1929. Concatenation of Array

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Problem Description

The problem presents us with a challenge to create a new array that is exactly twice the length of the given integer array nums. The new array should have a structure such that the first half is a copy of the original nums array, and the second half is an exact repeat of the same nums array. This means for each element nums[i] where i is within the bounds of the original array, we want to set ans[i] and ans[i + n] (where n is the length of nums) to be equal to nums[i]. In essence, we are concatenating nums onto itself to create the resultant array ans.

Intuition

To arrive at the solution, we can realize that Python's list concatenation operator + performs the task of combining two lists end-to-end to form a single list. In this case, since we need to repeat the array nums twice, we can make use of this operator to concatenate nums with itself - nums + nums. This straightforward operation yields a new array which has all elements of nums followed by nums again, and thus fulfills the problem's requirement with minimal steps and in an easy-to-understand manner.

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Solution Approach

The solution makes use of the in-built functionality of Python's list type to solve the problem. Here’s a step-by-step walk-through of the implementation:

  1. Concatenation Operation (+): In Python, the + operator is used to concatenate two lists. So, in this case, we use nums + nums to create a new list. When the + operator is used between two lists, Python creates a new list by taking the elements of the first list (on the left-hand side) and appending the elements of the second list (on the right-hand side) in the same order.

  2. Returning the New List: The concatenated list is then returned as the output. The beauty of this approach is its simplicity and efficiency; no explicit loops or additional storage is needed, and thus the code is concise and easy to read. Since Python internally manages the creation of the new list during the concatenation, we also do not need to be concerned with the underlying details of how the new memory is allocated and how the elements are copied over.

The Python interpreter handles the concatenation operation efficiently, and since it is a single line of code, this solution is both performant and elegant for this problem.

Here’s the code snippet encapsulating this approach:

1class Solution:
2    def getConcatenation(self, nums: List[int]) -> List[int]:
3        return nums + nums

This code defines a method in a class Solution, which takes a list nums as input and returns the concatenated list by simply adding nums to itself.

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Example Walkthrough

Let's consider an example to illustrate the solution approach. Assume we have the following array nums:

1nums = [1, 2, 3]

With this array, we want to create a new array ans that contains each element of nums repeated once, resulting in ans being twice as long as nums.

Following the solution approach:

  1. Concatenation Operation (+): We apply the concatenation operator to nums, effectively doubling it by appending nums to the end of itself.

    Before operation:

    1nums = [1, 2, 3]

    After operation using nums + nums:

    1ans = [1, 2, 3] + [1, 2, 3]

  2. Returning the New List: The concatenated list result for ans after using the operation nums + nums would be:

    1ans = [1, 2, 3, 1, 2, 3]

The new list ans is now [1, 2, 3, 1, 2, 3], which is exactly what is expected according to the problem description. It retains the order of elements as per the original list and successfully repeats the sequence to create a list of twice the length.

This demonstrates that the solution provided is effective for the given problem, producing the desired result with simplicity and efficiency.

Solution Implementation

1# Definition of the Solution class
2class Solution:
3    # Method to get the concatenation of the list 'nums'
4    # It takes a list of integers as input and returns a new list
5    def getConcatenation(self, nums: List[int]) -> List[int]:
6        # The list 'nums' is concatenated with itself
7        # and the result is returned
8        ans = nums * 2  # replicates 'nums' twice
9      
10        # Return the concatenated list
11        return ans
12
1class Solution {
2    // Method to concatenate the array with itself
3    public int[] getConcatenation(int[] nums) {
4        // Get the length of the input array
5        int n = nums.length;
6        // Create a new array that is twice the length of the input array
7        int[] result = new int[n * 2];
8        // Loop through the new array to fill it with elements from the input array
9        for (int i = 0; i < n * 2; ++i) {
10            // Since we are concatenating the input array with itself,
11            // we use the modulo operator to wrap around the index for array 'nums'
12            // This is because 'nums' has a length of 'n' and 'result' has a length of 'n * 2'
13            // Therefore, when 'i' is greater than or equal to 'n', 
14            // 'i % n' will start from 0 again, essentially repeating the array 'nums'
15            result[i] = nums[i % n];
16        }
17        // Return the concatenated array
18        return result;
19    }
20}
21
1#include <vector> // Include the vector header for using the std::vector class
2
3class Solution {
4public:
5    // Function to concatenate an array with itself
6    // @param nums: The original array of integers
7    // @return: Returns a new vector containing two copies of the original array
8    vector<int> getConcatenation(vector<int>& nums) {
9        int original_size = nums.size(); // Store the original size of nums
10        nums.reserve(2 * original_size); // Reserve total space in advance for efficiency
11
12        // Loop through the original array and append each element to the end of it
13        for (int i = 0; i < original_size; ++i) {
14            nums.push_back(nums[i]);
15        }
16
17        return nums; // Return the modified vector containing two concatenated copies
18    }
19};
20
1// Function to concatenate a given array with itself
2function getConcatenation(nums: number[]): number[] {
3    // Use the spread operator to concatenate the array 'nums' with itself
4    const concatenatedArray: number[] = [...nums, ...nums];
5
6    // Return the resulting concatenated array
7    return concatenatedArray;
8}
9
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1void search(Node root) {
2  if (!root) return;
3  visit(root);
4  root.visited = true;
5  for (Node node in root.adjacent) {
6    if (!node.visited) {
7      search(node);
8    }
9  }
10}

Time and Space Complexity

The time complexity of the code is O(n), where n is the length of the input list nums. This is because the code concatenates the list nums with itself, an operation which requires iterating over the length of nums once for the concatenation process.

The space complexity of the code is also O(n). In Python, concatenating two lists using the + operator creates a new list with a combined length of the two lists, thus requiring additional space proportional to the sum of the lengths of the two lists. Here, since both lists are the same, the total space required for the new list is proportional to twice the length of the input list nums.

Learn more about how to find time and space complexity quickly using problem constraints.

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What's the output of running the following function using the following tree as input? 

1def serialize(root):
2    res = []
3    def dfs(root):
4        if not root:
5            res.append('x')
6            return
7        res.append(root.val)
8        dfs(root.left)
9        dfs(root.right)
10    dfs(root)
11    return ' '.join(res)
12
1import java.util.StringJoiner;
2
3public static String serialize(Node root) {
4    StringJoiner res = new StringJoiner(" ");
5    serializeDFS(root, res);
6    return res.toString();
7}
8
9private static void serializeDFS(Node root, StringJoiner result) {
10    if (root == null) {
11        result.add("x");
12        return;
13    }
14    result.add(Integer.toString(root.val));
15    serializeDFS(root.left, result);
16    serializeDFS(root.right, result);
17}
18
1function serialize(root) {
2    let res = [];
3    serialize_dfs(root, res);
4    return res.join(" ");
5}
6
7function serialize_dfs(root, res) {
8    if (!root) {
9        res.push("x");
10        return;
11    }
12    res.push(root.val);
13    serialize_dfs(root.left, res);
14    serialize_dfs(root.right, res);
15}
16

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