Problem Description

You are given an integer array nums with length n. Your task is to split this array into subarrays to maximize the total cost.

The cost of a subarray nums[l..r] (where 0 <= l <= r < n) is calculated using an alternating sum pattern:

• cost(l, r) = nums[l] - nums[l+1] + nums[l+2] - nums[l+3] + ...
• More formally: cost(l, r) = nums[l] - nums[l + 1] + ... + nums[r] * (-1)^(r - l)

The sign of each element alternates based on its position within the subarray:

• The first element is positive
• The second element is negative
• The third element is positive
• And so on...

When you split the array into k subarrays at indices i₁, i₂, ..., i_(k-1):

• The first subarray is from index 0 to i₁
• The second subarray is from index i₁+1 to i₂
• And so on...
• The last subarray is from index i_(k-1)+1 to n-1

The total cost is the sum of all individual subarray costs: cost(0, i₁) + cost(i₁ + 1, i₂) + ... + cost(i_(k-1) + 1, n - 1)

Each element must belong to exactly one subarray. You can choose not to split the array at all (k=1), in which case the total cost is simply cost(0, n-1).

Your goal is to find the optimal way to split the array that results in the maximum possible total cost.

For example, if nums = [1, 2, 3, 4]:

• Without splitting: cost = 1 - 2 + 3 - 4 = -2
• Splitting at index 0: cost(0,0) + cost(1,3) = 1 + (2 - 3 + 4) = 1 + 3 = 4
• Splitting at index 1: cost(0,1) + cost(2,3) = (1 - 2) + (3 - 4) = -1 + (-1) = -2
• And so on...

The key insight is that by strategically placing splits, you can control which elements get positive or negative signs in the alternating sum, potentially increasing the total cost.

Intuition

Let's think about what happens when we split the array. Each split essentially "resets" the alternating pattern. The key observation is that when we start a new subarray, the first element of that subarray always gets a positive sign.

Consider a simple example: if we have [5, -3] and don't split, we get 5 - (-3) = 8. But if we split after 5, we get 5 + (-3) = 2. Notice how the -3 changed from being subtracted (double negative making it positive) to being added (staying negative).

This leads us to realize that splitting is essentially a way to control the signs of elements. When we encounter a negative number that would normally get a positive coefficient (due to the alternating pattern), we might want to split before it to make it the start of a new subarray (giving it a positive sign but keeping its negative value).

Now, let's think about this problem differently. Instead of thinking about where to split, we can think about whether each element should:

1. Continue the alternating pattern from the previous element (get flipped)
2. Start a new subarray (not get flipped, always positive coefficient)

This transforms the problem into a decision at each position: should we "flip" the current element's sign (continue the pattern) or not (start fresh)?

The dynamic programming insight comes from recognizing that our decision at position i depends on what happened at position i-1:

• If the previous element was not flipped (had a positive sign), the current element can either be flipped (continue pattern with negative sign) or not flipped (start new subarray)
• If the previous element was flipped (had a negative sign), the current element must not be flipped (it either continues the pattern with positive sign, or starts a new subarray with positive sign - both result in the same operation)

This leads to a state representation: dfs(i, j) where:

• i is the current position
• j indicates whether the current element can be flipped (1 if yes, 0 if no)

The value j=0 means the previous element was flipped, so the current one cannot be. The value j=1 means the previous element was not flipped, so we have a choice for the current one.

Starting with dfs(0, 0) makes sense because the first element of the entire array cannot be flipped - it must have a positive coefficient.

Learn more about Dynamic Programming patterns.

Solution Approach

The solution uses dynamic programming with memoization to solve this problem efficiently. We implement a recursive function dfs(i, j) that explores all possible ways to split the array.

State Definition:

• dfs(i, j) represents the maximum cost we can get starting from index i to the end of the array
• j indicates whether the element at index i can be flipped:
  • j = 0: the element cannot be flipped (must have positive coefficient)
  • j = 1: the element can be flipped (we can choose to give it negative coefficient)

Base Case:

• If i >= len(nums), we've processed all elements, so return 0

Recursive Cases:

1. Element not flipped (starts new subarray or continues with positive coefficient):

   • We add nums[i] with a positive coefficient
   • The next element can potentially be flipped, so we call dfs(i + 1, 1)
   • Cost: nums[i] + dfs(i + 1, 1)

2. Element flipped (only when j = 1):

   • We add nums[i] with a negative coefficient
   • The next element cannot be flipped (must have positive coefficient in the alternating pattern)
   • Cost: -nums[i] + dfs(i + 1, 0)
   • We take the maximum between flipping and not flipping

Implementation Details:

@cache  # Memoization to avoid recomputation
def dfs(i: int, j: int) -> int:
    if i >= len(nums):
        return 0
  
    # Option 1: Don't flip current element
    ans = nums[i] + dfs(i + 1, 1)
  
    # Option 2: Flip current element (only if allowed)
    if j == 1:
        ans = max(ans, -nums[i] + dfs(i + 1, 0))
  
    return ans

Starting Point:

• We call dfs(0, 0) because the first element of the array must have a positive coefficient (cannot be flipped from its initial state)

Time Complexity: O(n) where n is the length of the array, as each state (i, j) is computed at most once due to memoization.

Space Complexity: O(n) for the recursion stack and memoization cache.

The beauty of this approach is that it implicitly handles all possible splits by deciding at each position whether to continue the alternating pattern or start fresh, effectively determining where the subarray boundaries should be placed to maximize the total cost.

Example Walkthrough

Let's walk through the solution with nums = [1, -2, 3, -4].

We start with dfs(0, 0) - processing index 0, where the element cannot be flipped.

Step 1: dfs(0, 0)

• At index 0, value is 1
• Since j = 0, we cannot flip, so we must use +1
• Next call: 1 + dfs(1, 1)

Step 2: dfs(1, 1)

• At index 1, value is -2
• Since j = 1, we have two choices:
  • Don't flip: -2 + dfs(2, 1)
  • Flip: -(-2) + dfs(2, 0) = 2 + dfs(2, 0)
• We'll explore both paths

Path A: Don't flip at index 1

• dfs(2, 1): value is 3
  • Don't flip: 3 + dfs(3, 1)
  • Flip: -3 + dfs(3, 0)

Continuing Path A (don't flip 3):

• dfs(3, 1): value is -4
  • Don't flip: -4 + dfs(4, 1) = -4 + 0 = -4
  • Flip: 4 + dfs(4, 0) = 4 + 0 = 4
  • Maximum = 4
• So dfs(2, 1) = 3 + 4 = 7
• Path A total: 1 + (-2) + 7 = 6

Path B: Flip at index 1

• dfs(2, 0): value is 3
  • Since j = 0, must not flip: 3 + dfs(3, 1)
• dfs(3, 1): value is -4
  • Don't flip: -4
  • Flip: 4
  • Maximum = 4
• So dfs(2, 0) = 3 + 4 = 7
• Path B total: 1 + 2 + 7 = 10

Final Result:

• dfs(1, 1) = max(6, 10) = 10
• dfs(0, 0) = 1 + 10 = 11

The optimal solution gives us a total cost of 11, which corresponds to:

• Taking 1 with positive sign
• Flipping -2 to get +2 (this effectively starts a new subarray at index 1)
• Taking 3 with positive sign (continuing from the "reset")
• Flipping -4 to get +4 (alternating pattern within the subarray)

This translates to splitting the array as [1] | [-2, 3, -4]:

• First subarray: cost = 1
• Second subarray: cost = -2 - 3 + (-4) = -2 + 3 + 4 = 10
• Total: 1 + 10 = 11

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the array nums. This is because the recursive function dfs(i, j) has at most 2n distinct states - for each index i from 0 to n-1, the parameter j can only be 0 or 1. Due to the @cache decorator (memoization), each state is computed at most once. Since each state performs O(1) operations (excluding recursive calls), the total time complexity is O(n).

The space complexity is O(n), where n is the length of the array nums. This consists of two components:

1. The recursion stack depth, which can go up to O(n) in the worst case when we traverse through all elements sequentially
2. The memoization cache, which stores at most 2n states (for all combinations of i and j), requiring O(n) space

Therefore, the overall space complexity is O(n).

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

Common Pitfalls

Pitfall 1: Misunderstanding the State Representation

The Problem: A common mistake is misinterpreting what the can_subtract (or j) parameter represents. Developers often think:

• j = 0 means "this element must be positive"
• j = 1 means "this element must be negative"

This leads to incorrect implementations like:

# WRONG interpretation
def dfs(i, must_be_negative):
    if must_be_negative:
        return -nums[i] + dfs(i + 1, 0)  # Force negative
    else:
        return nums[i] + dfs(i + 1, 1)   # Force positive

The Reality:

• j = 0 means "this element CANNOT be flipped" (must stay positive as the first element of its subarray)
• j = 1 means "this element CAN be flipped" (we have a choice to make it negative or keep it positive)

The Solution: Always remember that j = 1 gives us a choice - we can either:

1. Continue the alternating pattern (make it negative)
2. Start a new subarray (keep it positive)

The correct implementation explores both options when j = 1:

def dfs(i, can_flip):
    # Always try keeping positive (starting new subarray)
    result = nums[i] + dfs(i + 1, 1)
  
    # If we can flip, also try making it negative
    if can_flip == 1:
        result = max(result, -nums[i] + dfs(i + 1, 0))
  
    return result

Pitfall 2: Incorrect Initial Call

The Problem: Starting the recursion with dfs(0, 1) instead of dfs(0, 0):

# WRONG
return dfs(0, 1)  # This allows flipping the first element

This suggests the very first element of the array could be negative, which violates the problem constraints. The first element of any subarray must always be positive in the alternating sum pattern.

The Solution: Always start with dfs(0, 0) because the first element of the entire array is also the first element of its subarray and must have a positive coefficient:

# CORRECT
return dfs(0, 0)  # First element cannot be flipped

Pitfall 3: Forgetting Memoization

The Problem: Implementing the recursive solution without memoization:

# WRONG - No memoization
def dfs(i, j):
    if i >= len(nums):
        return 0
    ans = nums[i] + dfs(i + 1, 1)
    if j == 1:
        ans = max(ans, -nums[i] + dfs(i + 1, 0))
    return ans

This leads to exponential time complexity O(2^n) as we recompute the same states multiple times.

The Solution: Use @cache decorator or implement manual memoization:

# Using @cache decorator
from functools import cache

@cache
def dfs(i, j):
    # ... implementation

# Or manual memoization
memo = {}
def dfs(i, j):
    if (i, j) in memo:
        return memo[(i, j)]
    # ... compute result
    memo[(i, j)] = result
    return result

Pitfall 4: Confusion About Subarray Boundaries

The Problem: Thinking that choosing to "not flip" always means continuing the current subarray, and "flipping" means starting a new one. This mental model is backwards.

The Correct Understanding:

• When we add an element positively (nums[i]), we're either:

  • Starting a NEW subarray (if previous was negative)
  • Continuing with the alternating pattern (if previous was positive)

• When we subtract an element (-nums[i]), we're:

  • ALWAYS continuing the current subarray's alternating pattern
  • This is why the next element after a subtraction cannot be flipped (j = 0)

The algorithm cleverly handles subarray boundaries implicitly through these state transitions rather than explicitly tracking where splits occur.