1470. Shuffle the Array

EasyArray
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Problem Description

The problem provides an array nums that has 2n elements in it. The elements are structured in a way that the first n elements are grouped together (let's call them x group) and the next n elements are grouped together (let's call them y group). The objective is to rearrange this array by combining the elements from x group and y group alternatively, starting with the first element of the x group. In the end, the array should look like [x1, y1, x2, y2, ..., xn, yn], which means we pick one element from the x group and then pick the corresponding element from the y group and continue this pattern until the array is fully rearranged.

Intuition

To arrive at the solution, we can realize that we need to build a new array by taking one element from the x group (first n elements of nums) and then taking the corresponding element from the y group (last n elements of nums). The approach is quite straightforward:

  1. Initialize an empty list ans where we will store our answer.
  2. Iterate over the range from 0 to n-1 (since we have 2n elements, and n denotes the number of pairs we need to create).
  3. In each iteration, append the element from the x group (nums[i]) to ans.
  4. Then, append the corresponding element from the y group (nums[i + n]) to ans.
  5. After the loop ends, ans will have 2n elements in the desired order [x1, y1, x2, y2, ..., xn, yn].
  6. Return the ans array which is now correctly shuffled.

This solution is easy to understand and implement. It uses extra space for the new array and takes linear time relative to the size of the original array, which is an efficient way to achieve the task.

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Solution Approach

The implemented solution makes use of a simple algorithm and a basic array data structure to achieve the desired result. The algorithm can be considered as a single-pass approach, which can be broken down into these main steps:

  1. Initialization: An empty list called ans is initialized. This is where the shuffled array will be stored.
  2. Single-Pass Loop:
    • A for loop is initiated to iterate n times, where n is half the size of the input array nums. This is because of the input array's structure, housing two groups of n elements each (x group and y group).
    • Inside the loop, a simple pattern is followed:
      • The element nums[i] from the x group (the first half of the array) is appended to the ans list.
      • Following that, the element nums[i + n] from the y group (the second half of the array) is appended to the ans list.
      • This pattern is based on an observation that for any index i in the first half of the array, its corresponding pair from the second half can be accessed at index i + n.
  3. Building the Result: The steps inside the loop ensures that ans grows by two elements after each iteration, maintaining the desired alternate sequence of elements from both x and y groups.
  4. Returning the Result: Once the loop finishes, the list ans now contains all elements required for the output array in the correct order. It is then returned as the final shuffled array.

In terms of data structures, this approach only relies on the list data structure, which is intrinsic to Python. No additional or specialized data structures are used. The algorithm's simplicity and absence of nested loops or recursive calls mean that the time complexity is linear, O(n), where n is the number of elements in half of the input list. The space complexity is also O(n) to account for the ans list which stores the reshuffled elements.

This approach effectively demonstrates how algorithmic patterns can emerge from the inherent structure of a problem, enabling a straightforward implementation that efficiently yields the desired result.

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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

Example Walkthrough

Let's take an example to illustrate the solution approach where the nums array is [1, 3, 2, 4], which represents 2n elements, with n being 2. We have the first n elements forming group x (which are [1, 3]) and the second n elements forming group y ([2, 4]).

Following the steps from the solution approach:

  1. Initialization: We initialize an empty list ans where the shuffled array will be stored.

  2. Single-Pass Loop:

    • We start with a for loop that iterates from 0 to n-1, which in this case is from 0 to 1.
    • In the first iteration (i=0):
      • We append nums[0] (which is 1 from the x group) to ans, making ans now [1].
      • We then append nums[0 + n] (which is nums[2], or 2 from the y group) to ans. Now ans is [1, 2].
    • In the second iteration (i=1):
      • We append nums[1] (which is 3 from the x group) to ans, so ans becomes [1, 2, 3].
      • Next, we append nums[1 + n] (which is nums[3], or 4 from the y group) to ans. The ans list is now [1, 2, 3, 4].

  3. Building the Result: At the end of the loop, ans has grown to include all 2n elements in the desired alternate x, y pattern, which is [1, 2, 3, 4] for our example.

  4. Returning the Result: We return the ans list, which is [1, 2, 3, 4]. This is our final shuffled array, where the x group and the y group elements are alternatively arranged.

Through this example, we can observe the simplicity and effectiveness of the approach, which leads us to the correct shuffled array by performing the operation step by step.

Solution Implementation

1from typing import List
2
3class Solution:
4    def shuffle(self, nums: List[int], n: int) -> List[int]:
5        # Initialize an empty list for the shuffled result
6        shuffled_list = []
7
8        # Iterate over the range from 0 to n
9        for i in range(n):
10            # Append the element from the first half of the list
11            shuffled_list.append(nums[i])
12
13            # Append the corresponding element from the second half of the list
14            shuffled_list.append(nums[i + n])
15
16        # Return the shuffled list
17        return shuffled_list
18
1class Solution {
2    public int[] shuffle(int[] nums, int n) {
3        // Initialize a new array 'result' with size twice of 'n'.
4        int[] result = new int[2 * n];
5      
6        // Use a single loop to iterate through the first half of the 'nums' array.
7        for (int index = 0, resultIndex = 0; index < n; ++index) {
8            // Place element from the first half of 'nums' into the 'result' array.
9            result[resultIndex++] = nums[index];
10            // Place the corresponding element from the second half of 'nums'.
11            result[resultIndex++] = nums[index + n];
12        }
13      
14        // Return the shuffled 'result' array.
15        return result;
16    }
17}
18
1#include <vector>
2using std::vector;
3
4class Solution {
5public:
6    // Shuffle the `nums` vector in a specific pattern and return the result.
7    vector<int> shuffle(vector<int>& nums, int n) {
8        // Initialize an empty vector to hold the shuffled result.
9        vector<int> shuffledResult;
10
11        // Loop over the elements in the first half of `nums`.
12        for (int i = 0; i < n; ++i) {
13            // Add the element from the first half of `nums` to the result.
14            shuffledResult.push_back(nums[i]);
15            // Add the corresponding element from the second half of `nums` to the result.
16            shuffledResult.push_back(nums[i + n]);
17        }
18      
19        // Return the shuffled vector.
20        return shuffledResult;
21    }
22};
23
1/**
2 * Shuffles an array consisting of 2n elements in the form of [x1,x2,...,xn,y1,y2,...,yn]
3 * into the form [x1,y1,x2,y2,...,xn,yn].
4 *
5 * @param {number[]} nums - The array to shuffle with 2n elements.
6 * @param {number} n - Half the length of the array, representing the split point.
7 * @returns {number[]} - The shuffled array.
8 */
9function shuffle(nums: number[], n: number): number[] {
10    // Initialize an empty array to store the shuffled elements
11    let shuffledArray: number[] = [];
12
13    // Loop through the first half of the array
14    for (let i = 0; i < n; i++) {
15        // Push the element from the first half
16        shuffledArray.push(nums[i]);
17        // Push the corresponding element from the second half
18        shuffledArray.push(nums[n + i]);
19    }
20
21    // Return the shuffled array
22    return shuffledArray;
23}
24
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Time and Space Complexity

Time Complexity

The time complexity of the given code can be analyzed based on the for loop that iterates over n elements. Inside the loop, the code performs a constant amount of work (two append operations). Therefore, the time complexity is directly proportional to n.

The time complexity is O(n).

Space Complexity

The space complexity consists of the additional space required by the solution in addition to the input data. Since it creates a new list ans that eventually holds 2 * n elements, it uses additional space proportional to the size of the input.

Hence, the space complexity is O(n).

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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