Learn to Rotate Array to Left in Java A Key LeetCode DSA Interview Question

Introduction

In the world of coding interviews and competitive programming, mastering array manipulation is essential. One of the classic challenges you'll encounter is rotating an array. This problem is not only a staple on platforms like LeetCode but also a critical concept in Data Structures and Algorithms (DSA). In this article, we'll explore a robust solution to the left rotation of an array using Java, ensuring you're well-equipped for interviews and coding challenges.

Problem Statement

Given an array of integers and a number k, the task is to rotate the array to the left by k steps. For example, rotating the array {1, 2, 3, 4, 5, 6, 7, 8} left by 2 results in {3, 4, 5, 6, 7, 8, 1, 2}.

Why This Problem Matters

Understanding array rotation is crucial for several reasons:

• Fundamental Concept: It helps in grasping more complex data structures.
• Common Interview Question: Frequently asked in coding interviews.
• Real-World Applications: Useful in scenarios involving circular queues and scheduling algorithms.

Java Solution: Rotating an Array to the Left

Here’s a clean and efficient Java solution to the problem using the reversal algorithm, which operates in linear time, O(n).

import java.util.Arrays;

public class RotateArrayLeft {

public static void main(String[] args) {

int[] arr = {1, 2, 3, 4, 5, 6, 7, 8};

int k=2;

System.out.println("Original Array: "+Arrays.toString(arr));

rotateArrayLeftByKSteps(arr, k);

System.out.println("Array after k rotation to left: \n"

                                                    +Arrays.toString(arr));

}

private static void rotateArrayLeftByKSteps(int[] arr, int k) {

if (arr.length==0 || k<=0) {

return;

}

/*If k is greater than arrays length, then we adjust k

* to avoid extra rotation

*/

k = k%arr.length;

//first reverse k elements of array

int i=0;

int j=k-1;

while (i<j) {

int temp = arr[i];

arr[i] = arr[j];

arr[j]=temp;

i++;

j--;

}

//now reverse the remaining array

int p = k;

int q = arr.length-1;

while (p<q) {

int temp = arr[p];

arr[p] = arr[q];

arr[q] = temp;

p++;

q--;

}

//now reverse the complete array

int start=0;

int end = arr.length-1;

while (start<end) {

int temp = arr[start];

arr[start] = arr[end];

arr[end] = temp;

start++;

end--;

}

}

}

Explanation of the Code

1. Input Handling: The function checks if the array is empty or if k is less than or equal to zero.
2. Adjusting k: To optimize, we compute k % arr.length to avoid unnecessary rotations.
3. Reversals:
   • First, reverse the first k elements.
   • Then, reverse the remaining elements.
   • Finally, reverse the entire array to achieve the desired rotation.

This method is efficient, with a time complexity of O(n) and a space complexity of O(1) since it operates in place.

Conclusion

Rotating an array is an important problem that combines various DSA concepts, making it essential for anyone preparing for technical interviews. By mastering this solution in Java, you’ll be better equipped to tackle similar challenges on LeetCode and beyond.

Tags:

#LeetCode #Java #DSA #CodingChallenges #ArrayManipulation #Programming #InterviewPrep #CodingInterview #TechInterview #LearnToCode