Problem Description

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

Your task is to select exactly k disjoint (non-overlapping) subarrays from nums. These subarrays must be ordered such that the last element of subarray i appears before the first element of subarray i+1 in the original array.

The strength of your selection is calculated using an alternating sum formula:

strength = k * sum(sub₁) - (k-1) * sum(sub₂) + (k-2) * sum(sub₃) - ... - 2 * sum(subₖ₋₁) + sum(subₖ)

Where:

• sum(subᵢ) is the sum of all elements in the i-th subarray
• The coefficients alternate between positive and negative (since k is odd, it starts and ends with positive)
• Each subarray's sum is multiplied by a decreasing coefficient starting from k down to 1

The goal is to find the maximum possible strength value by strategically choosing which k subarrays to select from the array.

For example, if k = 3, the formula would be: strength = 3 * sum(sub₁) - 2 * sum(sub₂) + 1 * sum(sub₃)

Important constraints:

• The subarrays must be disjoint (cannot share elements)
• The subarrays must maintain their relative order in the original array
• You don't need to use all elements from nums - some elements can be left unselected

Intuition

Let's think about what each element contributes to the final strength. If element nums[i-1] is selected and belongs to the j-th subarray, its contribution to the strength depends on which subarray it's in.

Looking at the strength formula, the j-th subarray has a coefficient that follows a pattern:

• 1st subarray: coefficient is k (positive)
• 2nd subarray: coefficient is -(k-1) (negative)
• 3rd subarray: coefficient is (k-2) (positive)
• And so on...

So the coefficient for the j-th subarray is (k - j + 1) * (-1)^(j+1). This means each element nums[i-1] in the j-th subarray contributes nums[i-1] * (k - j + 1) * (-1)^(j+1) to the total strength.

Now, how do we decide which elements to include in which subarrays? This is a classic dynamic programming problem because:

1. We need to make decisions for each element (include it or not, and if included, which subarray it belongs to)
2. These decisions affect future choices (once we start a new subarray, we can't go back)
3. We want to optimize (maximize) the total strength

The key insight is to track two states for each position:

• Whether the current element is selected or not
• How many complete subarrays we've formed so far

We use f[i][j][0] to represent the maximum strength when we've processed the first i elements, formed j subarrays, and the i-th element is NOT selected.

We use f[i][j][1] to represent the maximum strength when we've processed the first i elements, formed j subarrays, and the i-th element IS selected.

For each element, we have choices:

• Skip it entirely (contributes nothing)
• Include it as part of the current subarray (if we're building one)
• Start a new subarray with it (if we haven't reached k subarrays yet)

The transitions between states naturally follow: if we select element i, it either extends the current subarray or starts a new one. If we don't select it, we can potentially end the current subarray and prepare to start a new one later.

Learn more about Dynamic Programming and Prefix Sum patterns.

Solution Approach

We implement a 3D dynamic programming solution where f[i][j][s] represents the maximum strength achievable using the first i elements, selecting exactly j subarrays, where s indicates whether the i-th element is selected (s=1) or not (s=0).

Initialization:

• Create a 3D DP array with dimensions (n+1) × (k+1) × 2, initialized to negative infinity
• Set f[0][0][0] = 0 as the base case (no elements processed, no subarrays selected, nothing selected)

State Transitions:

For each position i from 1 to n and each number of subarrays j from 0 to k:

1. When the i-th element is NOT selected (f[i][j][0]):

   • We take the maximum of the previous state where element i-1 was either selected or not selected
   • f[i][j][0] = max(f[i-1][j][0], f[i-1][j][1])

2. When the i-th element IS selected (f[i][j][1]):

   • First, calculate the sign: sign = 1 if j is odd, else sign = -1
   • The contribution of this element is sign × nums[i-1] × (k - j + 1)

   Two cases to consider:

   • Extending the current subarray: If element i-1 was also selected and in the same subarray:
     • f[i][j][1] = f[i-1][j][1] + sign × nums[i-1] × (k - j + 1)
   • Starting a new subarray: If this element starts the j-th subarray (requires j > 0):
     • f[i][j][1] = max(f[i-1][j-1][0], f[i-1][j-1][1]) + sign × nums[i-1] × (k - j + 1)
     • We take the maximum of both cases where the previous element was selected or not

Final Answer: After processing all elements, the maximum strength is max(f[n][k][0], f[n][k][1]), which represents the maximum value when we've selected exactly k subarrays from all n elements.

Time Complexity: O(n × k) where n is the length of the array and k is the number of subarrays to select.

Space Complexity: O(n × k) for the 3D DP array.

Example Walkthrough

Let's walk through a small example with nums = [1, 2, -1, 4] and k = 3.

We need to select exactly 3 disjoint subarrays. The strength formula is: strength = 3 × sum(sub₁) - 2 × sum(sub₂) + 1 × sum(sub₃)

DP Table Setup:

• We create a 3D array f[i][j][s] where i ranges from 0 to 4 (length of nums + 1), j ranges from 0 to 3 (k), and s is 0 or 1.
• Initialize all values to negative infinity except f[0][0][0] = 0.

Key State Transitions:

Let's trace through some important states:

i=1 (element 1):

• f[1][1][1]: Start first subarray with element 1
  • Sign for j=1: positive (odd)
  • Contribution: 1 × 3 = 3
  • f[1][1][1] = 0 + 3 = 3

i=2 (element 2):

• f[2][1][1]: Two options for first subarray
  • Extend: f[1][1][1] + 2 × 3 = 3 + 6 = 9 (subarray [1,2])
  • Start new: f[1][0][0] + 2 × 3 = 0 + 6 = 6 (subarray [2])
  • Take max: f[2][1][1] = 9
• f[2][2][1]: Start second subarray with element 2
  • Must come after first subarray ended
  • Sign for j=2: negative (even)
  • f[2][2][1] = f[1][1][0] + 2 × (-2) = 3 + (-4) = -1

i=3 (element -1):

• f[3][2][1]: Options for second subarray
  • Extend from f[2][2][1]: -1 + (-1) × (-2) = -1 + 2 = 1
  • Start new from f[2][1][0]: 9 + (-1) × (-2) = 9 + 2 = 11
  • Take max: f[3][2][1] = 11

i=4 (element 4):

• f[4][3][1]: Options for third subarray
  • Sign for j=3: positive (odd)
  • Start new from f[3][2][0]: 11 + 4 × 1 = 15
  • This gives us the selection: [1,2], [-1], [4]

Verification:

• Subarray 1: [1,2], sum = 3, contribution = 3 × 3 = 9
• Subarray 2: [-1], sum = -1, contribution = -2 × (-1) = 2
• Subarray 3: [4], sum = 4, contribution = 1 × 4 = 4
• Total strength = 9 + 2 + 4 = 15

The maximum strength is 15, achieved by selecting subarrays [1,2], [-1], and [4].

Solution Implementation

Time and Space Complexity

The time complexity of this code is O(n × k), where n is the length of the array nums and k is the number of subarrays to select.

This is because:

• The outer loop iterates through all elements in nums, which takes O(n) time
• For each element, we have a nested loop that iterates from 0 to k, which takes O(k) time
• Inside the nested loops, all operations (comparisons, assignments, arithmetic) are O(1)
• Therefore, the total time complexity is O(n) × O(k) = O(n × k)

The space complexity is O(n × k) as well.

This is because:

• We create a 3D array f with dimensions (n + 1) × (k + 1) × 2
• The first dimension has size n + 1
• The second dimension has size k + 1
• The third dimension has fixed size 2
• Therefore, the total space used is O((n + 1) × (k + 1) × 2) = O(n × k)

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

Common Pitfalls

1. Incorrect Sign Calculation for Alternating Pattern

A critical pitfall is miscalculating the sign for the alternating sum formula. The problem states that for odd k, the pattern starts and ends with positive coefficients.

The Pitfall:

# INCORRECT: Using subarray_count directly for sign
sign = 1 if subarray_count & 1 else -1  # This is wrong!

This incorrectly assigns signs based on the current count of subarrays being processed, but the formula requires signs based on the position in the final selection.

The Correct Approach:

# CORRECT: Sign should alternate starting with positive for first subarray
sign = 1 if (subarray_count % 2) == 1 else -1

Since we're counting subarrays from 1 to k, odd positions (1st, 3rd, 5th...) get positive signs, and even positions (2nd, 4th...) get negative signs.

2. Initialization Issues with Negative Infinity

The Pitfall: Not properly handling the initialization can lead to incorrect propagation of invalid states.

# Potential issue: Some Python versions or environments might not handle -inf properly
dp = [[[-float('inf'), -float('inf')] for _ in range(k + 1)] for _ in range(n + 1)]

Better Solution:

# Use a very large negative number instead
MIN_VAL = -10**15  # Large enough to act as negative infinity for this problem
dp = [[[MIN_VAL, MIN_VAL] for _ in range(k + 1)] for _ in range(n + 1)]
dp[0][0][0] = 0

3. Off-by-One Error in Coefficient Calculation

The Pitfall: The coefficient for each subarray should decrease from k to 1, but it's easy to miscalculate this.

# INCORRECT: Wrong coefficient calculation
strength_multiplier = subarray_count  # This would give 1, 2, 3... instead of k, k-1, k-2...

The Correct Approach:

# CORRECT: Coefficient decreases as we select more subarrays
strength_multiplier = k - subarray_count + 1

4. Not Handling the Transition for Starting New Subarrays Properly

The Pitfall: When starting a new subarray, forgetting to check if we have subarrays left to allocate:

# INCORRECT: Not checking if subarray_count > 0
dp[i][subarray_count][1] = max(
    dp[i][subarray_count][1],
    max(dp[i - 1][subarray_count - 1]) + contribution  # Could access negative index!
)

The Correct Approach:

# CORRECT: Only start new subarray if we haven't started all k subarrays yet
if subarray_count > 0:
    previous_best = max(dp[i - 1][subarray_count - 1])
    dp[i][subarray_count][1] = max(
        dp[i][subarray_count][1],
        previous_best + sign * current_value * strength_multiplier
    )

5. Memory Optimization Opportunity Missed

While not a bug, the 3D DP array uses O(n×k×2) space. Since we only need the previous row, we can optimize:

Space-Optimized Solution:

def maximumStrength(self, nums: List[int], k: int) -> int:
    n = len(nums)
    MIN_VAL = -10**15
  
    # Use only two rows instead of n+1 rows
    prev = [[MIN_VAL, MIN_VAL] for _ in range(k + 1)]
    curr = [[MIN_VAL, MIN_VAL] for _ in range(k + 1)]
  
    prev[0][0] = 0
  
    for i, val in enumerate(nums, 1):
        curr[0][0] = 0  # Reset base case
      
        for j in range(k + 1):
            sign = 1 if j & 1 else -1
            coef = k - j + 1
          
            # Not in subarray
            curr[j][0] = max(prev[j][0], prev[j][1])
          
            # In subarray - continue
            curr[j][1] = prev[j][1] + sign * val * coef
          
            # In subarray - start new
            if j > 0:
                curr[j][1] = max(curr[j][1], 
                                max(prev[j-1]) + sign * val * coef)
      
        prev, curr = curr, prev
  
    return max(prev[k])

This reduces space complexity from O(n×k) to O(k).