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Maximum Strength of K Disjoint Subarrays

Difficulty: Hard

Problem Description

You are given an array of integers nums with length n, and a positive odd integer k. Select exactly k disjoint subarrays 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 defined as:

strength = k * sum(sub_1) - (k - 1) * sum(sub_2) + (k - 2) * sum(sub_3) - ... - 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.

Key Insights

  • The problem requires selecting k disjoint subarrays, where the arrangement of their sums affects the overall strength due to alternating signs.
  • The strength formula gives more weight to earlier subarrays, thus larger sums are preferentially placed in earlier positions.
  • The dynamic programming approach can be utilized to explore different combinations of subarrays while keeping track of their sums and contributions to the overall strength.
  • Prefix sums can be employed to efficiently calculate the sum of any subarray.

Space and Time Complexity

Time Complexity: O(n^2 * k) Space Complexity: O(n * k)

Solution

The solution utilizes dynamic programming to keep track of the maximum strength obtainable by selecting up to k disjoint subarrays. We will maintain a DP table where dp[i][j] represents the maximum strength we can obtain using the first i elements of the array and selecting j disjoint subarrays.

We also use a prefix sum array to quickly calculate the sum of any subarray. The key idea is to iterate through potential ending indices of the last selected subarray and compute the contribution to the strength based on the current number of selected subarrays.

Code Solutions

def maxStrength(nums, k):
    n = len(nums)
    # Prefix sums array
    prefix = [0] * (n + 1)
    for i in range(n):
        prefix[i + 1] = prefix[i] + nums[i]

    # DP table
    dp = [[float('-inf')] * (k + 1) for _ in range(n + 1)]
    dp[0][0] = 0

    # Dynamic programming to fill the dp table
    for j in range(1, k + 1):
        for i in range(1, n + 1):
            for p in range(0, i):
                sum_sub = prefix[i] - prefix[p]
                sign = (k - j) % 2
                if sign == 0:
                    dp[i][j] = max(dp[i][j], dp[p][j - 1] + (j * sum_sub))
                else:
                    dp[i][j] = max(dp[i][j], dp[p][j - 1] - ((j - 1) * sum_sub))

    # Get the maximum strength from the last row of dp table
    return max(dp[i][k] for i in range(n + 1))
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