LeetCode #3473 — MEDIUM

Sum of K Subarrays With Length at Least M

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

Solve on LeetCode
The Problem

Problem Statement

You are given an integer array nums and two integers, k and m.

Return the maximum sum of k non-overlapping subarrays of nums, where each subarray has a length of at least m.

Example 1:

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

Output: 13

Explanation:

The optimal choice is:

  • Subarray nums[3..5] with sum 3 + 3 + 4 = 10 (length is 3 >= m).
  • Subarray nums[0..1] with sum 1 + 2 = 3 (length is 2 >= m).

The total sum is 10 + 3 = 13.

Example 2:

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

Output: -10

Explanation:

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
Patterns Used
📊Dynamic Programming→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums and two integers, k and m. Return the maximum sum of k non-overlapping subarrays of nums, where each subarray has a length of at least m.

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,-1,3,3,4]
2
2

Example 2

[-10,3,-1,-2]
4
1
Step 02

Core Insight

What unlocks the optimal approach

  • Dynamic Programming
  • Prefix Sum
  • Let <code>dp[i][j]</code> be the maximum sum with <code>i</code> subarrays for the first <code>j</code> elements
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. 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 #3473: Sum of K Subarrays With Length at Least M
class Solution {
  public int maxSum(int[] nums, int k, int m) {
    final int n = nums.length;
    int[] prefix = new int[n + 1];
    Integer[][][] mem = new Integer[n][2][k + 1];
    for (int i = 0; i < n; i++)
      prefix[i + 1] = prefix[i] + nums[i];
    return maxSum(nums, 0, 0, k, m, prefix, mem);
  }

  private static final int INF = 20_000_000;

  private int maxSum(int[] nums, int i, int ongoing, int k, int m, int[] prefix,
                     Integer[][][] mem) {
    if (k < 0)
      return -INF;
    if (i == nums.length)
      return k == 0 ? 0 : -INF;
    if (mem[i][ongoing][k] != null)
      return mem[i][ongoing][k];
    if (ongoing == 1)
      // 1. End the current subarray (transition to state 0, same index i)
      // 2. Extend the current subarray by picking nums[i] and move to i + 1.
      return mem[i][1][k] = Math.max(maxSum(nums, i, 0, k, m, prefix, mem),
                                     maxSum(nums, i + 1, 1, k, m, prefix, mem) + nums[i]);
    // ongoing == 0
    // 1. Skip nums[i].
    // 2. Pick nums[i..i + m - 1] (only if k > 0 and there're enough elements).
    int res = maxSum(nums, i + 1, 0, k, m, prefix, mem);
    if (i + m <= nums.length) // If we have enough elements for a new segment
      res = Math.max(res,
                     maxSum(nums, i + m, 1, k - 1, m, prefix, mem) + (prefix[i + m] - prefix[i]));

    return mem[i][0][k] = res;
  }
}
Step 06

Interactive Study Demo

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

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n × m)
Space
O(n × m)

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.

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