Problem Description

You are provided with an int array nums that is 0-indexed and a positive int x. The goal is to calculate the maximum total score you can obtain starting at position 0 of the array and moving to any subsequent position. The rules are outlined as follows:

• You can move from your current position i to any position j such that i < j.
• When you visit position i, you earn nums[i] points added to your score.
• If you move between positions i and j and nums[i] and nums[j] have different parities (one is odd, the other is even), you lose x points from your score.

You kick off with nums[0] points, and you have to figure out the maximum score that can be achieved under these conditions.

Intuition

The task is to maximize the score while taking into account that moving between numbers of different parity comes with a penalty of x points. To do this, we need to use dynamic programming to track the highest scores while considering the parity of the current number.

Here's an outline of the approach:

• Initialize a list f with two elements, set to negative infinity [-inf, -inf]. This list will keep track of the max scores for even and odd indices.
• Set the element of f corresponding to the parity of nums[0] (even or odd) to nums[0]. This represents the score starting at position 0.
• Iterate through the nums array starting from index 1. For each value v at index i:
  • Calculate the maximum score when staying on the same parity (f[v & 1] + v) and when changing parity (f[v & 1 ^ 1] + v - x).
  • Update f[v & 1] with the highest score from the above step.
• After processing all elements, the maximum score will be the maximum element from f.

The key intuition in this solution comes from recognizing that at any index i in the array, you have two scenarios to consider:

1. The last score came from an index with the same parity as i. In this case, you just add the current value to the previous score since no penalty is incurred.
2. The last score came from an index with different parity. Here, you add the current value to the previous score from the opposite parity and subtract the penalty x.

This process will lead us to the highest possible score, taking into account the penalty for switching parities.

Learn more about Dynamic Programming patterns.

Solution Approach

The implementation uses a dynamic programming approach to compute the maximum score. Here's a step-by-step walkthrough:

• Firstly, the algorithm initializes a list f with two elements, [-inf, -inf]. This record is to keep track of the two possible states for our score related to parity: even (0) and odd (1). In Python, -inf denotes negative infinity which is a useful placeholder for "not yet computed or improbably low score."

• The first element of nums is factored into our initial state. Since we always start at position 0, f[nums[0] & 1] is set to nums[0]. The expression nums[0] & 1 will be 0 if nums[0] is even, and 1 if it is odd, so it determines the index of f that gets updated.

• The algorithm then iterates through elements in nums starting from index 1. For each value v, it computes the two possible scenarios:

  • Staying on the same parity (f[v & 1] + v),
  • Switching parity (f[v & 1 ^ 1] + v - x). The ^ operator is a bitwise XOR, which flips the bit, effectively getting us the other parity.

  The update for the score at parity v & 1 chooses whichever of these two possibilities gives a higher score. This is done by the max function.

• After evaluating all elements in nums, the maximum score is the highest value in f, which can be obtained using Python's built-in max(f) function.

The code snippet provided succinctly translates this approach into a Python function as part of a Solution class:

1class Solution:
2    def maxScore(self, nums: List[int], x: int) -> int:
3        f = [-inf] * 2
4        f[nums[0] & 1] = nums[0]
5        for v in nums[1:]:
6            f[v & 1] = max(f[v & 1] + v, f[v & 1 ^ 1] + v - x)
7        return max(f)

Each iteration effectively represents a choice at every position i with a value v from nums: taking its score as part of the existing parity sequence or starting a new sequence of the opposite parity with an x penalty. The algorithm dynamically keeps track of the best choice by updating only the score of the relevant parity after each decision. This pattern avoids the need for recursive traversal through all potential positions and parities, significantly reducing the complexity of the problem.

Example Walkthrough

Let's illustrate the solution approach with a small example:

Suppose we have the array nums = [4, 5, 2, 7, 3] and the penalty x = 3.

We initiate the variable f with two elements [-inf, -inf] to keep track of the max scores for the even (f[0]) and odd indices (f[1]).

• f = [-inf, -inf]

Starting at position 0, nums[0] = 4, which is even, so we update f[0] with the value of nums[0].

• f = [4, -inf]

We move to nums[1] = 5, which is odd:

• If we stay with odd, f[1] would become 5 (since -inf + 5 is just 5), but there's a catch: we start from an even index so we must apply the penalty x. The new score would be 5 - 3 = 2.
• If we switch to even, f[0] + nums[1] would be 4 + 5 = 9. We take the max of both, which is 9, so we update f[1].
• f = [4, 9]

Next, nums[2] = 2, which is even:

• Staying with even parity, f[0] + 2 = 4 + 2 = 6.
• Switching to odd, f[1] + 2 - x = 9 + 2 - 3 = 8. The higher score is 8, so f[0] becomes 8.
• f = [8, 9]

With nums[3] = 7, which is odd:

• Staying odd, f[1] + 7 = 9 + 7 = 16.
• Switching to even, f[0] + 7 - x = 8 + 7 - 3 = 12. We take the max which is 16 and update f[1].
• f = [8, 16]

Lastly, nums[4] = 3, which is odd:

• Staying odd, f[1] + 3 = 16 + 3 = 19.
• Switching to even, f[0] + 3 - x = 8 + 3 - 3 = 8. The max is 19, so f[1] remains 19.
• f = [8, 19]

The maximum score we can get is max(f), which is 19.

The entire process demonstrates the dynamic programming algorithm's effectiveness in computing the maximum score by considering the penalty for switching between even and odd numbers. Each decision is based on whether to continue the sequence of the current parity or start a new one of the opposite parity with a penalty. The algorithm avoids the need to check each path separately, instead of using a running tally that gets updated in each step, which is considerably more efficient.

Solution Implementation

Time and Space Complexity

The given Python code snippet aims to calculate a certain "maximum score" by iterating through the input list nums and applying some operations based on the elements' parity (odd or even) and a given integer x. To analyze the time and space complexity of this code, let's consider n to be the length of the input list nums.

Time Complexity

The time complexity of this code is determined by the number of operations performed in the for-loop that iterates through the nums list:

• The for-loop runs (n - 1) times, as it starts from the second element in nums.
• Within the loop, a constant number of operations are executed: two bitwise AND operations, four direct accesses by index to the list f, up to two max operations, and a few arithmetic operations.

Since these operations inside the loop are all of constant time complexity, the overall time complexity of the loop is O(n - 1). Simplifying this, we get:

O(n - 1) = O(n).

Thus, the time complexity of the code is O(n).

Space Complexity

As for the space complexity:

• A new list f of fixed size 2 is created. This does not depend on the size of the input and is thus O(1).
• Variable v is a single integer that is used to iterate through nums, which is also O(1) space.
• There are no other data structures or recursive calls that use additional space that scales with the input size.

Therefore, the space complexity of the code is O(1).

Without a reference answer provided alongside the code, the analysis is based solely on the provided snippet.

Learn more about how to find time and space complexity quickly using problem constraints.