How do you solve Trionic Array I on LeetCode?

Trionic Array I (LeetCode #3637) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Trionic arrays have a specific structure of segments.

What is the time complexity of Trionic Array I?

The optimal solution for Trionic Array I runs in O(n) time and uses O(1) space. Single pass through the array, hence O(n). No extra space is used.

What is the brute force solution for Trionic Array I?

The brute force approach for Trionic Array I is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check all possible pairs of indices (p, q) to see if they satisfy the trionic conditions. This means iterating through all combinations of p and q and checking the segments. The optimal Optimal Solution approach improves this to O(n).

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

Trionic Array I uses the following concepts: Array. The recommended patterns to study are: Array.

What are the interview tips for Trionic Array I?

Clarify the problem requirements. Discuss edge cases with the interviewer. Explain your thought process clearly.

What are common mistakes when solving Trionic Array I?

Not checking boundary conditions for p and q. Confusing increasing and decreasing checks.

#3637

Trionic Array I

Easy

ArrayArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

We can traverse the array and identify the increasing and decreasing segments in a single pass, reducing the need for nested loops.

⚙️

Algorithm

3 steps

1Step 1: Traverse the array to find the first increasing segment until it stops.
2Step 2: Continue to find the decreasing segment until it stops.
3Step 3: Finally, check if the remaining segment is increasing.

solution.py9 lines

1def isTrionic(nums):
2    n = len(nums)
3    i = 0
4    while i < n - 1 and nums[i] < nums[i + 1]:
5        i += 1
6    if i == 0 or i == n - 1: return False
7    while i < n - 1 and nums[i] > nums[i + 1]:
8        i += 1
9    return i < n - 1 and all(nums[j] < nums[j + 1] for j in range(i, n - 1))

ℹ

Complexity note: Single pass through the array, hence O(n). No extra space is used.

1Trionic arrays have a specific structure of segments.
2Identifying boundaries of segments is key to solving the problem.

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