Sum of K Subarrays With Length at Least M

This problem requires selecting k non-overlapping subarrays of at least length m to achieve the maximum total sum. The optimal solution uses state transition dynamic programming combined with prefix sums to efficiently compute subarray sums and track the best choices for each state. Handling array boundaries and varying subarray lengths correctly is key to avoiding overlapping selections.

Problem Statement

Given an integer array nums and integers k and m, find the maximum total sum of k non-overlapping subarrays where each subarray has length at least m. Each subarray must be contiguous, and no elements can be reused across different subarrays.

Return an integer representing the maximum sum possible. For example, nums = [1,2,-1,3,3,4], k = 2, m = 2 should return 13, because the optimal subarrays are [1,2,-1,3] and [3,4] summing to 13.

Examples

Example 1

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

Output: 13

The optimal choice is: The total sum is 10 + 3 = 13 .

Example 2

Input: nums = [-10,3,-1,-2], k = 4, m = 1

Output: -10

The optimal choice is choosing each element as a subarray. The output is (-10) + 3 + (-1) + (-2) = -10 .

Constraints

• 1 <= nums.length <= 2000
• -104 <= nums[i] <= 104
• 1 <= k <= floor(nums.length / m)
• 1 <= m <= 3

Solution Approach

Prefix Sum Preprocessing

Compute prefix sums of nums to quickly calculate any subarray sum in O(1) time. This allows the DP states to efficiently consider every possible subarray of length at least m without recomputing sums repeatedly.

State Transition Dynamic Programming

Define dp[i][j] as the maximum sum using the first i elements with j subarrays selected. For each end index, transition by checking all valid subarrays of length at least m ending at i and update dp[i][j] = max(dp[i-len][j-1]+sum(nums[i-len+1..i])) for len >= m.

Iterate and Track Maximum

Iterate through the array and for each DP state, maintain the maximum sum achieved for j subarrays. The final answer is dp[n][k], ensuring subarrays do not overlap and each meets the minimum length requirement.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity is O(n k m) where n is array length, k is number of subarrays, and m is minimum length due to iterating possible subarrays for each DP state. Space complexity is O(n*k) for storing DP states, though can be optimized to O(k) with rolling arrays.

What Interviewers Usually Probe

• The candidate identifies the state transition for DP and the need for prefix sums.
• They consider minimum length constraints and non-overlapping requirements.
• They attempt optimizations to reduce redundant subarray sum calculations.

Common Pitfalls or Variants

Common pitfalls

• Not handling minimum length m correctly leads to invalid subarrays.
• Overlapping subarrays due to incorrect DP indexing.
• Inefficient recomputation of subarray sums without prefix sums.

Follow-up variants

• Maximize sum of k subarrays with exact length m.
• Allow overlapping subarrays and maximize sum.
• Find minimum sum of k non-overlapping subarrays with length at least m.

FAQ

What is the main strategy for Sum of K Subarrays With Length at Least M?

Use state transition dynamic programming combined with prefix sums to select k non-overlapping subarrays efficiently.

Why are prefix sums important in this problem?

Prefix sums allow O(1) calculation of any subarray sum, which is critical to efficiently update DP states.

How do I enforce subarrays of at least length m?

Only consider subarrays of length >= m in the DP transitions to ensure the minimum length requirement is respected.

Can subarrays overlap in this problem?

No, all selected subarrays must be non-overlapping, and DP indexing ensures this constraint is met.

What is the expected complexity of the optimal solution?

Time complexity is O(n k m) due to iterating over possible subarrays for DP, and space complexity is O(n*k) for DP states.