#2369

Medium

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Check if There is a Valid Partition For The Array

This problem challenges you to determine if an array can be partitioned into valid subarrays using dynamic programming.

Problem Statement

You are given an integer array nums, and your task is to partition it into contiguous subarrays. A partition is valid if each subarray meets one of the following conditions: 1) All elements in the subarray are the same, 2) All elements in the subarray are consecutive integers. The goal is to return true if there exists at least one valid partition of the array; otherwise, return false.

The problem is reduced to checking if there is a valid partition for a smaller array. Using dynamic programming, we maintain states representing whether a valid partition exists for subarrays ending at each index. Based on the values in the array, we transition between states to ensure the subarrays meet the given conditions.

Examples

Example 1

Input: nums = [4,4,4,5,6]

Output: true

The array can be partitioned into the subarrays [4,4] and [4,5,6]. This partition is valid, so we return true.

Example 2

Input: nums = [1,1,1,2]

Output: false

There is no valid partition for this array.

Constraints

2 <= nums.length <= 105
1 <= nums[i] <= 106

Solution Approach

Dynamic Programming with State Transitions

The solution involves using dynamic programming (DP) to track the possibility of valid partitions up to each index. A state transition occurs based on whether the previous two or three elements can form a valid partition. At each index, we check if the current element continues a valid subarray or if we should start a new subarray.

Base Case and Initialization

The base case of the DP solution initializes the first two elements of the array. If the first two elements form a valid subarray, we mark it as a valid state. The next steps involve extending this base case with each new element.

Iterative State Checking

Using a loop, we iteratively check each index and update the DP table. For each element, we verify if it can extend the partition of the previous two elements or form a valid subarray on its own, thereby updating the DP state.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time and space complexity depend on the approach used for dynamic programming. With an array of length n, the time complexity is O(n) due to the need to process each element once, and the space complexity is O(n) for storing the DP states.

What Interviewers Usually Probe

Candidates should demonstrate knowledge of dynamic programming state transitions and handling subarrays.
Look for candidates who can explain how smaller subarrays lead to valid partitions and how this problem is reduced to checking smaller subproblems.
Candidates should be able to identify the base case and how transitions occur from one state to another.

Common Pitfalls or Variants

Common pitfalls

Forgetting to initialize the base case properly can lead to incorrect results.
Overcomplicating the problem by not recognizing the simple state transitions between valid subarrays.
Failing to handle the edge case of very small arrays properly can lead to unexpected outcomes.

Follow-up variants

Changing the condition of valid subarrays from equality to some other rule.
Introducing a constraint that subarrays must not only be valid but also sorted in a particular order.
Extending the problem to check for multiple valid partitions rather than just one.

FAQ

What is the primary technique used in solving the 'Check if There is a Valid Partition For The Array' problem?

The primary technique used in this problem is state transition dynamic programming, where the state of a valid partition is updated as we progress through the array.

How can dynamic programming help in reducing the problem complexity in this problem?

Dynamic programming reduces the complexity by breaking the problem into smaller subproblems, allowing us to check valid partitions efficiently without redundant calculations.

What are the conditions for a valid partition in the 'Check if There is a Valid Partition For The Array' problem?

A valid partition occurs when each subarray consists of either identical elements or consecutive integers.

What is the time complexity of the solution for the 'Check if There is a Valid Partition For The Array' problem?

The time complexity of the solution is O(n), where n is the length of the array, since each element is processed only once.

How does the base case influence the dynamic programming solution?

The base case initializes the first two elements of the array, marking whether they form a valid partition. This sets the stage for subsequent state transitions in the dynamic programming process.

terminal

Solution

Solution 1: Memoization Search

We design a function $dfs(i)$, which represents whether there is a valid partition starting from index $i$. So the answer is $dfs(0)$.

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class Solution:
    def validPartition(self, nums: List[int]) -> bool:
        @cache
        def dfs(i: int) -> bool:
            if i >= n:
                return True
            a = i + 1 < n and nums[i] == nums[i + 1]
            b = i + 2 < n and nums[i] == nums[i + 1] == nums[i + 2]
            c = (
                i + 2 < n
                and nums[i + 1] - nums[i] == 1
                and nums[i + 2] - nums[i + 1] == 1
            )
            return (a and dfs(i + 2)) or ((b or c) and dfs(i + 3))

        n = len(nums)
        return dfs(0)

Solution 2: Dynamic Programming

We can convert the memoization search in Solution 1 into dynamic programming.

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class Solution:
    def validPartition(self, nums: List[int]) -> bool:
        @cache
        def dfs(i: int) -> bool:
            if i >= n:
                return True
            a = i + 1 < n and nums[i] == nums[i + 1]
            b = i + 2 < n and nums[i] == nums[i + 1] == nums[i + 2]
            c = (
                i + 2 < n
                and nums[i + 1] - nums[i] == 1
                and nums[i + 2] - nums[i + 1] == 1
            )
            return (a and dfs(i + 2)) or ((b or c) and dfs(i + 3))

        n = len(nums)
        return dfs(0)

Continue Practicing

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