How do you solve Check if There is a Valid Partition For The Array on LeetCode?

Check if There is a Valid Partition For The Array (LeetCode #2369) 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: Dynamic programming is a powerful technique for problems involving partitions or subsequences.

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

The optimal solution for Check if There is a Valid Partition For The Array runs in O(n) time and uses O(n) space. The time complexity is O(n) because we only make a single pass through the array, and the space complexity is O(n) due to the DP array we maintain.

What is the brute force solution for Check if There is a Valid Partition For The Array?

The brute force approach for Check if There is a Valid Partition For The Array is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks all possible partitions of the array to see if any of them meet the valid conditions. This is straightforward but inefficient as it explores all combinations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Check if There is a Valid Partition For The Array?

Check if There is a Valid Partition For The Array uses the following concepts: Array, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Array.

What are the interview tips for Check if There is a Valid Partition For The Array?

Always clarify the problem statement and constraints before jumping into coding. Think about edge cases, such as the smallest and largest possible inputs. Explain your thought process clearly while coding; it shows your understanding.

What are common mistakes when solving Check if There is a Valid Partition For The Array?

Failing to consider all valid partition conditions, especially the increasing sequence. Not initializing the DP array correctly or misunderstanding its purpose.

#2369

Check if There is a Valid Partition For The Array

Medium

ArrayDynamic ProgrammingDynamic ProgrammingArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

The optimal solution uses dynamic programming to build a solution incrementally. We maintain a DP array where each entry indicates if the subarray up to that index can be partitioned validly.

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Algorithm

4 steps

1Step 1: Initialize a DP array of size n+1 with false values, set DP[0] to true.
2Step 2: Iterate through the nums array, checking for valid partitions of size 2 and 3.
3Step 3: Update the DP array based on the conditions for valid partitions.
4Step 4: Return the value of DP[n] to determine if the entire array can be partitioned.

solution.py10 lines

1def hasValidPartition(nums):
2    n = len(nums)
3    dp = [False] * (n + 1)
4    dp[0] = True
5    for i in range(2, n + 1):
6        if nums[i - 1] == nums[i - 2]:
7            dp[i] = dp[i] or dp[i - 2]
8        if i > 2 and (nums[i - 1] == nums[i - 2] == nums[i - 3] or (nums[i - 1] == nums[i - 2] + 1 and nums[i - 2] == nums[i - 3] + 1)):
9            dp[i] = dp[i] or dp[i - 3]
10    return dp[n]

ℹ

Complexity note: The time complexity is O(n) because we only make a single pass through the array, and the space complexity is O(n) due to the DP array we maintain.

1Dynamic programming is a powerful technique for problems involving partitions or subsequences.
2Recognizing valid patterns in subarrays can significantly reduce the complexity of the solution.

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