Problem Description

You are given a 0-indexed array nums. The alternating sum of an array is calculated by taking the sum of elements at even indices and subtracting the sum of elements at odd indices.

For example, for the array [4,2,5,3], the alternating sum is (4 + 5) - (2 + 3) = 4, where elements at indices 0 and 2 are added, and elements at indices 1 and 3 are subtracted.

Your task is to find the maximum alternating sum of any subsequence from the array nums.

Key points to understand:

• A subsequence is formed by deleting some (possibly zero) elements from the original array without changing the relative order of remaining elements
• After selecting a subsequence, the elements are reindexed starting from 0
• You need to calculate the alternating sum of this reindexed subsequence
• Return the maximum possible alternating sum among all possible subsequences

For instance, if you have [4,2,3,7,2,1,4] and choose the subsequence [2,7,4], this subsequence gets reindexed as index 0, 1, 2. The alternating sum would be 2 - 7 + 4 = -1.

The challenge is to select the optimal subsequence that maximizes the alternating sum when the selected elements are reindexed and the alternating sum formula is applied.

Intuition

When we pick elements for our subsequence, we need to decide whether each element should be placed at an even index (added) or odd index (subtracted) in the reindexed subsequence. This leads us to think about the problem in terms of states.

At any point while traversing the array, we can be in one of two states:

1. We've selected an even number of elements so far (the next element would be at an even index if selected)
2. We've selected an odd number of elements so far (the next element would be at an odd index if selected)

For each element in the original array, we have three choices:

• Skip it entirely
• Include it as an element at an even index (add it)
• Include it as an element at an odd index (subtract it)

The key insight is that we want to track two values as we process each element:

• f[i]: Maximum alternating sum ending at position i where we've selected an odd number of elements (next element would be at an odd index)
• g[i]: Maximum alternating sum ending at position i where we've selected an even number of elements (next element would be at an even index)

For each new element x:

• If we want to place it at an odd index (subtract it), we must have had an even count before, so: f[i] = max(g[i-1] - x, f[i-1]) (either subtract x from previous even state, or skip x)
• If we want to place it at an even index (add it), we must have had an odd count before, so: g[i] = max(f[i-1] + x, g[i-1]) (either add x from previous odd state, or skip x)

Initially, we start with 0 elements selected (even count), so g[0] = 0 and f[0] = 0.

The beauty of this approach is that it naturally handles all the complex decisions about which elements to include and at what positions, by simply tracking these two states and making locally optimal decisions at each step.

Learn more about Dynamic Programming patterns.

Solution Approach

The solution uses dynamic programming with two state arrays to track the maximum alternating sum at each position.

Data Structures:

• f[i]: Array storing the maximum alternating sum up to index i when we've selected an odd number of elements (next element would be subtracted)
• g[i]: Array storing the maximum alternating sum up to index i when we've selected an even number of elements (next element would be added)

Implementation Steps:

1. Initialize arrays: Create two arrays f and g of size n+1 (where n is the length of nums), initialized with zeros. The extra space allows for 1-indexed processing.

2. Process each element: Iterate through the array using enumeration starting from index 1:

   for i, x in enumerate(nums, 1):

3. Update states: For each element x at position i:

   • Update f[i] (odd count state):

     f[i] = max(g[i - 1] - x, f[i - 1])

     This means either:
     • Transition from even count state by subtracting x: g[i-1] - x
     • Skip x and maintain the previous odd count state: f[i-1]
   • Update g[i] (even count state):

     g[i] = max(f[i - 1] + x, g[i - 1])

     This means either:
     • Transition from odd count state by adding x: f[i-1] + x
     • Skip x and maintain the previous even count state: g[i-1]

4. Return the result: The maximum alternating sum is the maximum of the two final states:

   return max(f[n], g[n])

Why this works:

• The DP recurrence ensures we consider all valid subsequences
• At each step, we make the optimal choice of either including the current element (at the appropriate position) or skipping it
• The two states capture whether the last selected element was at an even or odd index
• The final answer is the maximum between ending in either state

Time Complexity: O(n) - single pass through the array
Space Complexity: O(n) - two arrays of size n+1

Example Walkthrough

Let's walk through the solution with the array nums = [4, 2, 5, 3].

We'll track two states:

• f[i]: max alternating sum with odd count of elements (next element would be subtracted)
• g[i]: max alternating sum with even count of elements (next element would be added)

Initial State:

• f[0] = 0 (no elements selected, odd count impossible)
• g[0] = 0 (no elements selected, even count = 0)

Processing element 4 (index 0, i=1):

• f[1] = max(g[0] - 4, f[0]) = max(0 - 4, 0) = 0
  • Either subtract 4 (transition from even→odd): -4
  • Or skip 4 (stay in odd state): 0
  • Choose: skip (0 > -4)
• g[1] = max(f[0] + 4, g[0]) = max(0 + 4, 0) = 4
  • Either add 4 (transition from odd→even): 4
  • Or skip 4 (stay in even state): 0
  • Choose: add 4 (4 > 0)

Processing element 2 (index 1, i=2):

• f[2] = max(g[1] - 2, f[1]) = max(4 - 2, 0) = 2
  • Either subtract 2 after adding 4: 4 - 2 = 2
  • Or skip 2 (maintain previous odd state): 0
  • Choose: subtract 2 (2 > 0)
• g[2] = max(f[1] + 2, g[1]) = max(0 + 2, 4) = 4
  • Either add 2 (from odd state): 2
  • Or skip 2 (maintain previous even state): 4
  • Choose: skip (4 > 2)

Processing element 5 (index 2, i=3):

• f[3] = max(g[2] - 5, f[2]) = max(4 - 5, 2) = 2
  • Either subtract 5 from even state: -1
  • Or skip 5 (maintain odd state): 2
  • Choose: skip (2 > -1)
• g[3] = max(f[2] + 5, g[2]) = max(2 + 5, 4) = 7
  • Either add 5 to odd state: 7
  • Or skip 5 (maintain even state): 4
  • Choose: add 5 (7 > 4)

Processing element 3 (index 3, i=4):

• f[4] = max(g[3] - 3, f[3]) = max(7 - 3, 2) = 4
  • Either subtract 3 from even state: 4
  • Or skip 3 (maintain odd state): 2
  • Choose: subtract 3 (4 > 2)
• g[4] = max(f[3] + 3, g[3]) = max(2 + 3, 7) = 7
  • Either add 3 to odd state: 5
  • Or skip 3 (maintain even state): 7
  • Choose: skip (7 > 5)

Final Result: max(f[4], g[4]) = max(4, 7) = 7

The maximum alternating sum is 7, which corresponds to selecting the subsequence [4, 5]:

• After reindexing: index 0 gets 4, index 1 gets 5
• Alternating sum: 4 - 5 = -1... Wait, that's not right!

Let me reconsider. Looking at g[4] = 7, this represents selecting elements that sum to 7 with an even count. Tracing back:

• At g[3] = 7: came from f[2] + 5 = 2 + 5
• At f[2] = 2: came from g[1] - 2 = 4 - 2
• At g[1] = 4: came from adding 4

So the sequence is: add 4 (even index), subtract 2 (odd index), add 5 (even index) = 4 - 2 + 5 = 7

This corresponds to selecting subsequence [4, 2, 5] which has alternating sum 4 - 2 + 5 = 7.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n), where n is the length of the input array nums. The algorithm iterates through the array exactly once with a single for loop, performing constant-time operations (max comparisons and arithmetic operations) for each element.

Space Complexity: O(n), where n is the length of the input array. The algorithm uses two auxiliary arrays f and g, each of size n + 1, to store the dynamic programming states. Therefore, the total extra space used is 2 * (n + 1) = O(n).

Note: This space complexity can be optimized to O(1) since each state f[i] and g[i] only depends on f[i-1] and g[i-1]. By using just two variables instead of arrays, we could achieve constant space complexity while maintaining the same time complexity.

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

Common Pitfalls

Pitfall 1: Misunderstanding the State Definition

The Problem: Many developers incorrectly interpret the DP states as tracking whether we're at an even or odd index in the original array, rather than tracking whether we've selected an even or odd number of elements for our subsequence.

Incorrect Approach:

# WRONG: Thinking f[i] means "element at index i is at odd position"
for i in range(n):
    if i % 2 == 0:  # Even index in original array
        f[i] = g[i-1] + nums[i]
    else:  # Odd index in original array
        g[i] = f[i-1] - nums[i]

Why It Fails: This approach forces us to consider elements based on their original position, but subsequences get reindexed. For example, if we select elements at indices [1, 3, 5] from the original array, they become indices [0, 1, 2] in the subsequence.

Correct Understanding:

• f[i]: Maximum sum when we've selected an odd count of elements (1, 3, 5, ...) up to position i
• g[i]: Maximum sum when we've selected an even count of elements (0, 2, 4, ...) up to position i

The state tracks the parity of the count of selected elements, not their original positions.

Pitfall 2: Incorrect Base Case Initialization

The Problem: Initializing the DP arrays incorrectly, particularly setting f[0] to a non-zero value.

Incorrect Approach:

f = [float('-inf')] * (n + 1)  # WRONG: Can't have odd count with 0 elements
g = [0] * (n + 1)

Why It Fails: Setting f[0] = -infinity assumes it's impossible to have an odd count with zero elements selected, which is correct. However, this creates issues when we try to transition states because f[i] = max(g[i-1] - x, f[i-1]) would propagate negative infinity unnecessarily.

Correct Initialization:

f = [0] * (n + 1)  # Both start at 0
g = [0] * (n + 1)

Both arrays should start at 0 because:

• g[0] = 0: Selecting 0 elements gives sum 0 (even count)
• f[0] = 0: Acts as a placeholder; the first real odd-count state comes from adding the first element

Pitfall 3: Space Optimization Without Understanding Dependencies

The Problem: Attempting to optimize space by using only two variables instead of arrays, but updating them in the wrong order.

Incorrect Approach:

def maxAlternatingSum(self, nums):
    odd, even = 0, 0
    for x in nums:
        odd = max(even - x, odd)    # Uses updated 'even' if we update it first
        even = max(odd + x, even)   # WRONG: Uses the newly updated 'odd'
    return max(odd, even)

Why It Fails: The second update uses the newly computed odd value instead of the previous iteration's value, breaking the DP recurrence relation.

Correct Space-Optimized Solution:

def maxAlternatingSum(self, nums):
    odd, even = 0, 0
    for x in nums:
        # Store previous values to avoid dependency issues
        new_odd = max(even - x, odd)
        new_even = max(odd + x, even)
        odd, even = new_odd, new_even
    return max(odd, even)

Or use simultaneous assignment:

def maxAlternatingSum(self, nums):
    odd, even = 0, 0
    for x in nums:
        odd, even = max(even - x, odd), max(odd + x, even)
    return max(odd, even)