How do you solve Trionic Array II on LeetCode?

Trionic Array II (LeetCode #3640) 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: Trionic subarrays require careful index selection.

What is the time complexity of Trionic Array II?

The optimal solution for Trionic Array II runs in O(n) time and uses O(n) space. Single pass through the array for each DP phase leads to O(n) complexity.

What is the brute force solution for Trionic Array II?

The brute force approach for Trionic Array II is Brute Force, with O(n²) time complexity and O(1) space complexity. Try every possible combination of indices to find trionic subarrays. This involves checking all possible subarrays and their properties. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Trionic Array II?

Trionic Array II uses the following concepts: Array, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Array.

What are the interview tips for Trionic Array II?

Explain your thought process clearly. Use examples to illustrate your approach. Practice coding under time constraints.

What are common mistakes when solving Trionic Array II?

Not checking boundary conditions for indices. Forgetting to reset DP arrays for each new subarray.

#3640

Trionic Array II

Hard

ArrayDynamic ProgrammingDynamic ProgrammingArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

Use dynamic programming to track maximum sums at each phase of the trionic subarray, reducing redundant calculations.

⚙️

Algorithm

3 steps

1Step 1: Initialize four DP arrays for each phase of the trionic subarray.
2Step 2: Iterate through the array to fill these arrays based on increasing and decreasing conditions.
3Step 3: Calculate the maximum sum using the values from the DP arrays.

solution.py12 lines

1def maxTrionic(nums):
2    n = len(nums)
3    dp0, dp1, dp2, dp3 = [0]*n, [0]*n, [0]*n, [0]*n
4    for i in range(n):
5        dp0[i] = nums[i] if i == 0 else max(dp0[i-1] + nums[i], nums[i])
6    for i in range(n):
7        if i > 0 and nums[i] < nums[i-1]: dp1[i] = dp0[i-1]
8    for i in range(n):
9        if i > 0 and nums[i] > nums[i-1]: dp2[i] = dp1[i-1]
10    for i in range(n):
11        if i > 0 and nums[i] < nums[i-1]: dp3[i] = dp2[i-1]
12    return max(dp3)

ℹ

Complexity note: Single pass through the array for each DP phase leads to O(n) complexity.

1Trionic subarrays require careful index selection.
2Dynamic programming helps avoid redundant checks.

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