Left Rotate Array by One

Arrays Fundamentals Easy

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.

Examples:

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

Output: [2, 3, 4, 5, 1]

Explanation: Initially, nums = [1, 2, 3, 4, 5]

Rotating once to left -> nums = [2, 3, 4, 5, 1]

Input: nums = [-1, 0, 3, 6]

Output: [0, 3, 6, -1]

Explanation: Initially, nums = [-1, 0, 3, 6]

Rotating once to left -> nums = [0, 3, 6, -1]

Input: nums = [7, 6, 5, 4]

Constraints

1 <= nums.length <= 10⁵
-10⁴ <= nums[i] <= 10⁴

Hints

Focus on moving the first element of the array to the end while shifting all other elements one position to the left.
Think about how swapping or overwriting elements can achieve the desired rotation within the same array.

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Editorial

Intuition

To rotate an array by one position to the right, consider that the last element of the array will move to the first position, while all other elements shift one position to the right. The thought process involves first capturing the value of the first element in a temporary variable. Next, iterate through the array starting from the second element and shift each element to the position of its predecessor. Finally, place the initially captured value into the last position of the array. This approach ensures that the array is rotated by one position effectively.

Approach

Store the value of the first element of the array in a temporary variable.
Iterate through the array starting from the second element.
Shift each element one position to the left by assigning the current element to the position of its predecessor.
After completing the iteration, place the value from the temporary variable into the last position of the array.

Solution

#include<bits/stdc++.h>
using namespace std;

class Solution {
public:
    void rotateArrayByOne(vector<int>& nums) {
        // Store the first element in a temporary variable
        int temp = nums[0];
        
        // Shift elements to the left
        for (int i = 1; i < nums.size(); ++i) {
            nums[i - 1] = nums[i];
        }

        // Place the first element at the end
        nums[nums.size() - 1] = temp;
    }
};

int main() {
    Solution solution;
    vector<int> nums = {1, 2, 3, 4, 5};

    solution.rotateArrayByOne(nums);

    for (int num : nums) {
        cout << num << " "; // Output the rotated array
    }

    return 0;
}
class Solution {
    public void rotateArrayByOne(int[] nums) {
        // Store the first element in a temporary variable
        int temp = nums[0];
        
        // Shift elements to the left
        for (int i = 1; i < nums.length; i++) {
            nums[i - 1] = nums[i];
        }

        // Place the first element at the end
        nums[nums.length - 1] = temp;
    }

    // Main method for testing
    public static void main(String[] args) {
        Solution solution = new Solution();
        int[] nums = {1, 2, 3, 4, 5};

        solution.rotateArrayByOne(nums);

        // Output the rotated array
        for (int num : nums) {
            System.out.print(num + " ");
        }
    }
}
class Solution:
    def rotateArrayByOne(self, nums):
        # Store the first element in a temporary variable
        temp = nums[0]
        
        # Shift elements to the left
        for i in range(1, len(nums)):
            nums[i - 1] = nums[i]
        
        # Place the first element at the end
        nums[-1] = temp

# Main method for testing
if __name__ == "__main__":
    solution = Solution()
    nums = [1, 2, 3, 4, 5]
    
    solution.rotateArrayByOne(nums)
    
    print(nums)  # Output the rotated array
class Solution {
    rotateArrayByOne(nums) {
        // Store the first element in a temporary variable
        const temp = nums[0];
        
        // Shift elements to the left
        for (let i = 1; i < nums.length; i++) {
            nums[i - 1] = nums[i];
        }

        // Place the first element at the end
        nums[nums.length - 1] = temp;
    }
}

// Main method for testing
const solution = new Solution();
const nums = [1, 2, 3, 4, 5];

solution.rotateArrayByOne(nums);

console.log(nums);  // Output the rotated array

Complexity Analysis

Time Complexity: O(N), where N is the number of elements in the array. Each element is visited once during the iteration.

Space Complexity: O(1). The space used does not depend on the size of the input array and remains constant.

Code

Problem List