Rotate Array by K Positions

This snippet demonstrates how to rotate an array to the right by K positions using the reverse algorithm. Instead of shifting elements one by one, it reverses the entire array, then reverses the first K elements and finally the remaining elements. This approach is optimal and widely used in coding interviews due to its O(n) time complexity and O(1) extra space.

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Published

19 Jul 2026

Last Updated

19 Jul 2026

Java Implementation

Complete Java Solution

Production-ready Java implementation for quick revision, interview preparation, and real-world development.

39Lines

27Code Lines

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arraysCategory

Time: O(n) | Space: O(1)Complexity

Solution.java

Java

1public class RotateArray {
2 
3    public static void reverse(int[] arr, int left, int right) {
4        while (left < right) {
5            int temp = arr[left];
6            arr[left] = arr[right];
7            arr[right] = temp;
8 
9            left++;
10            right--;
11        }
12    }
13 
14    public static void rotate(int[] arr, int k) {
15 
16        int n = arr.length;
17 
18        k = k % n;
19 
20        reverse(arr, 0, n - 1);
21        reverse(arr, 0, k - 1);
22        reverse(arr, k, n - 1);
23    }
24 
25    public static void main(String[] args) {
26 
27        int[] numbers = {1, 2, 3, 4, 5, 6, 7};
28 
29        int k = 3;
30 
31        rotate(numbers, k);
32 
33        System.out.print("Rotated Array: ");
34 
35        for (int num : numbers) {
36            System.out.print(num + " ");
37        }
38    }
39}

39 lines768 chars

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arrays

Reverse an Array

This snippet demonstrates how to reverse an array efficiently using the two-pointer technique. It swaps elements from both ends of the array until the pointers meet in the middle. This approach is memory-efficient because it performs the reversal in place without requiring an additional array.

ArrayTwo PointersReverse

Time: O(n) | Space: O(1)