How do you solve Split and Merge Array Transformation on LeetCode?

Split and Merge Array Transformation (LeetCode #3690) 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: BFS is effective for state exploration.

What is the time complexity of Split and Merge Array Transformation?

The optimal solution for Split and Merge Array Transformation runs in O(n) time and uses O(n) space. BFS explores each state once, leading to linear complexity in terms of operations.

What is the brute force solution for Split and Merge Array Transformation?

The brute force approach for Split and Merge Array Transformation is Brute Force, with O(n²) time complexity and O(1) space complexity. Try every possible way to split and merge the array until it matches the target. This exhaustive search will eventually find the solution but is inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Split and Merge Array Transformation?

Split and Merge Array Transformation uses the following concepts: Array, Hash Table, Breadth-First Search. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Split and Merge Array Transformation?

Explain your thought process clearly. Discuss edge cases. Be ready to optimize your solution.

What are common mistakes when solving Split and Merge Array Transformation?

Not considering all possible positions for insertion. Overlooking duplicate states.

#3690

Split and Merge Array Transformation

Medium

ArrayHash TableBreadth-First SearchHash 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)

Use BFS to explore states of nums1, transforming it into nums2 by considering each split-and-merge operation as a state transition.

⚙️

Algorithm

3 steps

1Step 1: Initialize a queue with nums1 and a count of operations.
2Step 2: For each state, generate new states by removing subarrays and reinserting them in all possible positions.
3Step 3: If a new state matches nums2, return the operation count.

solution.py19 lines

1from collections import deque
2
3def min_operations(nums1, nums2):
4    queue = deque([(nums1, 0)])
5    seen = set()
6    while queue:
7        current, ops = queue.popleft()
8        if current == nums2:
9            return ops
10        for i in range(len(current)):
11            for j in range(i, len(current)):
12                sub = current[i:j+1]
13                remaining = current[:i] + current[j+1:]
14                for k in range(len(remaining) + 1):
15                    new_state = remaining[:k] + sub + remaining[k:]
16                    if tuple(new_state) not in seen:
17                        seen.add(tuple(new_state))
18                        queue.append((new_state, ops + 1))
19    return -1

ℹ

Complexity note: BFS explores each state once, leading to linear complexity in terms of operations.

1BFS is effective for state exploration.
2Tracking seen states prevents cycles.

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