LeetCode #3077 — HARD

Maximum Strength of K Disjoint Subarrays

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given an array of integers nums with length n, and a positive odd integer k.

Select exactly k disjoint subarrays sub₁, sub₂, ..., sub_k from nums such that the last element of sub_i appears before the first element of sub_{i+1} for all 1 <= i <= k-1. The goal is to maximize their combined strength.

The strength of the selected subarrays is defined as:

strength = k * sum(sub₁)- (k - 1) * sum(sub₂) + (k - 2) * sum(sub₃) - ... - 2 * sum(sub_{k-1}) + sum(sub_k)

where sum(sub_i) is the sum of the elements in the i-th subarray.

Return the maximum possible strength that can be obtained from selecting exactly k disjoint subarrays from nums.

Note that the chosen subarrays don't need to cover the entire array.

Example 1:

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

Output: 22

Explanation:

The best possible way to select 3 subarrays is: nums[0..2], nums[3..3], and nums[4..4]. The strength is calculated as follows:

strength = 3 * (1 + 2 + 3) - 2 * (-1) + 2 = 22

Example 2:

Input: nums = [12,-2,-2,-2,-2], k = 5

Output: 64

Explanation:

The only possible way to select 5 disjoint subarrays is: nums[0..0], nums[1..1], nums[2..2], nums[3..3], and nums[4..4]. The strength is calculated as follows:

strength = 5 * 12 - 4 * (-2) + 3 * (-2) - 2 * (-2) + (-2) = 64

Example 3:

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

Output: -1

Explanation:

The best possible way to select 1 subarray is: nums[0..0]. The strength is -1.

Constraints:

1 <= n <= 10⁴
-10⁹ <= nums[i] <= 10⁹
1 <= k <= n
1 <= n * k <= 10⁶
k is odd.

Step 01

Brute Force Baseline

Problem summary: You are given an array of integers nums with length n, and a positive odd integer k. Select exactly k disjoint subarrays sub1, sub2, ..., subk from nums such that the last element of subi appears before the first element of sub{i+1} for all 1 <= i <= k-1. The goal is to maximize their combined strength. The strength of the selected subarrays is defined as: strength = k * sum(sub1)- (k - 1) * sum(sub2) + (k - 2) * sum(sub3) - ... - 2 * sum(sub{k-1}) + sum(subk) where sum(subi) is the sum of the elements in the i-th subarray. Return the maximum possible strength that can be obtained from selecting exactly k disjoint subarrays from nums. Note that the chosen subarrays don't need to cover the entire array.

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,-1,2]
3

Example 2

[12,-2,-2,-2,-2]
5

Example 3

[-1,-2,-3]
1

Core Insight

What unlocks the optimal approach

Let <code>dp[i][j][x == 0/1]</code> be the maximum strength to select <code>j</code> disjoint subarrays from the original array’s suffix (<code>nums[i..(n - 1)]</code>), x denotes whether we select the element or not.
Initially <code>dp[n][0][0] == 0</code>.
We have <code>dp[i][j][1] = nums[i] * get(j) + max(dp[i + 1][j - 1][0], dp[i + 1][j][1])</code> where <code>get(j) = j</code> if <code>j</code> is odd, otherwise <code>-j</code>.
We can select <code>nums[i]</code> as a separate subarray or select at least <code>nums[i]</code> and <code>nums[i + 1]</code> as the first subarray. <code>dp[i][j][0] = max(dp[i + 1][j][0], dp[i][j][1])</code>.
The answer is <code>dp[0][k][0]</code>.

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

Largest constraint values

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 #3077: Maximum Strength of K Disjoint Subarrays
class Solution {
    public long maximumStrength(int[] nums, int k) {
        int n = nums.length;
        long[][][] f = new long[n + 1][k + 1][2];
        for (int i = 0; i <= n; i++) {
            for (int j = 0; j <= k; j++) {
                Arrays.fill(f[i][j], Long.MIN_VALUE / 2);
            }
        }
        f[0][0][0] = 0;
        for (int i = 1; i <= n; i++) {
            int x = nums[i - 1];
            for (int j = 0; j <= k; j++) {
                long sign = (j & 1) == 1 ? 1 : -1;
                long val = sign * x * (k - j + 1);
                f[i][j][0] = Math.max(f[i - 1][j][0], f[i - 1][j][1]);
                f[i][j][1] = Math.max(f[i][j][1], f[i - 1][j][1] + val);
                if (j > 0) {
                    long t = Math.max(f[i - 1][j - 1][0], f[i - 1][j - 1][1]) + val;
                    f[i][j][1] = Math.max(f[i][j][1], t);
                }
            }
        }
        return Math.max(f[n][k][0], f[n][k][1]);
    }
}

// Accepted solution for LeetCode #3077: Maximum Strength of K Disjoint Subarrays
func maximumStrength(nums []int, k int) int64 {
	n := len(nums)
	f := make([][][]int64, n+1)
	const inf int64 = math.MinInt64 / 2
	for i := range f {
		f[i] = make([][]int64, k+1)
		for j := range f[i] {
			f[i][j] = []int64{inf, inf}
		}
	}
	f[0][0][0] = 0
	for i := 1; i <= n; i++ {
		x := nums[i-1]
		for j := 0; j <= k; j++ {
			sign := int64(-1)
			if j&1 == 1 {
				sign = 1
			}
			val := sign * int64(x) * int64(k-j+1)
			f[i][j][0] = max(f[i-1][j][0], f[i-1][j][1])
			f[i][j][1] = max(f[i][j][1], f[i-1][j][1]+val)
			if j > 0 {
				t := max(f[i-1][j-1][0], f[i-1][j-1][1]) + val
				f[i][j][1] = max(f[i][j][1], t)
			}
		}
	}
	return max(f[n][k][0], f[n][k][1])
}

# Accepted solution for LeetCode #3077: Maximum Strength of K Disjoint Subarrays
class Solution:
    def maximumStrength(self, nums: List[int], k: int) -> int:
        n = len(nums)
        f = [[[-inf, -inf] for _ in range(k + 1)] for _ in range(n + 1)]
        f[0][0][0] = 0
        for i, x in enumerate(nums, 1):
            for j in range(k + 1):
                sign = 1 if j & 1 else -1
                f[i][j][0] = max(f[i - 1][j][0], f[i - 1][j][1])
                f[i][j][1] = max(f[i][j][1], f[i - 1][j][1] + sign * x * (k - j + 1))
                if j:
                    f[i][j][1] = max(
                        f[i][j][1], max(f[i - 1][j - 1]) + sign * x * (k - j + 1)
                    )
        return max(f[n][k])

// Accepted solution for LeetCode #3077: Maximum Strength of K Disjoint 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 #3077: Maximum Strength of K Disjoint Subarrays
// class Solution {
//     public long maximumStrength(int[] nums, int k) {
//         int n = nums.length;
//         long[][][] f = new long[n + 1][k + 1][2];
//         for (int i = 0; i <= n; i++) {
//             for (int j = 0; j <= k; j++) {
//                 Arrays.fill(f[i][j], Long.MIN_VALUE / 2);
//             }
//         }
//         f[0][0][0] = 0;
//         for (int i = 1; i <= n; i++) {
//             int x = nums[i - 1];
//             for (int j = 0; j <= k; j++) {
//                 long sign = (j & 1) == 1 ? 1 : -1;
//                 long val = sign * x * (k - j + 1);
//                 f[i][j][0] = Math.max(f[i - 1][j][0], f[i - 1][j][1]);
//                 f[i][j][1] = Math.max(f[i][j][1], f[i - 1][j][1] + val);
//                 if (j > 0) {
//                     long t = Math.max(f[i - 1][j - 1][0], f[i - 1][j - 1][1]) + val;
//                     f[i][j][1] = Math.max(f[i][j][1], t);
//                 }
//             }
//         }
//         return Math.max(f[n][k][0], f[n][k][1]);
//     }
// }

// Accepted solution for LeetCode #3077: Maximum Strength of K Disjoint Subarrays
function maximumStrength(nums: number[], k: number): number {
    const n: number = nums.length;
    const f: number[][][] = Array.from({ length: n + 1 }, () =>
        Array.from({ length: k + 1 }, () => [-Infinity, -Infinity]),
    );
    f[0][0][0] = 0;
    for (let i = 1; i <= n; i++) {
        const x: number = nums[i - 1];
        for (let j = 0; j <= k; j++) {
            const sign: number = (j & 1) === 1 ? 1 : -1;
            const val: number = sign * x * (k - j + 1);
            f[i][j][0] = Math.max(f[i - 1][j][0], f[i - 1][j][1]);
            f[i][j][1] = Math.max(f[i][j][1], f[i - 1][j][1] + val);
            if (j > 0) {
                f[i][j][1] = Math.max(f[i][j][1], Math.max(...f[i - 1][j - 1]) + val);
            }
        }
    }
    return Math.max(...f[n][k]);
}

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 × k)

Space

O(n × k)

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.

Maximum Strength of K Disjoint Subarrays

Problem Statement

Roadmap

Brute Force Baseline

Baseline thinking

Example 1

Example 2

Example 3

Related Problems

Core Insight

What unlocks the optimal approach

Algorithm Walkthrough

Iteration Checklist

Edge Cases

Full Annotated Code

Interactive Study Demo

Complexity Analysis

Approach Breakdown

Common Mistakes

Off-by-one on range boundaries

State misses one required dimension