LeetCode #3690 — MEDIUM

Split and Merge Array Transformation

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode
The Problem

Problem Statement

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:

  1. Choose a subarray nums1[L..R].
  2. Remove that subarray, leaving the prefix nums1[0..L-1] (empty if L = 0) and the suffix nums1[R+1..n-1] (empty if R = n - 1).
  3. Re-insert the removed subarray (in its original order) at any position in the remaining array (i.e., between any two elements, at the very start, or at the very end).

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:

  • Split out the subarray [3] (L = 0, R = 0); the remaining array is [1,2].
  • Insert [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:

  • Remove [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].
  • Remove [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].
  • Remove [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] <= 105
  • nums2 is a permutation of nums1.

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: 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: Choose a subarray nums1[L..R]. Remove that subarray, leaving the prefix nums1[0..L-1] (empty if L = 0) and the suffix nums1[R+1..n-1] (empty if R = n - 1). Re-insert the removed subarray (in its original order) at any position in the remaining array (i.e., between any two elements, at the very start, or at the very end). Return the minimum number of split-and-merge operations needed to transform nums1 into nums2.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map

Example 1

[3,1,2]
[1,2,3]

Example 2

[1,1,2,3,4,5]
[5,4,3,2,1,1]
Step 02

Core Insight

What unlocks the optimal approach

  • Use <code>BFS</code> over the space of array states, starting from <code>nums1</code> and aiming for <code>nums2</code>.
  • Represent each state as an array (or tuple) and enqueue it alongside its current operation count.
  • Maintain a visited set (e.g. a hash set or dictionary keyed by the state) to avoid revisiting the same configuration.
  • For each dequeued state, generate all possible "split-and-merge" successors by choosing every valid subarray <code>[L..R]</code>, removing it, and inserting it at every possible position.
  • Stop as soon as you dequeue <code>nums2</code>, and return its associated operation count.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Upper-end input sizes
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3690: Split and Merge Array Transformation
class Solution {
    public int minSplitMerge(int[] nums1, int[] nums2) {
        int n = nums1.length;
        List<Integer> target = toList(nums2);
        List<Integer> start = toList(nums1);
        List<List<Integer>> q = List.of(start);
        Set<List<Integer>> vis = new HashSet<>();
        vis.add(start);
        for (int ans = 0;; ++ans) {
            var t = q;
            q = new ArrayList<>();
            for (var cur : t) {
                if (cur.equals(target)) {
                    return ans;
                }
                for (int l = 0; l < n; ++l) {
                    for (int r = l; r < n; ++r) {
                        List<Integer> remain = new ArrayList<>();
                        for (int i = 0; i < l; ++i) {
                            remain.add(cur.get(i));
                        }
                        for (int i = r + 1; i < n; ++i) {
                            remain.add(cur.get(i));
                        }
                        List<Integer> sub = cur.subList(l, r + 1);
                        for (int i = 0; i <= remain.size(); ++i) {
                            List<Integer> nxt = new ArrayList<>();
                            for (int j = 0; j < i; ++j) {
                                nxt.add(remain.get(j));
                            }
                            for (int x : sub) {
                                nxt.add(x);
                            }
                            for (int j = i; j < remain.size(); ++j) {
                                nxt.add(remain.get(j));
                            }
                            if (vis.add(nxt)) {
                                q.add(nxt);
                            }
                        }
                    }
                }
            }
        }
    }

    private List<Integer> toList(int[] arr) {
        List<Integer> res = new ArrayList<>(arr.length);
        for (int x : arr) {
            res.add(x);
        }
        return res;
    }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n! × n^4)
Space
O(n! × n)

Approach Breakdown

BRUTE FORCE
O(n²) time
O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED
O(n) time
O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

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