How do you solve Check Array Formation Through Concatenation on LeetCode?

Check Array Formation Through Concatenation (LeetCode #1640) 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: Using a HashMap allows for efficient lookups, reducing the need for nested loops.

What is the brute force solution for Check Array Formation Through Concatenation?

The brute force approach for Check Array Formation Through Concatenation is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible way to concatenate the pieces to form the target array. This involves trying to match each piece in the order they appear in the target array, which can be inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Check Array Formation Through Concatenation?

Check Array Formation Through Concatenation uses the following concepts: Array, Hash Table. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Check Array Formation Through Concatenation?

Always clarify the constraints and requirements of the problem before diving into coding. Consider edge cases, such as when `arr` or `pieces` is empty. Explain your thought process as you code, which helps interviewers understand your reasoning.

What are common mistakes when solving Check Array Formation Through Concatenation?

Forgetting to check if the piece exists in the map before accessing it. Not handling the case where the piece does not match the expected segment in `arr`.

#1640

Check Array Formation Through Concatenation

Easy

ArrayHash TableHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

The optimal approach uses a HashMap to store the pieces for quick look-up, allowing us to efficiently check if the current segment of `arr` matches any piece.

⚙️

Algorithm

5 steps

1Step 1: Create a HashMap to store the pieces with their first element as the key.
2Step 2: Initialize a pointer for the target array `arr`.
3Step 3: While the pointer is within bounds, check if the current element exists in the HashMap.
4Step 4: If it exists, retrieve the corresponding piece and check if it matches the next elements in `arr`. Move the pointer accordingly.
5Step 5: If any piece does not match, return false. If the pointer reaches the end of `arr`, return true.

solution.py11 lines

1def canFormArray(arr, pieces):
2    piece_map = {piece[0]: piece for piece in pieces}
3    i = 0
4    while i < len(arr):
5        if arr[i] not in piece_map:
6            return False
7        piece = piece_map[arr[i]]
8        if arr[i:i + len(piece)] != piece:
9            return False
10        i += len(piece)
11    return True

ℹ

Complexity note: The time complexity is O(n) since we only traverse `arr` once and use a HashMap for constant time lookups. The space complexity is O(n) due to the storage of pieces in the HashMap.

1Using a HashMap allows for efficient lookups, reducing the need for nested loops.
2The problem emphasizes the importance of maintaining the order of elements within each piece.

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