Problem Description

The problem defines an alternating sum of an array as the sum of elements at even indices minus the sum of elements at odd indices. For instance, in an array [4,2,5,3], the alternating sum is (4 + 5) - (2 + 3) = 4.

The task is to find the maximum alternating sum of any subsequence in a given array nums. A subsequence can be obtained by deleting some elements (possibly none) from the original array but keeping the order of the remaining elements.

Intuition

To get the maximum alternating sum, we have to choose a subsequence where the difference between the sum of elements at even indices and the sum of elements at odd indices is maximized. This implies we want to include larger numbers at even indices and smaller numbers at odd indices, if possible.

To find the solution, we use dynamic programming to keep track of two values while iterating through the array: f and g. Here f represents the maximum alternating sum ending with an element at an even index, and g represents the maximum alternating sum ending with an element at an odd index.

1. When we are at a new number x, we have two choices: either include x in our subsequence or not. If the last included number was at an even index (and we are at an odd index now), f is updated by subtracting x from it, because we assume that including x would contribute to an alternating sum pattern. On the other hand, if we decide not to include x, then f remains the same. We choose the maximum of these two options to get the new value of f.

2. Similarly, when updating g (representing an odd index), we consider adding x (since we are now at an even index) to the previous g or keeping g as is. We select the maximum of these two choices.

3. As we loop through the array, f and g are updated in an alternating manner, reflecting the inclusion or exclusion of elements in the subsequence to maintain the pattern.

4. Finally, after considering all elements in nums, the maximum of f and g will be our answer, as it will represent the maximum alternating sum attainable by any subsequence of the original array.

Learn more about Dynamic Programming patterns.

Solution Approach

The solution to this problem is elegantly handled with dynamic programming, where two variables f and g represent the current best alternating sums ending on an even index and an odd index, respectively. Here's a step-by-step breakdown of how the algorithm is implemented:

1. Initialization: We initialize two variables f and g to zero. Here, f will be used to track the maximum alternating sum ending with an even index, and g will track the maximum alternating sum ending with an odd index.

2. Iterating through nums: We loop through each element x in the provided array nums. For each element, we have to decide whether including it in the subsequence would lead to a higher alternating sum.

3. Dynamic Programming Transition:

   • To update f, we take the maximum between the current value of f (which includes not picking x) and g - x (which accounts for picking x and adhering to the alternating sum rule). The equation can be written as f = max(f, g - x).
   • To update g, we take the maximum between the current value of g (which also includes not picking x) and f + x (which accounts for picking x and maintaining the alternating pattern). The equation is g = max(g, f + x).

4. Maximizing the Result: With each element processed, the variables f and g are updated to always reflect the highest possible alternating sums up to that point.

5. Returning the Result: After the loop is finished, the larger of f and g will represent the largest alternating sum of a subsequence that can be formed from the array nums. Since the subsequence can end with either an even or odd indexed element, we take max(f, g) as the final result.

Using only two variables to keep track of the state at each step makes the solution space-efficient. The dynamic programming technique utilized here is especially useful as it avoids the need to consider all possible subsequences explicitly, which would be computationally expensive. The given implementation thus runs in O(n) time, where n is the number of elements in nums, since it processes each element only once.

Example Walkthrough

Let's consider a small example using the array nums = [3, 1, 6, 4] to illustrate how the dynamic programming solution works with the given approach.

1. Initialization:

   • Initialize f and g to zero.
   • f (ends with even index) = 0
   • g (ends with odd index) = 0

2. Element 1 (3 at index 0 which is even):

   • To update f, we choose max between f (which is 0) and g - x (which is also 0 because g is 0).
   • To update g, it remains 0, since 0 + 3 is less than f.
   • Updated values: f = 3, g = 0

3. Element 2 (1 at index 1 which is odd):

   • To update f, we choose max between f (which is 3) and g - x (which is -1, because g is 0 and x is 1).
   • f doesn't change because 3 is greater than -1.
   • To update g, we choose max between g (which is 0) and f + x (which is 4, because f is 3 and x is 1).
   • Updated values: f = 3, g = 4

4. Element 3 (6 at index 2 which is even):

   • To update f, we choose max between f (which is 3) and g - x (which is -2, because g is 4 and x is 6).
   • f doesn't change because 3 is greater than -2.
   • To update g, we choose max between g (which is 4) and f + x (which is 9, because f is 3 and x is 6).
   • Updated values: f = 3, g = 9

5. Element 4 (4 at index 3 which is odd):

   • To update f, we choose max between f (which is 3) and g - x (which is 5, because g is 9 and x is 4).
   • Updated values: f = 5, g still 9, as f + x (which is 7) is less than g.

After considering all elements in nums, we have two final values for f and g. f holds the maximum alternating sum when we end on an even index and g when we end on an odd index.

• f is 5
• g is 9

The maximum between f and g is g. Therefore, the maximum alternating sum of any subsequence of the given array nums is 9.

This walkthrough captures the gradual update process of the dynamic programming approach where variables f and g facilitates the capture of the highest sums possible with alternating subsequence selection at every step without considering all subsequence combinations.

Solution Implementation

Time and Space Complexity

The given Python code snippet defines a function that calculates the maximum alternating sum of an array. To analyze its time and space complexity, consider the following:

Time Complexity:

• The function iterates through the list of numbers once. Let n be the length of nums.
• Within this single loop, it performs constant-time operations such as computing the maximum value between two numbers and summing or subtracting x from the variables f and g.
• There are no nested loops or recursive calls that would increase the complexity.
• Therefore, the time complexity is O(n).

Space Complexity:

• The function uses a fixed number of integer variables f, and g irrespective of the input size.
• No additional data structures are created that grow with the size of the input.
• Hence, the space complexity of the function is O(1).

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