Problem Statement:

Given an integer array nums, rotate the array to the left by one.

Note:

There is no need to return anything, just modify the given array.

Example 1:
Input:
 nums = [1, 2, 3, 4, 5]  
Output:
 [2, 3, 4, 5, 1]  
Explanation:
 Initially, nums = [1, 2, 3, 4, 5]  
Rotating once to the left results in nums = [2, 3, 4, 5, 1].

Example 2:
Input:
 nums = [-1, 0, 3, 6]  
Output:
 [0, 3, 6, -1]  
Explanation:
 Initially, nums = [-1, 0, 3, 6]  
Rotating once to the left results in nums = [0, 3, 6, -1].

2. Brute Force

Algorithm

• Create a dummy array of the same length as the original array.
• Shift all elements in the original array toward the left, copying them into the dummy array.
• After shifting, place the value of the 0th index of the original array into the last element of the dummy array.
• Finally, print the dummy array which now contains the left-shifted elements with the 0th element moved to the last position.

// Class containing the solve method
class Solution {

    // Function to solve and shift array elements left by one position
    public static void solve(int[] arr, int n) {
        int[] temp = new int[n];  // Temporary array to store shifted elements

        // Shift elements to the left by one position
        for (int i = 1; i < n; i++) {
            temp[i - 1] = arr[i];
        }
        temp[n - 1] = arr[0];  // First element moves to the last position

        // Print the rotated array
        for (int i = 0; i < n; i++) {
            System.out.print(temp[i] + " ");
        }
        System.out.println();
    }
}

// Main class
public class Main {
    public static void main(String[] args) {
        int n = 5;  // Size of the array
        int[] arr = {1, 2, 3, 4, 5};  // Original array

        Solution.solve(arr, n);  // Call the solve function from Solution class
    }
}

Complexity Analysis

Time Complexity: O(N), where N is the size of the array. This is because we traverse the array once to shift the elements.