LeetCode #3196 — MEDIUM

Maximize Total Cost of Alternating Subarrays

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given an integer array nums with length n.

The cost of a subarray nums[l..r], where 0 <= l <= r < n, is defined as:

cost(l, r) = nums[l] - nums[l + 1] + ... + nums[r] * (−1)^{r − l}

Your task is to split nums into subarrays such that the total cost of the subarrays is maximized, ensuring each element belongs to exactly one subarray.

Formally, if nums is split into k subarrays, where k > 1, at indices i₁, i₂, ..., i_{k − 1}, where 0 <= i₁ < i₂ < ... < i_{k - 1} < n - 1, then the total cost will be:

cost(0, i₁) + cost(i₁ + 1, i₂) + ... + cost(i_{k − 1} + 1, n − 1)

Return an integer denoting the maximum total cost of the subarrays after splitting the array optimally.

Note: If nums is not split into subarrays, i.e. k = 1, the total cost is simply cost(0, n - 1).

Example 1:

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

Output: 10

Explanation:

One way to maximize the total cost is by splitting [1, -2, 3, 4] into subarrays [1, -2, 3] and [4]. The total cost will be (1 + 2 + 3) + 4 = 10.

Example 2:

Input: nums = [1,-1,1,-1]

Output: 4

Explanation:

One way to maximize the total cost is by splitting [1, -1, 1, -1] into subarrays [1, -1] and [1, -1]. The total cost will be (1 + 1) + (1 + 1) = 4.

Example 3:

Input: nums = [0]

Output: 0

Explanation:

We cannot split the array further, so the answer is 0.

Example 4:

Input: nums = [1,-1]

Output: 2

Explanation:

Selecting the whole array gives a total cost of 1 + 1 = 2, which is the maximum.

Constraints:

1 <= nums.length <= 10⁵
-10⁹ <= nums[i] <= 10⁹

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums with length n. The cost of a subarray nums[l..r], where 0 <= l <= r < n, is defined as: cost(l, r) = nums[l] - nums[l + 1] + ... + nums[r] * (−1)r − l Your task is to split nums into subarrays such that the total cost of the subarrays is maximized, ensuring each element belongs to exactly one subarray. Formally, if nums is split into k subarrays, where k > 1, at indices i1, i2, ..., ik − 1, where 0 <= i1 < i2 < ... < ik - 1 < n - 1, then the total cost will be: cost(0, i1) + cost(i1 + 1, i2) + ... + cost(ik − 1 + 1, n − 1) Return an integer denoting the maximum total cost of the subarrays after splitting the array optimally. Note: If nums is not split into subarrays, i.e. k = 1, the total cost is simply cost(0, n - 1).

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming

Example 1

[1,-2,3,4]

Example 2

[1,-1,1,-1]

Example 3

[0]

Step 02

Core Insight

What unlocks the optimal approach

The problem can be solved using dynamic programming.
Since we can always start a new subarray, the problem is the same as selecting some elements in the array and flipping their signs to negative to maximize the sum. However, we cannot flip the signs of 2 consecutive elements, and the first element in the array cannot be negative.
Let <code>dp[i][0/1]</code> be the largest sum we can get for prefix <code>nums[0..i]</code>, where <code>dp[i][0]</code> is the maximum if the <code>ith</code> element wasn't flipped, and <code>dp[i][1]</code> is the maximum if the <code>ith</code> element was flipped.
Based on the restriction: <code>dp[i][0] = max(dp[i - 1][0], dp[i - 1][1]) + nums[i]</code> <code>dp[i][1] = dp[i - 1][0] - nums[i]</code>
The initial state is: <code>dp[1][0] = nums[0] + nums[1]</code> <code>dp[1][1] = nums[0] - nums[1]</code> and the answer is <code>max(dp[n - 1][0], dp[n - 1][1])</code>.
Can you optimize the space complexity?

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3196: Maximize Total Cost of Alternating Subarrays
class Solution {
    private Long[][] f;
    private int[] nums;
    private int n;

    public long maximumTotalCost(int[] nums) {
        n = nums.length;
        this.nums = nums;
        f = new Long[n][2];
        return dfs(0, 0);
    }

    private long dfs(int i, int j) {
        if (i >= n) {
            return 0;
        }
        if (f[i][j] != null) {
            return f[i][j];
        }
        f[i][j] = nums[i] + dfs(i + 1, 1);
        if (j == 1) {
            f[i][j] = Math.max(f[i][j], -nums[i] + dfs(i + 1, 0));
        }
        return f[i][j];
    }
}

// Accepted solution for LeetCode #3196: Maximize Total Cost of Alternating Subarrays
func maximumTotalCost(nums []int) int64 {
	n := len(nums)
	f := make([][2]int64, n)
	for i := range f {
		f[i] = [2]int64{-1e18, -1e18}
	}
	var dfs func(int, int) int64
	dfs = func(i, j int) int64 {
		if i >= n {
			return 0
		}
		if f[i][j] != -1e18 {
			return f[i][j]
		}
		f[i][j] = int64(nums[i]) + dfs(i+1, 1)
		if j > 0 {
			f[i][j] = max(f[i][j], int64(-nums[i])+dfs(i+1, 0))
		}
		return f[i][j]
	}
	return dfs(0, 0)
}

# Accepted solution for LeetCode #3196: Maximize Total Cost of Alternating Subarrays
class Solution:
    def maximumTotalCost(self, nums: List[int]) -> int:
        @cache
        def dfs(i: int, j: int) -> int:
            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

        return dfs(0, 0)

// Accepted solution for LeetCode #3196: Maximize Total Cost of Alternating Subarrays
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3196: Maximize Total Cost of Alternating Subarrays
// class Solution {
//     private Long[][] f;
//     private int[] nums;
//     private int n;
// 
//     public long maximumTotalCost(int[] nums) {
//         n = nums.length;
//         this.nums = nums;
//         f = new Long[n][2];
//         return dfs(0, 0);
//     }
// 
//     private long dfs(int i, int j) {
//         if (i >= n) {
//             return 0;
//         }
//         if (f[i][j] != null) {
//             return f[i][j];
//         }
//         f[i][j] = nums[i] + dfs(i + 1, 1);
//         if (j == 1) {
//             f[i][j] = Math.max(f[i][j], -nums[i] + dfs(i + 1, 0));
//         }
//         return f[i][j];
//     }
// }

// Accepted solution for LeetCode #3196: Maximize Total Cost of Alternating Subarrays
function maximumTotalCost(nums: number[]): number {
    const n = nums.length;
    const f: number[][] = Array.from({ length: n }, () => Array(2).fill(-Infinity));
    const dfs = (i: number, j: number): number => {
        if (i >= n) {
            return 0;
        }
        if (f[i][j] !== -Infinity) {
            return f[i][j];
        }
        f[i][j] = nums[i] + dfs(i + 1, 1);
        if (j) {
            f[i][j] = Math.max(f[i][j], -nums[i] + dfs(i + 1, 0));
        }
        return f[i][j];
    };
    return dfs(0, 0);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.