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Visit Array Positions to Maximize Score

Difficulty: Medium

Problem Description

You are given a 0-indexed integer array nums and a positive integer x. You are initially at position 0 in the array and can visit other positions. For each position you visit, you get a score of nums[i]. However, if you move between positions with different parities, you lose a score of x. Return the maximum total score you can achieve.

Key Insights

  • You can move from position i to any position j where i < j, allowing for a variety of paths through the array.
  • The score is affected by the parity of the numbers, as moving between positions with different parities incurs a penalty.
  • To maximize the score, it is essential to identify groups of numbers with the same parity and sum them efficiently while accounting for any penalties incurred.

Space and Time Complexity

Time Complexity: O(n)
Space Complexity: O(1)

Solution

The solution involves iterating through the array while maintaining two sums: one for even-indexed numbers and one for odd-indexed numbers. Starting from the initial position (index 0), we calculate the maximum possible score by considering the total scores of both even and odd groups, applying penalties where necessary based on the last visited position's parity. The algorithm ensures that we can quickly determine the optimal path without needing to check every possible combination.

Code Solutions

def maxScore(nums, x):
    even_sum = 0
    odd_sum = 0
    n = len(nums)
    
    # Calculate the total sum for even and odd indexed numbers
    for num in nums:
        if num % 2 == 0:
            even_sum += num
        else:
            odd_sum += num
    
    # Start with the score at position 0
    max_score = nums[0]
    
    # Determine if the first number is even or odd
    if nums[0] % 2 == 0:
        max_score += even_sum
    else:
        max_score += odd_sum
    
    # If moving from the first number causes a penalty
    if n > 1:
        if nums[0] % 2 == 0:
            max_score -= x
        else:
            max_score -= x
            
    return max_score
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