Problem Description

In this problem, we're given an integer array nums, and our goal is to partition this array into one or more contiguous subarrays. A partition is considered valid if each resulting subarray meets one of three specific conditions:

1. The subarray consists of exactly 2 equal elements, such as [2, 2].
2. The subarray consists of exactly 3 equal elements, like [4, 4, 4].
3. The subarray consists of exactly 3 consecutive increasing elements, which means the difference between adjacent elements is 1. An example of such a subarray is [3, 4, 5].

We need to determine if there's at least one such valid partition for the given array, and we should return true if there is, or false otherwise.

Intuition

The intuition behind the solution is to use Depth First Search (DFS) to explore all possible ways to partition the array. Starting from the beginning of the array, we recursively check if we can form a valid subarray according to the rules given. If we can, we proceed to explore partitions of the remaining part of the array.

Here's the thought process for DFS:

1. If we reach the end of the array (index i equals the length of the array n), it means we've successfully partitioned the array into valid subarrays, so we return true.

2. Starting from the current index i, we check if we can form a valid subarray with the next one or two elements based on the given conditions:

   • If the current and the next element are the same, we check the possibility of forming a valid partition with the subsequent elements by calling the DFS for the index i + 2.
   • If the current and the next two elements are the same, we again call the DFS for index i + 3.
   • If the current element and the next two elements form a sequence of three consecutive increasing numbers, we call the DFS for the index i + 3.

3. We consider a partition valid if any of these subarray conditions lead to a successful partition of the rest of the array.

4. To optimize the solution, we use memoization (@cache) to avoid re-computing the same state of the array multiple times.

By exploring all possible partitions in a DFS manner and using memoization to reduce duplicate work, we can determine if the array has at least one valid partition in an efficient way.

Learn more about Dynamic Programming patterns.

Solution Approach

The solution uses a recursive Depth First Search (DFS) approach with memoization to avoid redundant computations. In the code, we define a recursive function dfs that attempts to find a valid partition starting from index i. The base case for our recursive function is when i is equal to n, the length of the array, which means we have successfully reached the end of the array with a valid partition and thus return true.

Let's walk through the implementation details:

• Recursion and DFS: The core of the solution is the recursive function dfs(i) which is called with the starting index i to explore all possible valid subarrays from that index onwards.
• Memoization: This function is decorated with @cache, which is a feature in Python used for memoization. It stores the results of the function calls with particular arguments so that future calls with the same arguments can return the stored result immediately, without re-executing the whole function.
• Base Case: When i equals n, we've reached the end and hence return true, implying a successful partition.
• Recursive Calls:
  • If nums[i] is equal to nums[i + 1], we attempt to create a subarray of length 2 by making a recursive call to dfs(i + 2).
  • If nums[i], nums[i + 1], and nums[i + 2] are equal, we attempt to create a subarray of length 3 by making a recursive call to dfs(i + 3).
  • If nums[i], nums[i + 1], and nums[i + 2] form a sequence of three consecutive increasing numbers, we attempt to create a subarray of length 3 by making a recursive call to dfs(i + 3).
• Early Termination: The res variable is used to keep track of whether a valid partition has been found. As soon as we find a valid partition, res becomes true and subsequent calls won't be made since the or operator short-circuits.
• Efficient Checking: To evaluate the conditions, the implementation uses simple if checks and arithmetic comparisons. No additional data structures are required, which keeps the space complexity low.

Finally, dfs(0) is called to initiate the partitioning process starting from the first element of the array, and the result of this call determines whether at least one valid partition exists. If it returns true, the entire array can be partitioned according to the given rules. If not, the array does not have a valid partition configuration.

Example Walkthrough

Let's consider an example nums array to illustrate the solution approach: [3, 3, 2, 2, 1, 1, 2, 3, 4].

Here is a step-by-step walk-through:

1. Start with index i = 0, attempt to find valid subarrays beginning at that index.
2. Since nums[0] and nums[1] are equal (3, 3), try to make a subarray [3, 3] and call dfs(2) for the next index.
3. At index i = 2, nums[2] and nums[3] are also equal (2, 2), so take subarray [2, 2] and call dfs(4) for the next index.
4. At index i = 4, nums[4] and nums[5] are equal (1, 1), form subarray [1, 1] and call dfs(6) for the next index.
5. At index i = 6, nums[6], nums[7], and nums[8] form a sequence of three consecutive increasing numbers (2, 3, 4), so take subarray [2, 3, 4] and call dfs(9) for the next index.
6. Now i = 9 which equals n (the length of nums). According to our base case, we've successfully found a partition for the entire array and return true.

With this walk-through, you can see that the original array can be partitioned into valid subarrays [3, 3], [2, 2], [1, 1], and [2, 3, 4] conforming to the given rules. The DFS approach systematically checks for each subarray possibility at every index, and the use of memoization ensures that we do not perform redundant calculations, leading to an efficient solution.

Solution Implementation

Time and Space Complexity

The given code defines a recursive function dfs that uses memoization to return whether a valid partition exists starting from the i-th index of the array nums. The function recursively tries to partition the array into groups of two or three adjacent elements, either with the same value or forming a consecutive increasing sequence.

Time Complexity:

The time complexity of the dfs function primarily depends on the number of subproblems it needs to solve, which is directly related to the length of nums (n). Due to memoization (through the use of the @cache decorator), each subproblem (each starting index of the array) is solved only once. There are n possible starting indices.

For each index, there are at most three recursive calls to check for three different partitioning conditions. However, these recursive calls don't multiply the complexity because memoization ensures that we do not recompute results for the same inputs.

Therefore, the time complexity of the algorithm is O(n), as each index is processed a constant number of times.

Space Complexity:

The space complexity consists of the memory used for the memoization cache and the stack space used by the recursive calls.

1. Memoization cache: The cache will store results for each index of the array, leading to a space complexity of O(n) for the memoization cache.

2. Recursive stack: In the worst case, the function might end up calling itself recursively n/2 times if the partition consists entirely of pairs of identical numbers ([1, 1, 2, 2, ..., n/2, n/2]). This would lead to a stack depth of n/2. Hence, the worst-case space complexity for the call stack is O(n).

Considering both the memoization cache and the recursive call stack, the overall space complexity is also O(n).

Learn more about how to find time and space complexity quickly using problem constraints.