Concatenation of Array

Difficulty: Easy | Acceptance: 90.40% | Paid: No Topics: Array, Simulation

Given an integer array nums, return an array ans where ans[i] == nums[i] and ans[i + n] == nums[i] for 0 <= i < n, where n is the length of nums. Specifically, ans is the concatenation of two nums arrays.

• Examples
• Constraints
• Simple Iteration
• Built-in Concatenation
• Pre-allocation

Examples

Example 1:

Input: nums = [1,2,1]
Output: [1,2,1,1,2,1]
Explanation: The array ans is formed as follows:
- ans = [nums[0],nums[1],nums[2],nums[0],nums[1],nums[2]]
- ans = [1,2,1,1,2,1]

Example 2:

Input: nums = [1,3,2,1]
Output: [1,3,2,1,1,3,2,1]
Explanation: The array ans is formed as follows:
- ans = [nums[0],nums[1],nums[2],nums[3],nums[0],nums[1],nums[2],nums[3]]
- ans = [1,3,2,1,1,3,2,1]

Constraints

n == nums.length
1 <= n <= 1000
1 <= nums[i] <= 1000

Simple Iteration

Intuition Iterate through the input array twice and append each element to the result array.

Steps

• Create an empty result array
• Loop through nums twice
• Append each element to result

class Solution:
    def getConcatenation(self, nums: List[int]) -> List[int]:
        ans = []
        for _ in range(2):
            for num in nums:
                ans.append(num)
        return ans

Complexity

• Time: O(n)
• Space: O(n)
• Notes: Simple and intuitive approach with clear logic

Built-in Concatenation

Intuition Use language-specific built-in methods to concatenate arrays directly.

Steps

• Use array concatenation operation
• Return the concatenated result

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

Complexity

• Time: O(n)
• Space: O(n)
• Notes: Most concise solution using built-in methods

Pre-allocation

Intuition Pre-allocate the result array with the exact required size and fill it in a single pass.

Steps

• Create array of size 2n
• Fill first half with nums
• Fill second half with nums

class Solution:
    def getConcatenation(self, nums: List[int]) -> List[int]:
        n = len(nums)
        ans = [0] * (2 * n)
        for i in range(n):
            ans[i] = nums[i]
            ans[i + n] = nums[i]
        return ans

Complexity

• Time: O(n)
• Space: O(n)
• Notes: Efficient memory allocation with single loop iteration