How do you solve Concatenation of Array on LeetCode?

Concatenation of Array (LeetCode #1929) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: The problem requires understanding how to manipulate array indices effectively.

What is the brute force solution for Concatenation of Array?

The brute force approach for Concatenation of Array is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves creating a new array of double the size and manually filling it with the elements of the original array twice. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Concatenation of Array?

Concatenation of Array uses the following concepts: Array, Simulation. The recommended patterns to study are: Array, Simulation.

What are the interview tips for Concatenation of Array?

Always think about the size of the input and how that affects your approach. Practice writing out your thought process before jumping into coding. Be mindful of edge cases, such as the smallest and largest possible inputs.

What are common mistakes when solving Concatenation of Array?

Not initializing the output array correctly. Overcomplicating the solution instead of using a direct approach.

#1929

Concatenation of Array

Easy

ArraySimulationArraySimulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

The optimal approach involves directly filling the new array in a single pass. This is efficient and leverages the structure of the problem, reducing the need for multiple loops.

⚙️

Algorithm

2 steps

1Step 1: Create an empty array 'ans' of size 2 * n.
2Step 2: Loop through the original array 'nums' once, and for each index i, set ans[i] = nums[i] and ans[i + n] = nums[i].

solution.py6 lines

1nums = [1, 2, 1]
2ans = [0] * (2 * len(nums))
3for i in range(len(nums)):
4    ans[i] = nums[i]
5    ans[i + len(nums)] = nums[i]
6print(ans)

ℹ

Complexity note: This complexity is linear because we only loop through the input array once. The space complexity is O(n) due to the creation of the new array of size 2n.

1The problem requires understanding how to manipulate array indices effectively.
2Recognizing patterns in array manipulation can lead to more efficient solutions.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.