LeetCode 3196 - Maximize Total Cost of Alternating Subarrays

The problem gives us an integer array nums, and we must divide the array into one or more contiguous subarrays. Every element must belong to exactly one subarray.

leetcodemediumarraydynamic-programming

LeetCode Problem 3196

Difficulty: 🟡 Medium
Topics: Array, Dynamic Programming

Solution

LeetCode 3196 - Maximize Total Cost of Alternating Subarrays

Problem Understanding

The problem gives us an integer array nums, and we must divide the array into one or more contiguous subarrays. Every element must belong to exactly one subarray.

For each subarray, we compute an alternating sum:

$$\text{cost}(l, r) = nums[l] - nums[l+1] + nums[l+2] - nums[l+3] + \dots$$

The sign alternates between positive and negative, starting with positive at the beginning of every subarray.

For example:

  • [1, -2, 3] becomes:

$$1 - (-2) + 3 = 6$$

  • [4] becomes:

$$4$$

The total score is the sum of the costs of all chosen subarrays.

The key detail is that whenever we start a new subarray, the alternating pattern resets. That reset can be beneficial because an element that would normally receive a negative sign can instead become the first element of a new subarray and receive a positive sign.

The constraints are large:

  • n can be up to 10^5
  • Values can be as large as 10^9 in magnitude

This immediately tells us that any algorithm worse than roughly O(n log n) will likely time out. Quadratic dynamic programming or brute force partition generation is not feasible.

Several edge cases are important:

  • A single element array, where no split is possible.
  • Arrays with all positive numbers.
  • Arrays with all negative numbers.
  • Alternating positive and negative values.
  • Very large values that require 64-bit integer arithmetic.

Because values can reach 10^9 and there can be 10^5 elements, the answer may exceed 32-bit integer range. We must use 64-bit integers.

Approaches

Brute Force Approach

A brute force solution would try every possible way to partition the array.

For an array of length n, each gap between adjacent elements can either contain a split or not contain a split. Since there are n - 1 gaps, there are:

$$2^{n-1}$$

possible partitions.

For each partition, we would compute the alternating cost of every subarray and track the maximum total.

This works because it explicitly checks every valid partition configuration, guaranteeing the optimal answer is found.

However, the complexity is exponential, which is completely infeasible for n = 10^5.

Key Insight

The important observation is that the sign of an element depends only on its position inside the current subarray.

At every index, we effectively have two choices:

  1. Continue the current subarray.
  2. Start a new subarray.

If we continue the current subarray, the sign alternates.

If we start a new subarray, the current element receives a positive sign again.

This naturally leads to dynamic programming.

We can track two states:

  • The best total when the current element is taken with a positive sign.
  • The best total when the current element is taken with a negative sign.

Transitions become extremely simple:

  • A positive state can come from:

  • starting a new subarray

  • or continuing after a negative state

  • A negative state can only come from continuing after a positive state

This gives an O(n) solution.

Approach Comparison

Approach Time Complexity Space Complexity Notes
Brute Force O(2^n * n) O(1) Tries every partition configuration
Optimal Dynamic Programming O(n) O(1) Tracks best positive and negative ending states

Algorithm Walkthrough

State Definition

We maintain two DP values while scanning the array:

  • positive

The maximum total cost where the current element is assigned a positive sign.

  • negative

The maximum total cost where the current element is assigned a negative sign.

Step-by-Step Process

  1. Initialize the states using the first element.

The first element must begin a subarray, so it always has a positive sign.

Therefore:

$$positive = nums[0]$$

A negative state is impossible initially. 2. Iterate through the remaining elements.

For each nums[i], compute the next states. 3. Compute the next positive state.

The current element can become positive in two ways:

  • Start a new subarray at index i
  • Continue after a previous negative sign

So:

$$nextPositive = \max(positive, negative) + nums[i]$$ 4. Compute the next negative state.

A negative sign only happens if we continue a subarray whose previous element had a positive sign.

So:

$$nextNegative = positive - nums[i]$$ 5. Update the DP states.

Replace the old values with the newly computed ones. 6. After processing all elements, return the larger of the two states.

The final subarray may end with either sign.

Why It Works

The DP maintains the best achievable total for every valid sign configuration at each position.

At every index, the algorithm considers all legal ways to either:

  • continue the current alternating pattern
  • or reset the pattern by starting a new subarray

Because every partition corresponds to a sequence of these choices, and the DP always keeps the optimal value for each state, the final result is guaranteed to be optimal.

Python Solution

from typing import List

class Solution:
    def maximumTotalCost(self, nums: List[int]) -> int:
        positive = nums[0]
        negative = float("-inf")

        for i in range(1, len(nums)):
            current = nums[i]

            next_positive = max(positive, negative) + current
            next_negative = positive - current

            positive = next_positive
            negative = next_negative

        return max(positive, negative)

The implementation directly follows the DP transitions described earlier.

positive stores the best total when the current element is positive inside its subarray.

negative stores the best total when the current element is negative inside its subarray.

For every new element:

  • next_positive considers both starting a new subarray and continuing from a negative state.
  • next_negative represents continuing an alternating pattern from a positive state.

The algorithm only keeps the previous states, so the memory usage remains constant.

The use of float("-inf") ensures invalid states are never accidentally chosen during maximization.

Go Solution

func maximumTotalCost(nums []int) int64 {
    positive := int64(nums[0])
    negative := int64(-1 << 60)

    for i := 1; i < len(nums); i++ {
        current := int64(nums[i])

        nextPositive := max64(positive, negative) + current
        nextNegative := positive - current

        positive = nextPositive
        negative = nextNegative
    }

    return max64(positive, negative)
}

func max64(a, b int64) int64 {
    if a > b {
        return a
    }
    return b
}

The Go version is structurally identical to the Python implementation.

The primary difference is explicit use of int64, since the result may exceed 32-bit integer range.

Go does not provide a built-in max function for integers, so we define max64.

The large negative initialization value acts like negative infinity for invalid states.

Worked Examples

Example 1

Input:

nums = [1, -2, 3, 4]

Initialization

Index Value positive negative
0 1 1 -inf

Process Index 1

Current value = -2

$$nextPositive = \max(1, -inf) + (-2) = -1$$

$$nextNegative = 1 - (-2) = 3$$

Index Value positive negative
1 -2 -1 3

Process Index 2

Current value = 3

$$nextPositive = \max(-1, 3) + 3 = 6$$

$$nextNegative = -1 - 3 = -4$$

Index Value positive negative
2 3 6 -4

Process Index 3

Current value = 4

$$nextPositive = \max(6, -4) + 4 = 10$$

$$nextNegative = 6 - 4 = 2$$

Index Value positive negative
3 4 10 2

Final answer:

$$\max(10, 2) = 10$$

Example 2

Input:

nums = [1, -1, 1, -1]
Index Value positive negative
0 1 1 -inf
1 -1 0 2
2 1 3 -1
3 -1 2 4

Final answer:

$$4$$

Example 3

Input:

nums = [0]

Initialization:

Index Value positive negative
0 0 0 -inf

Final answer:

$$0$$

Example 4

Input:

nums = [1, -1]
Index Value positive negative
0 1 1 -inf
1 -1 0 2

Final answer:

$$2$$

Complexity Analysis

Measure Complexity Explanation
Time O(n) Each element is processed exactly once
Space O(1) Only two DP states are stored

The algorithm performs a constant amount of work for every array element, leading to linear time complexity.

No auxiliary arrays or recursion are used. The DP state is compressed into two variables, so the extra memory usage remains constant regardless of input size.

Test Cases

from typing import List

class Solution:
    def maximumTotalCost(self, nums: List[int]) -> int:
        positive = nums[0]
        negative = float("-inf")

        for i in range(1, len(nums)):
            current = nums[i]

            next_positive = max(positive, negative) + current
            next_negative = positive - current

            positive = next_positive
            negative = next_negative

        return max(positive, negative)

sol = Solution()

assert sol.maximumTotalCost([1, -2, 3, 4]) == 10  # provided example
assert sol.maximumTotalCost([1, -1, 1, -1]) == 4  # alternating signs
assert sol.maximumTotalCost([0]) == 0  # single zero
assert sol.maximumTotalCost([1, -1]) == 2  # no split needed

assert sol.maximumTotalCost([5]) == 5  # single positive
assert sol.maximumTotalCost([-5]) == -5  # single negative

assert sol.maximumTotalCost([1, 2, 3]) == 6  # best to split all
assert sol.maximumTotalCost([-1, -2, -3]) == 0  # beneficial splits

assert sol.maximumTotalCost([10, -10, 10, -10]) == 40  # perfect alternation
assert sol.maximumTotalCost([1000000000, -1000000000]) == 2000000000  # large values

assert sol.maximumTotalCost([4, -1, 2, -3, 5]) == 15  # mixed pattern
assert sol.maximumTotalCost([7, 6, 5, 4]) == 22  # all positives

print("All test cases passed!")

Test Summary

Test Why
[1, -2, 3, 4] Validates standard mixed case
[1, -1, 1, -1] Tests alternating signs
[0] Smallest possible input
[1, -1] Tests two-element alternating case
[5] Single positive value
[-5] Single negative value
[1, 2, 3] Verifies optimal splitting for positives
[-1, -2, -3] Tests behavior with all negatives
[10, -10, 10, -10] Strong alternating benefit
[1000000000, -1000000000] Ensures 64-bit safety
[4, -1, 2, -3, 5] General mixed scenario
[7, 6, 5, 4] Tests repeated beneficial resets

Edge Cases

Single Element Array

If the array contains only one element, no split is possible. The answer is simply that element itself.

This case can easily cause indexing mistakes in DP transitions because there are no remaining elements to iterate through. The implementation handles it naturally because the loop starts at index 1, which is skipped for a single-element array.

All Positive Numbers

For arrays like [1, 2, 3, 4], it is often optimal to split frequently so every number receives a positive sign.

A naive implementation may incorrectly assume continuing the alternating sequence is always beneficial. The DP correctly compares both options at every position and chooses the larger value.

Large Negative Numbers

Arrays containing large negative values can produce situations where resetting the subarray is extremely valuable.

For example:

[1000000000, -1000000000]

If handled incorrectly with 32-bit integers, overflow occurs. Both implementations use 64-bit arithmetic to safely store the result.

Impossible Negative State at Start

At index 0, a negative sign is impossible because every subarray starts with a positive sign.

If this state were initialized incorrectly, invalid transitions could contaminate later DP values. The implementation prevents this by initializing negative to negative infinity.