How do you solve Get Maximum in Generated Array on LeetCode?

Get Maximum in Generated Array (LeetCode #1646) 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: The generated array follows a specific pattern based on the index, allowing for efficient calculation.

What is the brute force solution for Get Maximum in Generated Array?

The brute force approach for Get Maximum in Generated Array is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating the entire array based on the given rules and then simply finding the maximum value in that array. This is straightforward but inefficient for larger values of n. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Get Maximum in Generated Array?

Get Maximum in Generated Array uses the following concepts: Array, Simulation. The recommended patterns to study are: Array, Simulation.

What are the interview tips for Get Maximum in Generated Array?

Always clarify the problem statement and constraints before jumping into coding. Think about edge cases, especially small values of n. Explain your thought process while coding; it shows your understanding and helps the interviewer follow along.

What are common mistakes when solving Get Maximum in Generated Array?

Failing to handle the base case where n = 0 correctly. Overcomplicating the solution by trying to optimize without understanding the generation rules.

#1646

Get Maximum in Generated Array

Easy

ArraySimulationArraySimulation

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)

The optimal solution builds the array in a single pass while calculating the maximum value on-the-fly. This approach reduces unnecessary iterations and leverages the properties of the generated array.

⚙️

Algorithm

3 steps

1Step 1: Initialize an array nums of length n + 1.
2Step 2: Set nums[0] = 0 and nums[1] = 1.
3Step 3: For each i from 1 to n // 2, fill nums[2 * i] and nums[2 * i + 1] as specified, while keeping track of the maximum value encountered.

solution.py14 lines

1# Full working Python code
2
3def getMaximumGenerated(n):
4    if n == 0:
5        return 0
6    nums = [0] * (n + 1)
7    nums[1] = 1
8    max_value = 1
9    for i in range(1, (n // 2) + 1):
10        nums[2 * i] = nums[i]
11        if 2 * i + 1 <= n:
12            nums[2 * i + 1] = nums[i] + nums[i + 1]
13            max_value = max(max_value, nums[2 * i + 1])
14    return max_value

ℹ

Complexity note: The time complexity is O(n) because we only traverse the array once to fill it, and the space complexity is O(n) due to the storage of the nums array.

1The generated array follows a specific pattern based on the index, allowing for efficient calculation.
2Understanding the recursive nature of the array generation helps in optimizing the solution.

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