[int], k: int) -> list[int]:
    if not nums:
        return []
    k %= len(nums)
    return nums[-k:] + nums[:-k] if k else nums[:]

<?php

function rotateRight(array $nums, int $k): array
{
    $n = count($nums);
    if ($n === 0) {
        return [];
    }
    $k %= $n;
    return array_merge(array_slice($nums, $n - $k), array_slice($nums, 0, $n - $k));
}

class Solution {
    public int[] rotateRight(int[] nums, int k) {
        int n = nums.length;
        int[] out = new int[n];
        for (int i = 0; i < n; i++) out[(i + k) % n] = nums[i];
        return out;
    }
}

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