How do you solve Sum of K Subarrays With Length at Least M on LeetCode?

Sum of K Subarrays With Length at Least M (LeetCode #3473) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * k) time complexity and O(n * k) space complexity. Key insight: Subarrays must be non-overlapping.

What is the time complexity of Sum of K Subarrays With Length at Least M?

The optimal solution for Sum of K Subarrays With Length at Least M runs in O(n * k) time and uses O(n * k) space. The DP approach iterates through n and k, leading to O(n * k) complexity.

What is the brute force solution for Sum of K Subarrays With Length at Least M?

The brute force approach for Sum of K Subarrays With Length at Least M is Brute Force, with O(n²) time complexity and O(1) space complexity. We can generate all possible combinations of k subarrays of length at least m and calculate their sums. This approach is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n * k).

What algorithm or data structure is used to solve Sum of K Subarrays With Length at Least M?

Sum of K Subarrays With Length at Least M uses the following concepts: Array, Dynamic Programming, Prefix Sum. The recommended patterns to study are: Dynamic Programming, Prefix Sum.

What are the interview tips for Sum of K Subarrays With Length at Least M?

Explain your thought process clearly. Discuss edge cases and constraints. Practice similar DP problems.

What are common mistakes when solving Sum of K Subarrays With Length at Least M?

Not considering subarray overlaps. Incorrectly calculating prefix sums.

#3473

Sum of K Subarrays With Length at Least M

Medium

ArrayDynamic ProgrammingPrefix SumDynamic ProgrammingPrefix Sum

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * k)
Space	O(1)	O(n * k)

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Intuition

Time O(n * k)Space O(n * k)

We can use dynamic programming and prefix sums to efficiently calculate the maximum sum of k non-overlapping subarrays. This reduces redundant calculations.

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Algorithm

3 steps

1Step 1: Calculate prefix sums to quickly get subarray sums.
2Step 2: Use a DP array where dp[i][j] represents the max sum of j subarrays using the first i elements.
3Step 3: For each position, update the DP table considering the last subarray ends at that position.

solution.py11 lines

1def maxSum(nums, k, m):
2    n = len(nums)
3    prefix = [0] * (n + 1)
4    for i in range(n): prefix[i + 1] = prefix[i] + nums[i]
5    dp = [[float('-inf')] * (k + 1) for _ in range(n + 1)]
6    dp[0][0] = 0
7    for j in range(1, k + 1):
8        for i in range(m, n + 1):
9            for p in range(i - m + 1):
10                dp[i][j] = max(dp[i][j], dp[p][j - 1] + prefix[i] - prefix[p])
11    return max(dp[n])

ℹ

Complexity note: The DP approach iterates through n and k, leading to O(n * k) complexity.

1Subarrays must be non-overlapping.
2Prefix sums allow quick sum calculations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.