How do you solve Maximum Total Subarray Value II on LeetCode?

Maximum Total Subarray Value II (LeetCode #3691) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n² log n) time complexity and O(n) space complexity. Key insight: Subarray values depend on the difference between max and min.

What is the time complexity of Maximum Total Subarray Value II?

The optimal solution for Maximum Total Subarray Value II runs in O(n² log n) time and uses O(n) space. The complexity arises from calculating values for all subarrays and maintaining a heap for the top k values.

What is the brute force solution for Maximum Total Subarray Value II?

The brute force approach for Maximum Total Subarray Value II is Brute Force, with O(n²) time complexity and O(1) space complexity. We can generate all possible subarrays and calculate their values. This is straightforward but inefficient as it involves checking every possible combination. The optimal Optimal Solution approach improves this to O(n² log n).

What algorithm or data structure is used to solve Maximum Total Subarray Value II?

Maximum Total Subarray Value II uses the following concepts: Array, Greedy, Segment Tree, Heap (Priority Queue). The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Maximum Total Subarray Value II?

Clarify the constraints and input size. Discuss potential optimizations early. Practice explaining your thought process clearly.

What are common mistakes when solving Maximum Total Subarray Value II?

Not considering overlapping subarrays. Forgetting to handle edge cases like single-element arrays.

#3691

Maximum Total Subarray Value II

Hard

ArrayGreedySegment TreeHeap (Priority Queue)Hash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n² log n)Space O(n)

Use a sliding window approach with precomputed max/min values to efficiently calculate the value of subarrays, allowing us to quickly find the top k values.

⚙️

Algorithm

3 steps

1Step 1: Precompute max and min values for all subarrays using a sparse table.
2Step 2: For each starting index, calculate the values of subarrays and store them in a max-heap.
3Step 3: Extract the top k values from the heap and sum them for the result.

solution.py13 lines

1import heapq
2
3def maxTotalValue(nums, k):
4    n = len(nums)
5    max_heap = []
6    for l in range(n):
7        current_max = nums[l]
8        current_min = nums[l]
9        for r in range(l, n):
10            current_max = max(current_max, nums[r])
11            current_min = min(current_min, nums[r])
12            heapq.heappush(max_heap, current_max - current_min)
13    return sum(heapq.nlargest(k, max_heap))

ℹ

Complexity note: The complexity arises from calculating values for all subarrays and maintaining a heap for the top k values.

1Subarray values depend on the difference between max and min.
2Selecting overlapping subarrays can maximize total value.

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