How do you solve Visit Array Positions to Maximize Score on LeetCode?

Visit Array Positions to Maximize Score (LeetCode #2786) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n) space complexity. Key insight: Understanding parity is crucial for calculating score adjustments.

What is the brute force solution for Visit Array Positions to Maximize Score?

The brute force approach for Visit Array Positions to Maximize Score is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we explore every possible path from the starting position to all other positions in the array. We calculate the score for each path and keep track of the maximum score found. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Visit Array Positions to Maximize Score?

Visit Array Positions to Maximize Score uses the following concepts: Array, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Greedy Algorithms.

What are the interview tips for Visit Array Positions to Maximize Score?

Clarify the problem requirements and constraints before coding. Think aloud while solving to demonstrate your thought process. Test edge cases and provide examples to validate your solution.

What are common mistakes when solving Visit Array Positions to Maximize Score?

Not considering the parity when calculating scores. Overlooking the need to reset scores when moving to a new position.

#2786

Visit Array Positions to Maximize Score

Medium

ArrayDynamic ProgrammingDynamic ProgrammingGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n²)Space O(n)

The optimal solution uses dynamic programming to maintain the maximum score achievable up to each position in the array. By considering the parity of the current position and the previous maximum scores, we can efficiently calculate the best score without redundant calculations.

⚙️

Algorithm

5 steps

1Step 1: Initialize a dp array where dp[i] represents the maximum score achievable at position i.
2Step 2: Set dp[0] = nums[0] since we start at the first position.
3Step 3: Iterate through the array, and for each position i, check all previous positions j (where j < i).
4Step 4: Update dp[i] based on whether the parities of nums[i] and nums[j] are the same or different, adjusting for score loss accordingly.
5Step 5: Return the maximum value in the dp array.

solution.py12 lines

1def maxScore(nums, x):
2    n = len(nums)
3    dp = [0] * n
4    dp[0] = nums[0]
5    for i in range(1, n):
6        dp[i] = nums[i]
7        for j in range(i):
8            if (nums[i] % 2) != (nums[j] % 2):
9                dp[i] = max(dp[i], dp[j] + nums[i] - x)
10            else:
11                dp[i] = max(dp[i], dp[j] + nums[i])
12    return max(dp)

ℹ

Complexity note: The time complexity remains O(n²) due to the nested loops, but we gain clarity and structure with dynamic programming. The space complexity is O(n) due to the dp array used to store intermediate results.

1Understanding parity is crucial for calculating score adjustments.
2Dynamic programming helps break down the problem into manageable subproblems.

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