Watch 9 video solutions for Split and Merge Array Transformation, a medium level problem involving Array, Hash Table, Breadth-First Search. This walkthrough by Sanyam IIT Guwahati has 2,144 views views. Want to try solving it yourself? Practice on FleetCode or read the detailed text solution.
You are given two integer arrays nums1 and nums2, each of length n. You may perform the following split-and-merge operation on nums1 any number of times:
nums1[L..R].nums1[0..L-1] (empty if L = 0) and the suffix nums1[R+1..n-1] (empty if R = n - 1).Return the minimum number of split-and-merge operations needed to transform nums1 into nums2.
Example 1:
Input: nums1 = [3,1,2], nums2 = [1,2,3]
Output: 1
Explanation:
[3] (L = 0, R = 0); the remaining array is [1,2].[3] at the end; the array becomes [1,2,3].Example 2:
Input: nums1 = [1,1,2,3,4,5], nums2 = [5,4,3,2,1,1]
Output: 3
Explanation:
[1,1,2] at indices 0 - 2; remaining is [3,4,5]; insert [1,1,2] at position 2, resulting in [3,4,1,1,2,5].[4,1,1] at indices 1 - 3; remaining is [3,2,5]; insert [4,1,1] at position 3, resulting in [3,2,5,4,1,1].[3,2] at indices 0 - 1; remaining is [5,4,1,1]; insert [3,2] at position 2, resulting in [5,4,3,2,1,1].
Constraints:
2 <= n == nums1.length == nums2.length <= 6-105 <= nums1[i], nums2[i] <= 105nums2 is a permutation of nums1.Problem Overview: You are given an array and a set of allowed operations that split elements or merge adjacent elements. The goal is to determine whether one configuration can be transformed into another using these operations. Each transformation creates a new array state, so the task becomes exploring valid states efficiently without revisiting duplicates.
Approach 1: Brute Force State Simulation (Exponential Time, High Space)
The most direct idea is to simulate every possible sequence of split and merge operations. At each step, iterate through the array, attempt all valid splits and merges, and generate new arrays. This quickly explodes because each operation creates multiple branches. Without tracking visited states, the same configuration appears repeatedly. Time complexity grows exponentially with the number of operations, and space complexity is also exponential due to storing intermediate states.
Approach 2: Breadth-First Search with Hash Set (Optimal BFS Traversal)
Treat each array configuration as a node in a graph. An edge exists between two nodes if one can be obtained from the other via a single split or merge operation. Use Breadth-First Search to explore states level by level starting from the initial array. Each generated configuration is encoded (for example as a string or tuple) and stored in a visited hash table to prevent reprocessing duplicates.
During BFS, generate neighbors by iterating through the array and applying valid split or merge operations. Push unseen states into a queue. If the target configuration appears, the transformation is possible. BFS guarantees the shortest sequence of operations if the problem asks for minimum steps. Time complexity depends on the number of reachable states but is typically expressed as O(S * N), where S is the number of unique states explored and N is the array length. Space complexity is O(S) for the queue and visited set.
This approach works well because each state is processed once and duplicate configurations are eliminated using hashing. Arrays are naturally modeled as nodes in a state graph, making BFS the most natural traversal strategy when exploring transformation sequences involving arrays.
Recommended for interviews: Start by explaining the brute force simulation to demonstrate understanding of the operations and the search space. Then transition to the BFS state‑graph model with a visited hash set. Interviewers usually expect the BFS approach because it systematically explores transformations while avoiding repeated work.
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Brute Force Simulation | Exponential | Exponential | Useful for understanding all possible operations or when constraints are extremely small |
| BFS with Hash Set (State Graph) | O(S * N) | O(S) | General case where many transformations exist and duplicate states must be avoided |
| BFS with Pruning | O(S * N) | O(S) | When additional constraints allow pruning impossible states early |